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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,687,915,948
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 46
| 0
|
k,r = map(int,input().split())
c=1
z=k
i=2
while str(z)[-1]!="0" or str(z)[-1] != str(r):
if str(z)[-1]=="0" or str(z)[-1] == str(r):
break
else:
c+=1
z=k*i
i+=1
continue
print(c)
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.
|
```python
k,r = map(int,input().split())
c=1
z=k
i=2
while str(z)[-1]!="0" or str(z)[-1] != str(r):
if str(z)[-1]=="0" or str(z)[-1] == str(r):
break
else:
c+=1
z=k*i
i+=1
continue
print(c)
```
| 3
|
|
797
|
A
|
k-Factorization
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math",
"number theory"
] | null | null |
Given a positive integer *n*, find *k* integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to *n*.
|
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=20).
|
If it's impossible to find the representation of *n* as a product of *k* numbers, print -1.
Otherwise, print *k* integers in any order. Their product must be equal to *n*. If there are multiple answers, print any of them.
|
[
"100000 2\n",
"100000 20\n",
"1024 5\n"
] |
[
"2 50000 \n",
"-1\n",
"2 64 2 2 2 \n"
] |
none
| 0
|
[
{
"input": "100000 2",
"output": "2 50000 "
},
{
"input": "100000 20",
"output": "-1"
},
{
"input": "1024 5",
"output": "2 64 2 2 2 "
},
{
"input": "100000 10",
"output": "2 2 2 2 2 5 5 5 5 5 "
},
{
"input": "99999 3",
"output": "3 813 41 "
},
{
"input": "99999 4",
"output": "3 3 41 271 "
},
{
"input": "99999 5",
"output": "-1"
},
{
"input": "1024 10",
"output": "2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "1024 11",
"output": "-1"
},
{
"input": "2048 11",
"output": "2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "2 3",
"output": "-1"
},
{
"input": "2 4",
"output": "-1"
},
{
"input": "2 5",
"output": "-1"
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "3 1",
"output": "3 "
},
{
"input": "3 2",
"output": "-1"
},
{
"input": "349 2",
"output": "-1"
},
{
"input": "8 1",
"output": "8 "
},
{
"input": "66049 2",
"output": "257 257 "
},
{
"input": "6557 2",
"output": "83 79 "
},
{
"input": "9 2",
"output": "3 3 "
},
{
"input": "4 2",
"output": "2 2 "
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "12 1",
"output": "12 "
},
{
"input": "17 1",
"output": "17 "
},
{
"input": "8 2",
"output": "2 4 "
},
{
"input": "14 2",
"output": "7 2 "
},
{
"input": "99991 1",
"output": "99991 "
},
{
"input": "30 2",
"output": "3 10 "
},
{
"input": "97 1",
"output": "97 "
},
{
"input": "92 2",
"output": "2 46 "
},
{
"input": "4 1",
"output": "4 "
},
{
"input": "4 3",
"output": "-1"
},
{
"input": "30 4",
"output": "-1"
},
{
"input": "2 6",
"output": "-1"
},
{
"input": "3 1",
"output": "3 "
},
{
"input": "3 2",
"output": "-1"
},
{
"input": "3 3",
"output": "-1"
},
{
"input": "3 4",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "3 6",
"output": "-1"
},
{
"input": "4 1",
"output": "4 "
},
{
"input": "4 2",
"output": "2 2 "
},
{
"input": "4 3",
"output": "-1"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "4 5",
"output": "-1"
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{
"input": "4 6",
"output": "-1"
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{
"input": "5 1",
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{
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"output": "-1"
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{
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"output": "-1"
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{
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{
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"output": "-1"
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{
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"output": "-1"
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{
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"output": "6 "
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{
"input": "6 2",
"output": "3 2 "
},
{
"input": "6 3",
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},
{
"input": "6 4",
"output": "-1"
},
{
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"output": "-1"
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{
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"output": "-1"
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{
"input": "7 1",
"output": "7 "
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{
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"output": "-1"
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{
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{
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{
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{
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{
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{
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{
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{
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"output": "-1"
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{
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"output": "-1"
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{
"input": "8 6",
"output": "-1"
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{
"input": "9 1",
"output": "9 "
},
{
"input": "9 2",
"output": "3 3 "
},
{
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"output": "-1"
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{
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{
"input": "9 5",
"output": "-1"
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{
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"output": "-1"
},
{
"input": "10 1",
"output": "10 "
},
{
"input": "10 2",
"output": "5 2 "
},
{
"input": "10 3",
"output": "-1"
},
{
"input": "10 4",
"output": "-1"
},
{
"input": "10 5",
"output": "-1"
},
{
"input": "10 6",
"output": "-1"
},
{
"input": "11 1",
"output": "11 "
},
{
"input": "11 2",
"output": "-1"
},
{
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"output": "-1"
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{
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"output": "-1"
},
{
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"output": "-1"
},
{
"input": "11 6",
"output": "-1"
},
{
"input": "12 1",
"output": "12 "
},
{
"input": "12 2",
"output": "2 6 "
},
{
"input": "12 3",
"output": "2 2 3 "
},
{
"input": "12 4",
"output": "-1"
},
{
"input": "12 5",
"output": "-1"
},
{
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"output": "-1"
},
{
"input": "13 1",
"output": "13 "
},
{
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"output": "-1"
},
{
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"output": "-1"
},
{
"input": "13 4",
"output": "-1"
},
{
"input": "13 5",
"output": "-1"
},
{
"input": "13 6",
"output": "-1"
},
{
"input": "14 1",
"output": "14 "
},
{
"input": "14 2",
"output": "7 2 "
},
{
"input": "14 3",
"output": "-1"
},
{
"input": "14 4",
"output": "-1"
},
{
"input": "14 5",
"output": "-1"
},
{
"input": "14 6",
"output": "-1"
},
{
"input": "15 1",
"output": "15 "
},
{
"input": "15 2",
"output": "5 3 "
},
{
"input": "15 3",
"output": "-1"
},
{
"input": "15 4",
"output": "-1"
},
{
"input": "15 5",
"output": "-1"
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"output": "-1"
},
{
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"output": "16 "
},
{
"input": "16 2",
"output": "2 8 "
},
{
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"output": "2 4 2 "
},
{
"input": "16 4",
"output": "2 2 2 2 "
},
{
"input": "16 5",
"output": "-1"
},
{
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"output": "-1"
},
{
"input": "17 1",
"output": "17 "
},
{
"input": "17 2",
"output": "-1"
},
{
"input": "17 3",
"output": "-1"
},
{
"input": "17 4",
"output": "-1"
},
{
"input": "17 5",
"output": "-1"
},
{
"input": "17 6",
"output": "-1"
},
{
"input": "18 1",
"output": "18 "
},
{
"input": "18 2",
"output": "3 6 "
},
{
"input": "18 3",
"output": "3 2 3 "
},
{
"input": "18 4",
"output": "-1"
},
{
"input": "18 5",
"output": "-1"
},
{
"input": "18 6",
"output": "-1"
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{
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"output": "19 "
},
{
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{
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{
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{
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"output": "-1"
},
{
"input": "19 6",
"output": "-1"
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{
"input": "20 1",
"output": "20 "
},
{
"input": "20 2",
"output": "2 10 "
},
{
"input": "20 3",
"output": "2 2 5 "
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{
"input": "20 4",
"output": "-1"
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{
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{
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{
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{
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{
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},
{
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"output": "-1"
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{
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"output": "-1"
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{
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"output": "95477 "
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{
"input": "95477 2",
"output": "311 307 "
},
{
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"output": "-1"
},
{
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"output": "-1"
},
{
"input": "95477 5",
"output": "-1"
},
{
"input": "35557 1",
"output": "35557 "
},
{
"input": "35557 2",
"output": "31 1147 "
},
{
"input": "35557 3",
"output": "31 31 37 "
},
{
"input": "35557 4",
"output": "-1"
},
{
"input": "35557 5",
"output": "-1"
},
{
"input": "42439 1",
"output": "42439 "
},
{
"input": "42439 2",
"output": "37 1147 "
},
{
"input": "42439 3",
"output": "37 31 37 "
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{
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"output": "-1"
},
{
"input": "42439 5",
"output": "-1"
}
] | 1,644,490,145
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
import math
n, k = map(int, input().split())
factors = []
while n % 2 == 0 and n:
n /= 2
factors.append(2)
while n % 3 == 0 and n:
n /= 3
factors.append(3)
while n % 5 == 0 and n:
n /= 5
factors.append(5)
if n and n != 1:
factors.append(n)
if len(factors) < k:
print(-1)
elif len(factors) > k:
for i in range(k-1):
print(factors[i], end=' ')
for i in range(k, len(factors)):
factors[k] *= factors[i]
print(factors[k])
else:
print(factors)
|
Title: k-Factorization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Given a positive integer *n*, find *k* integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to *n*.
Input Specification:
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=20).
Output Specification:
If it's impossible to find the representation of *n* as a product of *k* numbers, print -1.
Otherwise, print *k* integers in any order. Their product must be equal to *n*. If there are multiple answers, print any of them.
Demo Input:
['100000 2\n', '100000 20\n', '1024 5\n']
Demo Output:
['2 50000 \n', '-1\n', '2 64 2 2 2 \n']
Note:
none
|
```python
import math
n, k = map(int, input().split())
factors = []
while n % 2 == 0 and n:
n /= 2
factors.append(2)
while n % 3 == 0 and n:
n /= 3
factors.append(3)
while n % 5 == 0 and n:
n /= 5
factors.append(5)
if n and n != 1:
factors.append(n)
if len(factors) < k:
print(-1)
elif len(factors) > k:
for i in range(k-1):
print(factors[i], end=' ')
for i in range(k, len(factors)):
factors[k] *= factors[i]
print(factors[k])
else:
print(factors)
```
| 0
|
|
260
|
A
|
Adding Digits
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
|
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
|
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
|
[
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] |
[
"524848\n",
"121\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 21 73839",
"output": "9999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99991 623 36438",
"output": "9999150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 334 94854",
"output": "9999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99252 9827 84849",
"output": "9925270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99313 9833 10561",
"output": "9931330000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "94885 55815 11417",
"output": "9488550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99492 58525 53481",
"output": "9949250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99858 28531 79193",
"output": "9985850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99136 47208 42607",
"output": "9913680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "63270 19953 5555",
"output": "-1"
},
{
"input": "10240 128 100000",
"output": "1024000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "12 11 3",
"output": "12100"
},
{
"input": "14 12 99998",
"output": "1440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "1 11 3",
"output": "1100"
},
{
"input": "3 40 1",
"output": "-1"
},
{
"input": "150 100 10",
"output": "1500000000000"
},
{
"input": "5 10 1",
"output": "50"
},
{
"input": "1 15 10",
"output": "15000000000"
},
{
"input": "3 13 2",
"output": "390"
}
] | 1,529,825,802
| 702
|
Python 3
|
OK
|
TESTS
| 25
| 77
| 307,200
|
a,b,n=map(int,input().strip().split())
if a%b==0:
s='0'
s=s*n
print(str(a)+s)
else:
for i in range(0,10):
if (a*10+i)%b==0:
s='0'
s=s*(n-1)
a=a*10+i
print(str(a)+s)
break
else:
print(-1)
|
Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none
|
```python
a,b,n=map(int,input().strip().split())
if a%b==0:
s='0'
s=s*n
print(str(a)+s)
else:
for i in range(0,10):
if (a*10+i)%b==0:
s='0'
s=s*(n-1)
a=a*10+i
print(str(a)+s)
break
else:
print(-1)
```
| 3
|
|
868
|
A
|
Bark to Unlock
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
|
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
|
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
|
[
"ya\n4\nah\noy\nto\nha\n",
"hp\n2\nht\ntp\n",
"ah\n1\nha\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring.
| 250
|
[
{
"input": "ya\n4\nah\noy\nto\nha",
"output": "YES"
},
{
"input": "hp\n2\nht\ntp",
"output": "NO"
},
{
"input": "ah\n1\nha",
"output": "YES"
},
{
"input": "bb\n4\nba\nab\naa\nbb",
"output": "YES"
},
{
"input": "bc\n4\nca\nba\nbb\ncc",
"output": "YES"
},
{
"input": "ba\n4\ncd\nad\ncc\ncb",
"output": "YES"
},
{
"input": "pg\n4\nzl\nxs\ndi\nxn",
"output": "NO"
},
{
"input": "bn\n100\ndf\nyb\nze\nml\nyr\nof\nnw\nfm\ndw\nlv\nzr\nhu\nzt\nlw\nld\nmo\nxz\ntp\nmr\nou\nme\npx\nvp\nes\nxi\nnr\nbx\nqc\ngm\njs\nkn\ntw\nrq\nkz\nuc\nvc\nqr\nab\nna\nro\nya\nqy\ngu\nvk\nqk\ngs\nyq\nop\nhw\nrj\neo\nlz\nbh\nkr\nkb\nma\nrd\nza\nuf\nhq\nmc\nmn\nti\nwn\nsh\nax\nsi\nnd\ntz\ndu\nfj\nkl\nws\now\nnf\nvr\nye\nzc\niw\nfv\nkv\noo\nsm\nbc\nrs\nau\nuz\nuv\ngh\nsu\njn\ndz\nrl\nwj\nbk\nzl\nas\nms\nit\nwu",
"output": "YES"
},
{
"input": "bb\n1\naa",
"output": "NO"
},
{
"input": "qm\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm",
"output": "NO"
},
{
"input": "mq\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm",
"output": "YES"
},
{
"input": "aa\n1\naa",
"output": "YES"
},
{
"input": "bb\n1\nbb",
"output": "YES"
},
{
"input": "ba\n1\ncc",
"output": "NO"
},
{
"input": "ha\n1\nha",
"output": "YES"
},
{
"input": "aa\n1\naa",
"output": "YES"
},
{
"input": "ez\n1\njl",
"output": "NO"
},
{
"input": "aa\n2\nab\nba",
"output": "YES"
},
{
"input": "aa\n2\nca\ncc",
"output": "NO"
},
{
"input": "dd\n2\nac\ndc",
"output": "NO"
},
{
"input": "qc\n2\nyc\nkr",
"output": "NO"
},
{
"input": "aa\n3\nba\nbb\nab",
"output": "YES"
},
{
"input": "ca\n3\naa\nbb\nab",
"output": "NO"
},
{
"input": "ca\n3\nbc\nbd\nca",
"output": "YES"
},
{
"input": "dd\n3\nmt\nrg\nxl",
"output": "NO"
},
{
"input": "be\n20\nad\ncd\ncb\ndb\ndd\naa\nab\nca\nae\ned\ndc\nbb\nba\nda\nee\nea\ncc\nac\nec\neb",
"output": "YES"
},
{
"input": "fc\n20\nca\nbb\nce\nfd\nde\nfa\ncc\nec\nfb\nfc\nff\nbe\ncf\nba\ndb\ned\naf\nae\nda\nef",
"output": "YES"
},
{
"input": "ca\n20\ndc\naf\ndf\neg\naa\nbc\nea\nbd\nab\ndb\ngc\nfb\nba\nbe\nee\ngf\ncf\nag\nga\nca",
"output": "YES"
},
{
"input": "ke\n20\nzk\nra\nbq\nqz\nwt\nzg\nmz\nuk\nge\nuv\nud\nfd\neh\ndm\nsk\nki\nfv\ntp\nat\nfb",
"output": "YES"
},
{
"input": "hh\n50\nag\nhg\ndg\nfh\neg\ngh\ngd\nda\nbh\nab\nhf\ndc\nhb\nfe\nad\nec\nac\nfd\nca\naf\ncg\nhd\neb\nce\nhe\nha\ngb\nea\nae\nfb\nff\nbe\nch\nhh\nee\nde\nge\ngf\naa\ngg\neh\ned\nbf\nfc\nah\nga\nbd\ncb\nbg\nbc",
"output": "YES"
},
{
"input": "id\n50\nhi\ndc\nfg\nee\ngi\nhc\nac\nih\ndg\nfc\nde\ned\nie\neb\nic\ncf\nib\nfa\ngc\nba\nbe\nga\nha\nhg\nia\ndf\nab\nei\neh\nad\nii\nci\ndh\nec\nif\ndi\nbg\nag\nhe\neg\nca\nae\ndb\naa\nid\nfh\nhh\ncc\nfb\ngb",
"output": "YES"
},
{
"input": "fe\n50\nje\nbi\nbg\ngc\nfb\nig\ndf\nji\ndg\nfe\nfc\ncf\ngf\nai\nhe\nac\nch\nja\ngh\njf\nge\ncb\nij\ngb\ncg\naf\neh\nee\nhd\njd\njb\nii\nca\nci\nga\nab\nhi\nag\nfj\nej\nfi\nie\ndj\nfg\nef\njc\njg\njh\nhf\nha",
"output": "YES"
},
{
"input": "rn\n50\nba\nec\nwg\nao\nlk\nmz\njj\ncf\nfa\njk\ndy\nsz\njs\nzr\nqv\ntx\nwv\nrd\nqw\nls\nrr\nvt\nrx\nkc\neh\nnj\niq\nyi\nkh\nue\nnv\nkz\nrn\nes\nua\nzf\nvu\nll\neg\nmj\ncz\nzj\nxz\net\neb\nci\nih\nig\nam\nvd",
"output": "YES"
},
{
"input": "ee\n100\nah\nfb\ncd\nbi\nii\nai\nid\nag\nie\nha\ndi\nec\nae\nce\njb\ndg\njg\ngd\ngf\nda\nih\nbd\nhj\ngg\nhb\ndf\ned\nfh\naf\nja\nci\nfc\nic\nji\nac\nhi\nfj\nch\nbc\njd\naa\nff\nad\ngj\nej\nde\nee\nhe\ncf\nga\nia\ncg\nbb\nhc\nbe\ngi\njf\nbg\naj\njj\nbh\nfe\ndj\nef\ngb\nge\ndb\nig\ncj\ndc\nij\njh\nei\ndd\nib\nhf\neg\nbf\nfg\nab\ngc\nfd\nhd\ngh\neh\njc\neb\nhh\nca\nje\nbj\nif\nea\nhg\nfa\ncc\nba\ndh\ncb\nfi",
"output": "YES"
},
{
"input": "if\n100\njd\nbc\nje\nhi\nga\nde\nkb\nfc\ncd\ngd\naj\ncb\nei\nbf\ncf\ndk\ndb\ncg\nki\ngg\nkg\nfa\nkj\nii\njf\njg\ngb\nbh\nbg\neh\nhj\nhb\ndg\ndj\njc\njb\nce\ndi\nig\nci\ndf\nji\nhc\nfk\naf\nac\ngk\nhd\nae\nkd\nec\nkc\neb\nfh\nij\nie\nca\nhh\nkf\nha\ndd\nif\nef\nih\nhg\nej\nfe\njk\nea\nib\nck\nhf\nak\ngi\nch\ndc\nba\nke\nad\nka\neg\njh\nja\ngc\nfd\ncc\nab\ngj\nik\nfg\nbj\nhe\nfj\nge\ngh\nhk\nbk\ned\nid\nfi",
"output": "YES"
},
{
"input": "kd\n100\nek\nea\nha\nkf\nkj\ngh\ndl\nfj\nal\nga\nlj\nik\ngd\nid\ncb\nfh\ndk\nif\nbh\nkb\nhc\nej\nhk\ngc\ngb\nef\nkk\nll\nlf\nkh\ncl\nlh\njj\nil\nhh\nci\ndb\ndf\ngk\njg\nch\nbd\ncg\nfg\nda\neb\nlg\ndg\nbk\nje\nbg\nbl\njl\ncj\nhb\nei\naa\ngl\nka\nfa\nfi\naf\nkc\nla\ngi\nij\nib\nle\ndi\nck\nag\nlc\nca\nge\nie\nlb\nke\nii\nae\nig\nic\nhe\ncf\nhd\nak\nfb\nhi\ngf\nad\nba\nhg\nbi\nkl\nac\ngg\ngj\nbe\nlk\nld\naj",
"output": "YES"
},
{
"input": "ab\n1\nab",
"output": "YES"
},
{
"input": "ya\n1\nya",
"output": "YES"
},
{
"input": "ay\n1\nyb",
"output": "NO"
},
{
"input": "ax\n2\nii\nxa",
"output": "YES"
},
{
"input": "hi\n1\nhi",
"output": "YES"
},
{
"input": "ag\n1\nag",
"output": "YES"
},
{
"input": "th\n1\nth",
"output": "YES"
},
{
"input": "sb\n1\nsb",
"output": "YES"
},
{
"input": "hp\n1\nhp",
"output": "YES"
},
{
"input": "ah\n1\nah",
"output": "YES"
},
{
"input": "ta\n1\nta",
"output": "YES"
},
{
"input": "tb\n1\ntb",
"output": "YES"
},
{
"input": "ab\n5\nca\nda\nea\nfa\nka",
"output": "NO"
},
{
"input": "ac\n1\nac",
"output": "YES"
},
{
"input": "ha\n2\nha\nzz",
"output": "YES"
},
{
"input": "ok\n1\nok",
"output": "YES"
},
{
"input": "bc\n1\nbc",
"output": "YES"
},
{
"input": "az\n1\nzz",
"output": "NO"
},
{
"input": "ab\n2\nba\ntt",
"output": "YES"
},
{
"input": "ah\n2\nap\nhp",
"output": "NO"
},
{
"input": "sh\n1\nsh",
"output": "YES"
},
{
"input": "az\n1\nby",
"output": "NO"
},
{
"input": "as\n1\nas",
"output": "YES"
},
{
"input": "ab\n2\nab\ncd",
"output": "YES"
},
{
"input": "ab\n2\nxa\nza",
"output": "NO"
},
{
"input": "ab\n2\net\nab",
"output": "YES"
},
{
"input": "ab\n1\naa",
"output": "NO"
},
{
"input": "ab\n2\nab\nde",
"output": "YES"
},
{
"input": "ah\n2\nba\nha",
"output": "YES"
},
{
"input": "ha\n3\ndd\ncc\nha",
"output": "YES"
},
{
"input": "oo\n1\nox",
"output": "NO"
},
{
"input": "ab\n2\nax\nbx",
"output": "NO"
},
{
"input": "ww\n4\nuw\now\npo\nko",
"output": "NO"
},
{
"input": "ay\n1\nay",
"output": "YES"
},
{
"input": "yo\n1\nyo",
"output": "YES"
},
{
"input": "ba\n1\nba",
"output": "YES"
},
{
"input": "qw\n1\nqw",
"output": "YES"
},
{
"input": "la\n1\nla",
"output": "YES"
},
{
"input": "ab\n2\nbb\nbc",
"output": "NO"
},
{
"input": "aa\n2\nab\nac",
"output": "NO"
},
{
"input": "ah\n2\nbb\nha",
"output": "YES"
},
{
"input": "ya\n42\nab\nac\nad\nae\naf\nag\nah\nai\nak\naj\nba\nbc\nbd\nbe\nbf\nbg\nbh\nbi\nbk\nbj\ncb\nca\ncd\nce\ncf\ncg\nch\nci\nck\ncj\ndb\ndc\nda\nde\ndf\ndg\ndh\ndi\ndk\ndj\nef\nek",
"output": "NO"
},
{
"input": "ab\n3\nab\nxx\nyy",
"output": "YES"
},
{
"input": "ab\n2\nab\ncc",
"output": "YES"
},
{
"input": "sa\n2\nxx\nas",
"output": "YES"
},
{
"input": "ma\n1\nma",
"output": "YES"
},
{
"input": "ba\n1\nbb",
"output": "NO"
},
{
"input": "bc\n1\nab",
"output": "NO"
},
{
"input": "fa\n1\nfa",
"output": "YES"
},
{
"input": "ap\n1\nap",
"output": "YES"
},
{
"input": "ab\n1\nbb",
"output": "NO"
},
{
"input": "bk\n1\nbk",
"output": "YES"
},
{
"input": "xy\n2\nxy\naa",
"output": "YES"
},
{
"input": "ab\n2\nza\nbz",
"output": "YES"
}
] | 1,556,541,098
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 90
| 124
| 0
|
X = input()
N = int(input())
First, Second = False, False
for i in range(N):
Temp = input()
if X == Temp:
print("YES")
exit()
if X[0] == Temp[1]:
First = True
if X[1] == Temp[0]:
Second = True
if First and Second:
print("YES")
exit()
print("NO")
|
Title: Bark to Unlock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
Input Specification:
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Output Specification:
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['ya\n4\nah\noy\nto\nha\n', 'hp\n2\nht\ntp\n', 'ah\n1\nha\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring.
|
```python
X = input()
N = int(input())
First, Second = False, False
for i in range(N):
Temp = input()
if X == Temp:
print("YES")
exit()
if X[0] == Temp[1]:
First = True
if X[1] == Temp[0]:
Second = True
if First and Second:
print("YES")
exit()
print("NO")
```
| 3
|
|
467
|
A
|
George and Accommodation
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
|
Print a single integer — the number of rooms where George and Alex can move in.
|
[
"3\n1 1\n2 2\n3 3\n",
"3\n1 10\n0 10\n10 10\n"
] |
[
"0\n",
"2\n"
] |
none
| 500
|
[
{
"input": "3\n1 1\n2 2\n3 3",
"output": "0"
},
{
"input": "3\n1 10\n0 10\n10 10",
"output": "2"
},
{
"input": "2\n36 67\n61 69",
"output": "2"
},
{
"input": "3\n21 71\n10 88\n43 62",
"output": "3"
},
{
"input": "3\n1 2\n2 3\n3 4",
"output": "0"
},
{
"input": "10\n0 10\n0 20\n0 30\n0 40\n0 50\n0 60\n0 70\n0 80\n0 90\n0 100",
"output": "10"
},
{
"input": "13\n14 16\n30 31\n45 46\n19 20\n15 17\n66 67\n75 76\n95 97\n29 30\n37 38\n0 2\n36 37\n8 9",
"output": "4"
},
{
"input": "19\n66 67\n97 98\n89 91\n67 69\n67 68\n18 20\n72 74\n28 30\n91 92\n27 28\n75 77\n17 18\n74 75\n28 30\n16 18\n90 92\n9 11\n22 24\n52 54",
"output": "12"
},
{
"input": "15\n55 57\n95 97\n57 59\n34 36\n50 52\n96 98\n39 40\n13 15\n13 14\n74 76\n47 48\n56 58\n24 25\n11 13\n67 68",
"output": "10"
},
{
"input": "17\n68 69\n47 48\n30 31\n52 54\n41 43\n33 35\n38 40\n56 58\n45 46\n92 93\n73 74\n61 63\n65 66\n37 39\n67 68\n77 78\n28 30",
"output": "8"
},
{
"input": "14\n64 66\n43 44\n10 12\n76 77\n11 12\n25 27\n87 88\n62 64\n39 41\n58 60\n10 11\n28 29\n57 58\n12 14",
"output": "7"
},
{
"input": "38\n74 76\n52 54\n78 80\n48 49\n40 41\n64 65\n28 30\n6 8\n49 51\n68 70\n44 45\n57 59\n24 25\n46 48\n49 51\n4 6\n63 64\n76 78\n57 59\n18 20\n63 64\n71 73\n88 90\n21 22\n89 90\n65 66\n89 91\n96 98\n42 44\n1 1\n74 76\n72 74\n39 40\n75 76\n29 30\n48 49\n87 89\n27 28",
"output": "22"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "26\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2",
"output": "0"
},
{
"input": "68\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2",
"output": "68"
},
{
"input": "7\n0 1\n1 5\n2 4\n3 5\n4 6\n5 6\n6 8",
"output": "5"
},
{
"input": "1\n0 0",
"output": "0"
},
{
"input": "1\n100 100",
"output": "0"
},
{
"input": "44\n0 8\n1 11\n2 19\n3 5\n4 29\n5 45\n6 6\n7 40\n8 19\n9 22\n10 18\n11 26\n12 46\n13 13\n14 27\n15 48\n16 25\n17 20\n18 29\n19 27\n20 45\n21 39\n22 29\n23 39\n24 42\n25 37\n26 52\n27 36\n28 43\n29 35\n30 38\n31 70\n32 47\n33 38\n34 61\n35 71\n36 51\n37 71\n38 59\n39 77\n40 70\n41 80\n42 77\n43 73",
"output": "42"
},
{
"input": "3\n1 3\n2 7\n8 9",
"output": "2"
},
{
"input": "53\n0 1\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53",
"output": "0"
},
{
"input": "55\n0 0\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20\n21 21\n22 22\n23 23\n24 24\n25 25\n26 26\n27 27\n28 28\n29 29\n30 30\n31 31\n32 32\n33 33\n34 34\n35 35\n36 36\n37 37\n38 38\n39 39\n40 40\n41 41\n42 42\n43 43\n44 44\n45 45\n46 46\n47 47\n48 48\n49 49\n50 50\n51 51\n52 52\n53 53\n54 54",
"output": "0"
},
{
"input": "51\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 62\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 73\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 82\n55 68\n55 70\n55 63\n55 55\n55 55\n55 55\n55 75\n55 75\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 73\n55 55\n55 82\n55 99\n55 60",
"output": "12"
},
{
"input": "14\n1 1\n1 1\n1 55\n1 16\n1 1\n1 1\n1 55\n1 62\n1 53\n1 26\n1 1\n1 36\n1 2\n1 3",
"output": "8"
}
] | 1,688,754,396
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 46
| 0
|
x = int(input())
y = 0
for i in range(x):
l = [int(j) for j in input().split()]
if l[0]+1<l[1]:
f+=1
if l[0]==l[1]:
y+=0
print(y)
|
Title: George and Accommodation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Output Specification:
Print a single integer — the number of rooms where George and Alex can move in.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n']
Demo Output:
['0\n', '2\n']
Note:
none
|
```python
x = int(input())
y = 0
for i in range(x):
l = [int(j) for j in input().split()]
if l[0]+1<l[1]:
f+=1
if l[0]==l[1]:
y+=0
print(y)
```
| -1
|
|
894
|
B
|
Ralph And His Magic Field
|
PROGRAMMING
| 1,800
|
[
"combinatorics",
"constructive algorithms",
"math",
"number theory"
] | null | null |
Ralph has a magic field which is divided into *n*<=×<=*m* blocks. That is to say, there are *n* rows and *m* columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to *k*, where *k* is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007<==<=109<=+<=7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
|
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=1018, *k* is either 1 or -1).
|
Print a single number denoting the answer modulo 1000000007.
|
[
"1 1 -1\n",
"1 3 1\n",
"3 3 -1\n"
] |
[
"1\n",
"1\n",
"16\n"
] |
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
| 1,000
|
[
{
"input": "1 1 -1",
"output": "1"
},
{
"input": "1 3 1",
"output": "1"
},
{
"input": "3 3 -1",
"output": "16"
},
{
"input": "2 7 1",
"output": "64"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 4 -1",
"output": "8"
},
{
"input": "173 69 -1",
"output": "814271739"
},
{
"input": "110 142 1",
"output": "537040244"
},
{
"input": "162 162 -1",
"output": "394042552"
},
{
"input": "49 153 -1",
"output": "412796600"
},
{
"input": "94 182 1",
"output": "33590706"
},
{
"input": "106666666 233333333 1",
"output": "121241754"
},
{
"input": "2 2 1",
"output": "2"
},
{
"input": "146 34 -1",
"output": "742752757"
},
{
"input": "94 86 -1",
"output": "476913727"
},
{
"input": "2529756051797760 2682355969139391 -1",
"output": "0"
},
{
"input": "3126690179932000 2474382898739836 -1",
"output": "917305624"
},
{
"input": "3551499873841921 2512677762780671 -1",
"output": "350058339"
},
{
"input": "3613456196418270 2872267429531501 1",
"output": "223552863"
},
{
"input": "2886684369091916 3509787933422130 1",
"output": "341476979"
},
{
"input": "3536041043537343 2416093514489183 1",
"output": "394974516"
},
{
"input": "2273134852621270 2798005122439669 1",
"output": "901406364"
},
{
"input": "2870150496178092 3171485931753811 -1",
"output": "0"
},
{
"input": "999999999999999999 1000000000000000000 1",
"output": "102810659"
},
{
"input": "987654321987654321 666666666666666666 1",
"output": "279028602"
},
{
"input": "1 2 -1",
"output": "0"
},
{
"input": "2 1 -1",
"output": "0"
},
{
"input": "1000000000000000000 1 1",
"output": "1"
},
{
"input": "1000000006 100000000000000000 1",
"output": "123624987"
}
] | 1,590,080,013
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 29
| 171
| 307,200
|
import sys
import math
from collections import defaultdict,deque
import heapq
n,m,k=map(int,sys.stdin.readline().split())
mod=10**9+7
ans=pow(2,(n-1)*(m-1),mod)
if m%2!=0 and n%2==0 and k==-1:
ans=0
if m%2==0 and n%2!=0 and k==-1:
ans=0
print(ans)
|
Title: Ralph And His Magic Field
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ralph has a magic field which is divided into *n*<=×<=*m* blocks. That is to say, there are *n* rows and *m* columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to *k*, where *k* is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007<==<=109<=+<=7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input Specification:
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=1018, *k* is either 1 or -1).
Output Specification:
Print a single number denoting the answer modulo 1000000007.
Demo Input:
['1 1 -1\n', '1 3 1\n', '3 3 -1\n']
Demo Output:
['1\n', '1\n', '16\n']
Note:
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
```python
import sys
import math
from collections import defaultdict,deque
import heapq
n,m,k=map(int,sys.stdin.readline().split())
mod=10**9+7
ans=pow(2,(n-1)*(m-1),mod)
if m%2!=0 and n%2==0 and k==-1:
ans=0
if m%2==0 and n%2!=0 and k==-1:
ans=0
print(ans)
```
| 3
|
|
596
|
B
|
Wilbur and Array
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Wilbur the pig is tinkering with arrays again. He has the array *a*1,<=*a*2,<=...,<=*a**n* initially consisting of *n* zeros. At one step, he can choose any index *i* and either add 1 to all elements *a**i*,<=*a**i*<=+<=1,<=... ,<=*a**n* or subtract 1 from all elements *a**i*,<=*a**i*<=+<=1,<=...,<=*a**n*. His goal is to end up with the array *b*1,<=*b*2,<=...,<=*b**n*.
Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the array *a**i*. Initially *a**i*<==<=0 for every position *i*, so this array is not given in the input.
The second line of the input contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109).
|
Print the minimum number of steps that Wilbur needs to make in order to achieve *a**i*<==<=*b**i* for all *i*.
|
[
"5\n1 2 3 4 5\n",
"4\n1 2 2 1\n"
] |
[
"5",
"3"
] |
In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.
In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1.
| 1,000
|
[
{
"input": "5\n1 2 3 4 5",
"output": "5"
},
{
"input": "4\n1 2 2 1",
"output": "3"
},
{
"input": "3\n1 2 4",
"output": "4"
},
{
"input": "6\n1 2 3 6 5 4",
"output": "8"
},
{
"input": "10\n2 1 4 3 6 5 8 7 10 9",
"output": "19"
},
{
"input": "7\n12 6 12 13 4 3 2",
"output": "36"
},
{
"input": "15\n15 14 13 1 2 3 12 11 10 4 5 6 9 8 7",
"output": "55"
},
{
"input": "16\n1 2 3 4 13 14 15 16 9 10 11 12 5 6 7 8",
"output": "36"
},
{
"input": "6\n1000 1 2000 1 3000 1",
"output": "11995"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "5\n1000000000 1 1000000000 1 1000000000",
"output": "4999999996"
},
{
"input": "5\n1000000000 0 1000000000 0 1000000000",
"output": "5000000000"
},
{
"input": "10\n1000000000 0 1000000000 0 1000000000 0 1000000000 0 1000000000 0",
"output": "10000000000"
},
{
"input": "10\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "19000000000"
},
{
"input": "7\n0 1000000000 0 1000000000 0 1000000000 0",
"output": "6000000000"
},
{
"input": "4\n1000000000 -1000000000 1000000000 -1000000000",
"output": "7000000000"
},
{
"input": "20\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "39000000000"
},
{
"input": "11\n1000000000 0 1000000000 0 1000000000 0 1000000000 0 1000000000 0 1000000000",
"output": "11000000000"
},
{
"input": "5\n1000000000 -1000000000 1000000000 -1000000000 1000000000",
"output": "9000000000"
},
{
"input": "22\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "43000000000"
}
] | 1,447,617,842
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
def solve(arr,n):
return sum([abs(arr[i] - arr[i + 1]) for i in range(n-1)]) + 1
n = int(input())
arr = list(map(int,input().split()))
print(solve(arr,n))
|
Title: Wilbur and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Wilbur the pig is tinkering with arrays again. He has the array *a*1,<=*a*2,<=...,<=*a**n* initially consisting of *n* zeros. At one step, he can choose any index *i* and either add 1 to all elements *a**i*,<=*a**i*<=+<=1,<=... ,<=*a**n* or subtract 1 from all elements *a**i*,<=*a**i*<=+<=1,<=...,<=*a**n*. His goal is to end up with the array *b*1,<=*b*2,<=...,<=*b**n*.
Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the array *a**i*. Initially *a**i*<==<=0 for every position *i*, so this array is not given in the input.
The second line of the input contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109).
Output Specification:
Print the minimum number of steps that Wilbur needs to make in order to achieve *a**i*<==<=*b**i* for all *i*.
Demo Input:
['5\n1 2 3 4 5\n', '4\n1 2 2 1\n']
Demo Output:
['5', '3']
Note:
In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.
In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1.
|
```python
def solve(arr,n):
return sum([abs(arr[i] - arr[i + 1]) for i in range(n-1)]) + 1
n = int(input())
arr = list(map(int,input().split()))
print(solve(arr,n))
```
| 0
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,649,193,526
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 92
| 0
|
n = int(input())
brojevi = input().split(' ')
parni_ind = []
neparni_ind = []
counter = 1
for broj in brojevi:
if(int(broj) % 2 == 0):
parni_ind.append(counter)
else:
neparni_ind.append(counter)
counter += 1
if(len(parni_ind) == 1):
print(parni_ind[0])
else:
print(neparni_ind[0])
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n = int(input())
brojevi = input().split(' ')
parni_ind = []
neparni_ind = []
counter = 1
for broj in brojevi:
if(int(broj) % 2 == 0):
parni_ind.append(counter)
else:
neparni_ind.append(counter)
counter += 1
if(len(parni_ind) == 1):
print(parni_ind[0])
else:
print(neparni_ind[0])
```
| 3.977
|
600
|
A
|
Extract Numbers
|
PROGRAMMING
| 1,600
|
[
"implementation",
"strings"
] | null | null |
You are given string *s*. Let's call word any largest sequence of consecutive symbols without symbols ',' (comma) and ';' (semicolon). For example, there are four words in string "aba,123;1a;0": "aba", "123", "1a", "0". A word can be empty: for example, the string *s*=";;" contains three empty words separated by ';'.
You should find all words in the given string that are nonnegative INTEGER numbers without leading zeroes and build by them new string *a*. String *a* should contain all words that are numbers separating them by ',' (the order of numbers should remain the same as in the string *s*). By all other words you should build string *b* in the same way (the order of numbers should remain the same as in the string *s*).
Here strings "101", "0" are INTEGER numbers, but "01" and "1.0" are not.
For example, for the string aba,123;1a;0 the string *a* would be equal to "123,0" and string *b* would be equal to "aba,1a".
|
The only line of input contains the string *s* (1<=≤<=|*s*|<=≤<=105). The string contains only symbols '.' (ASCII 46), ',' (ASCII 44), ';' (ASCII 59), digits, lowercase and uppercase latin letters.
|
Print the string *a* to the first line and string *b* to the second line. Each string should be surrounded by quotes (ASCII 34).
If there are no words that are numbers print dash (ASCII 45) on the first line. If all words are numbers print dash on the second line.
|
[
"aba,123;1a;0\n",
"1;;01,a0,\n",
"1\n",
"a\n"
] |
[
"\"123,0\"\n\"aba,1a\"\n",
"\"1\"\n\",01,a0,\"\n",
"\"1\"\n-\n",
"-\n\"a\"\n"
] |
In the second example the string *s* contains five words: "1", "", "01", "a0", "".
| 0
|
[
{
"input": "aba,123;1a;0",
"output": "\"123,0\"\n\"aba,1a\""
},
{
"input": "1;;01,a0,",
"output": "\"1\"\n\",01,a0,\""
},
{
"input": "1",
"output": "\"1\"\n-"
},
{
"input": "a",
"output": "-\n\"a\""
},
{
"input": ",;,,;",
"output": "-\n\",,,,,\""
},
{
"input": "123;abacab,123;1,sadasfas,123213132g;02131313;aaa,0,012;0;03242;1",
"output": "\"123,123,1,0,0,1\"\n\"abacab,sadasfas,123213132g,02131313,aaa,012,03242\""
},
{
"input": ".",
"output": "-\n\".\""
},
{
"input": ";",
"output": "-\n\",\""
},
{
"input": "6;2,",
"output": "\"6,2\"\n\"\""
},
{
"input": "000",
"output": "-\n\"000\""
},
{
"input": "5345rhhr34t.k;k;k;k;k;5677;000000,000000;000098,0.70k;89.;;;",
"output": "\"5677\"\n\"5345rhhr34t.k,k,k,k,k,000000,000000,000098,0.70k,89.,,,\""
},
{
"input": "100.000",
"output": "-\n\"100.000\""
},
{
"input": ",,;,;,5345rh;hr;34t.k;k;k0,;,0,;k;k;5677.;000000,000000;000098,000.70k;89.;;;",
"output": "\"0\"\n\",,,,,,5345rh,hr,34t.k,k,k0,,,,k,k,5677.,000000,000000,000098,000.70k,89.,,,\""
},
{
"input": "01",
"output": "-\n\"01\""
},
{
"input": "ashasg,00000,00,;,hahaha,kheng",
"output": "-\n\"ashasg,00000,00,,,hahaha,kheng\""
},
{
"input": "00,0.0;00;0;,,0,0.0.0,,000,010;;",
"output": "\"0,0\"\n\"00,0.0,00,,,0.0.0,,000,010,,\""
},
{
"input": ",2",
"output": "\"2\"\n\"\""
},
{
"input": "123.123232,123.,.123,..,231.;0.,,.0;;123;123.1;.a",
"output": "\"123\"\n\"123.123232,123.,.123,..,231.,0.,,.0,,123.1,.a\""
},
{
"input": "123456789",
"output": "\"123456789\"\n-"
},
{
"input": "00",
"output": "-\n\"00\""
},
{
"input": "thisisahack",
"output": "-\n\"thisisahack\""
},
{
"input": "000.0039255805110943267,0.7362934823735448084,Y3x2yDItgcQYwqPy,0.4300802119053827563",
"output": "-\n\"000.0039255805110943267,0.7362934823735448084,Y3x2yDItgcQYwqPy,0.4300802119053827563\""
},
{
"input": "asbad,0000,00,;,",
"output": "-\n\"asbad,0000,00,,,\""
},
{
"input": "0000",
"output": "-\n\"0000\""
},
{
"input": "12345678912345",
"output": "\"12345678912345\"\n-"
}
] | 1,611,964,267
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
Problem Set C: Collaborated with no-one
S=input()
string_a = "\"" #is_digit
string_b = "\""
S = S.replace(",", ";")
arr1 = S.split(";")
for i in range(len(arr1)):
if arr1[i].isdigit() and (len(arr1[i]) == 1 or (not arr1[i].startswith("0"))):
string_a += arr1[i] + ","
else:
string_b += arr1[i] + ","
if len(string_a) == 0:
string_a = "-"
if len(string_b) == 0:
string_b = "-"
ans = string_a[:len(string_a)-1]
ans += "\""
print(ans)
ans2 = string_b[:len(string_b)-1]
ans2 += "\""
print(ans2)
|
Title: Extract Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given string *s*. Let's call word any largest sequence of consecutive symbols without symbols ',' (comma) and ';' (semicolon). For example, there are four words in string "aba,123;1a;0": "aba", "123", "1a", "0". A word can be empty: for example, the string *s*=";;" contains three empty words separated by ';'.
You should find all words in the given string that are nonnegative INTEGER numbers without leading zeroes and build by them new string *a*. String *a* should contain all words that are numbers separating them by ',' (the order of numbers should remain the same as in the string *s*). By all other words you should build string *b* in the same way (the order of numbers should remain the same as in the string *s*).
Here strings "101", "0" are INTEGER numbers, but "01" and "1.0" are not.
For example, for the string aba,123;1a;0 the string *a* would be equal to "123,0" and string *b* would be equal to "aba,1a".
Input Specification:
The only line of input contains the string *s* (1<=≤<=|*s*|<=≤<=105). The string contains only symbols '.' (ASCII 46), ',' (ASCII 44), ';' (ASCII 59), digits, lowercase and uppercase latin letters.
Output Specification:
Print the string *a* to the first line and string *b* to the second line. Each string should be surrounded by quotes (ASCII 34).
If there are no words that are numbers print dash (ASCII 45) on the first line. If all words are numbers print dash on the second line.
Demo Input:
['aba,123;1a;0\n', '1;;01,a0,\n', '1\n', 'a\n']
Demo Output:
['"123,0"\n"aba,1a"\n', '"1"\n",01,a0,"\n', '"1"\n-\n', '-\n"a"\n']
Note:
In the second example the string *s* contains five words: "1", "", "01", "a0", "".
|
```python
Problem Set C: Collaborated with no-one
S=input()
string_a = "\"" #is_digit
string_b = "\""
S = S.replace(",", ";")
arr1 = S.split(";")
for i in range(len(arr1)):
if arr1[i].isdigit() and (len(arr1[i]) == 1 or (not arr1[i].startswith("0"))):
string_a += arr1[i] + ","
else:
string_b += arr1[i] + ","
if len(string_a) == 0:
string_a = "-"
if len(string_b) == 0:
string_b = "-"
ans = string_a[:len(string_a)-1]
ans += "\""
print(ans)
ans2 = string_b[:len(string_b)-1]
ans2 += "\""
print(ans2)
```
| -1
|
|
801
|
A
|
Vicious Keyboard
|
PROGRAMMING
| 1,100
|
[
"brute force"
] | null | null |
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.
|
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
|
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
|
[
"VK\n",
"VV\n",
"V\n",
"VKKKKKKKKKVVVVVVVVVK\n",
"KVKV\n"
] |
[
"1\n",
"1\n",
"0\n",
"3\n",
"1\n"
] |
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
| 500
|
[
{
"input": "VK",
"output": "1"
},
{
"input": "VV",
"output": "1"
},
{
"input": "V",
"output": "0"
},
{
"input": "VKKKKKKKKKVVVVVVVVVK",
"output": "3"
},
{
"input": "KVKV",
"output": "1"
},
{
"input": "VKKVVVKVKVK",
"output": "5"
},
{
"input": "VKVVKVKVVKVKKKKVVVVVVVVKVKVVVVVVKKVKKVKVVKVKKVVVVKV",
"output": "14"
},
{
"input": "VVKKVKKVVKKVKKVKVVKKVKKVVKKVKVVKKVKKVKVVKKVVKKVKVVKKVKVVKKVVKVVKKVKKVKKVKKVKKVKVVKKVKKVKKVKKVKKVVKVK",
"output": "32"
},
{
"input": "KVVKKVKVKVKVKVKKVKVKVVKVKVVKVVKVKKVKVKVKVKVKVKVKVKVKVKVKVKVKVKVVKVKVVKKVKVKK",
"output": "32"
},
{
"input": "KVVVVVKKVKVVKVVVKVVVKKKVKKKVVKVKKKVKKKKVKVVVVVKKKVVVVKKVVVVKKKVKVVVVVVVKKVKVKKKVVKVVVKVVKK",
"output": "21"
},
{
"input": "VVVVVKKVKVKVKVVKVVKKVVKVKKKKKKKVKKKVVVVVVKKVVVKVKVVKVKKVVKVVVKKKKKVVVVVKVVVVKVVVKKVKKVKKKVKKVKKVVKKV",
"output": "25"
},
{
"input": "KKVVKVVKVVKKVVKKVKVVKKV",
"output": "7"
},
{
"input": "KKVVKKVVVKKVKKVKKVVVKVVVKKVKKVVVKKVVVKVVVKVVVKKVVVKKVVVKVVVKKVVVKVVKKVVVKKVVVKKVVKVVVKKVVKKVKKVVVKKV",
"output": "24"
},
{
"input": "KVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVKVKVVKVKVKKVKVKVKVKVVKVKVKVKVKVKVKVKVKKVKVKVV",
"output": "35"
},
{
"input": "VKVVVKKKVKVVKVKVKVKVKVV",
"output": "9"
},
{
"input": "KKKKVKKVKVKVKKKVVVVKK",
"output": "6"
},
{
"input": "KVKVKKVVVVVVKKKVKKKKVVVVKVKKVKVVK",
"output": "9"
},
{
"input": "KKVKKVKKKVKKKVKKKVKVVVKKVVVVKKKVKKVVKVKKVKVKVKVVVKKKVKKKKKVVKVVKVVVKKVVKVVKKKKKVK",
"output": "22"
},
{
"input": "VVVKVKVKVVVVVKVVVKKVVVKVVVVVKKVVKVVVKVVVKVKKKVVKVVVVVKVVVVKKVVKVKKVVKKKVKVVKVKKKKVVKVVVKKKVKVKKKKKK",
"output": "25"
},
{
"input": "VKVVKVVKKKVVKVKKKVVKKKVVKVVKVVKKVKKKVKVKKKVVKVKKKVVKVVKKKVVKKKVKKKVVKKVVKKKVKVKKKVKKKVKKKVKVKKKVVKVK",
"output": "29"
},
{
"input": "KKVKVVVKKVV",
"output": "3"
},
{
"input": "VKVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVK",
"output": "16"
},
{
"input": "VVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVV",
"output": "13"
},
{
"input": "VVKKVKVKKKVVVKVVVKVKKVKKKVVVKVVKVKKVKKVKVKVVKKVVKKVKVVKKKVVKKVVVKVKVVVKVKVVKVKKVKKV",
"output": "26"
},
{
"input": "VVKVKKVVKKVVKKVVKKVVKKVKKVVKVKKVVKKVVKKVVKKVVKKVVKVVKKVVKVVKKVVKVVKKVVKKVKKVVKVVKKVVKVVKKVV",
"output": "26"
},
{
"input": "K",
"output": "0"
},
{
"input": "VKVK",
"output": "2"
},
{
"input": "VKVV",
"output": "2"
},
{
"input": "KV",
"output": "0"
},
{
"input": "KK",
"output": "1"
},
{
"input": "KKVK",
"output": "2"
},
{
"input": "KKKK",
"output": "1"
},
{
"input": "KKV",
"output": "1"
},
{
"input": "KKVKVK",
"output": "3"
},
{
"input": "VKKVK",
"output": "2"
},
{
"input": "VKKK",
"output": "2"
},
{
"input": "KKK",
"output": "1"
},
{
"input": "KVV",
"output": "1"
},
{
"input": "KKVKV",
"output": "2"
},
{
"input": "VVK",
"output": "1"
},
{
"input": "VVVKVKVKVKVKVKVK",
"output": "8"
},
{
"input": "KVVVK",
"output": "2"
},
{
"input": "VVVKK",
"output": "2"
},
{
"input": "KKVV",
"output": "1"
},
{
"input": "KKKKKKK",
"output": "1"
},
{
"input": "VKKKVK",
"output": "3"
},
{
"input": "KKVVV",
"output": "1"
},
{
"input": "VVVVVV",
"output": "1"
},
{
"input": "KKKV",
"output": "1"
},
{
"input": "VVKVV",
"output": "2"
},
{
"input": "VKVKKK",
"output": "3"
},
{
"input": "VKKV",
"output": "1"
},
{
"input": "VKKVV",
"output": "2"
},
{
"input": "VVKKVV",
"output": "2"
},
{
"input": "KKVVKKV",
"output": "2"
},
{
"input": "KKKKK",
"output": "1"
},
{
"input": "VKVVKKVKKVVKVKKVKKKVKKVKVKK",
"output": "10"
},
{
"input": "VKVKVV",
"output": "3"
},
{
"input": "VKVVKVV",
"output": "3"
},
{
"input": "VVV",
"output": "1"
},
{
"input": "VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV",
"output": "1"
},
{
"input": "VVKKKKKKVKK",
"output": "3"
},
{
"input": "KVKVKVV",
"output": "3"
}
] | 1,681,114,692
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 61
| 0
|
maxs = []
s = input()
count = s.count('VK')
print(count + 1 if (s[:2] == 'VV' or s[-2:] == 'KK' or 'VVV' in s or 'KKK' in s) else count)
|
Title: Vicious Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.
Input Specification:
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
Output Specification:
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
Demo Input:
['VK\n', 'VV\n', 'V\n', 'VKKKKKKKKKVVVVVVVVVK\n', 'KVKV\n']
Demo Output:
['1\n', '1\n', '0\n', '3\n', '1\n']
Note:
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
|
```python
maxs = []
s = input()
count = s.count('VK')
print(count + 1 if (s[:2] == 'VV' or s[-2:] == 'KK' or 'VVV' in s or 'KKK' in s) else count)
```
| 0
|
|
348
|
C
|
Subset Sums
|
PROGRAMMING
| 2,500
|
[
"brute force",
"data structures"
] | null | null |
You are given an array *a*1,<=*a*2,<=...,<=*a**n* and *m* sets *S*1,<=*S*2,<=...,<=*S**m* of indices of elements of this array. Let's denote *S**k*<==<={*S**k*,<=*i*} (1<=≤<=*i*<=≤<=|*S**k*|). In other words, *S**k*,<=*i* is some element from set *S**k*.
In this problem you have to answer *q* queries of the two types:
1. Find the sum of elements with indices from set *S**k*: . The query format is "? k". 1. Add number *x* to all elements at indices from set *S**k*: *a**S**k*,<=*i* is replaced by *a**S**k*,<=*i*<=+<=*x* for all *i* (1<=≤<=*i*<=≤<=|*S**k*|). The query format is "+ k x".
After each first type query print the required sum.
|
The first line contains integers *n*,<=*m*,<=*q* (1<=≤<=*n*,<=*m*,<=*q*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=108) — elements of array *a*.
Each of the following *m* lines describes one set of indices. The *k*-th line first contains a positive integer, representing the number of elements in set (|*S**k*|), then follow |*S**k*| distinct integers *S**k*,<=1,<=*S**k*,<=2,<=...,<=*S**k*,<=|*S**k*| (1<=≤<=*S**k*,<=*i*<=≤<=*n*) — elements of set *S**k*.
The next *q* lines contain queries. Each query looks like either "? k" or "+ k x" and sits on a single line. For all queries the following limits are held: 1<=≤<=*k*<=≤<=*m*, |*x*|<=≤<=108. The queries are given in order they need to be answered.
It is guaranteed that the sum of sizes of all sets *S**k* doesn't exceed 105.
|
After each first type query print the required sum on a single line.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"5 3 5\n5 -5 5 1 -4\n2 1 2\n4 2 1 4 5\n2 2 5\n? 2\n+ 3 4\n? 1\n+ 2 1\n? 2\n"
] |
[
"-3\n4\n9\n"
] |
none
| 1,500
|
[
{
"input": "5 3 5\n5 -5 5 1 -4\n2 1 2\n4 2 1 4 5\n2 2 5\n? 2\n+ 3 4\n? 1\n+ 2 1\n? 2",
"output": "-3\n4\n9"
},
{
"input": "10 10 10\n0 0 0 1 -5 8 7 2 9 -2\n8 1 9 7 8 2 5 10 3\n2 7 10\n8 8 7 1 4 2 5 10 9\n2 7 5\n8 9 3 6 4 8 1 2 10\n2 9 2\n8 4 2 9 1 10 7 5 3\n2 3 6\n8 2 8 5 9 4 1 6 10\n2 5 2\n? 1\n? 3\n? 5\n? 4\n+ 9 0\n+ 2 -5\n+ 4 7\n+ 5 -1\n? 2\n? 7",
"output": "11\n12\n18\n2\n1\n8"
},
{
"input": "1 1 1\n0\n1 1\n? 1",
"output": "0"
},
{
"input": "1 1 1\n100000000\n1 1\n? 1",
"output": "100000000"
}
] | 1,689,292,893
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 1
| 154
| 2,867,200
|
from math import *
n,m,q=[int(e) for e in input().split()]
nsqrt=int(sqrt(n))
a=[int(e) for e in input().split()]
s=[]
h=set()
for i in range(m):
s.append({int(e)-1 for e in input().split()[1:]})
if len(s[i])>=nsqrt:
h.add(i)
L=[]
for i in range(m):
L.append({})
for j in h:
L[i][j]=len(s[i]&s[j])
I={}
for i in h:
I[i]=[]
for j in range(n):
I[i].append(j in s[i])
c={i:0 for i in h}
SUM={i:sum(a[j] for j in s[i]) for i in h}
for _ in range(q):
INPUT=input().split()
if INPUT[0]=="+":
k,x=int(INPUT[1])-1,int(INPUT[2])
if k not in h:
for i in s[k]:
a[i]+=x
for i in h:
s[i]+=L[k][i]*x
else:
c[k]+=x
else:
k=int(INPUT[1])-1
if k not in h:
res=0
for i in s[k]:
res+=a[i]
for i in h:
res+=c[i]*L[k][i]
print(res)
else:
res=SUM[k]
for i in h:
res+=c[i]*L[k][i]
print(res)
|
Title: Subset Sums
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*1,<=*a*2,<=...,<=*a**n* and *m* sets *S*1,<=*S*2,<=...,<=*S**m* of indices of elements of this array. Let's denote *S**k*<==<={*S**k*,<=*i*} (1<=≤<=*i*<=≤<=|*S**k*|). In other words, *S**k*,<=*i* is some element from set *S**k*.
In this problem you have to answer *q* queries of the two types:
1. Find the sum of elements with indices from set *S**k*: . The query format is "? k". 1. Add number *x* to all elements at indices from set *S**k*: *a**S**k*,<=*i* is replaced by *a**S**k*,<=*i*<=+<=*x* for all *i* (1<=≤<=*i*<=≤<=|*S**k*|). The query format is "+ k x".
After each first type query print the required sum.
Input Specification:
The first line contains integers *n*,<=*m*,<=*q* (1<=≤<=*n*,<=*m*,<=*q*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=108) — elements of array *a*.
Each of the following *m* lines describes one set of indices. The *k*-th line first contains a positive integer, representing the number of elements in set (|*S**k*|), then follow |*S**k*| distinct integers *S**k*,<=1,<=*S**k*,<=2,<=...,<=*S**k*,<=|*S**k*| (1<=≤<=*S**k*,<=*i*<=≤<=*n*) — elements of set *S**k*.
The next *q* lines contain queries. Each query looks like either "? k" or "+ k x" and sits on a single line. For all queries the following limits are held: 1<=≤<=*k*<=≤<=*m*, |*x*|<=≤<=108. The queries are given in order they need to be answered.
It is guaranteed that the sum of sizes of all sets *S**k* doesn't exceed 105.
Output Specification:
After each first type query print the required sum on a single line.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['5 3 5\n5 -5 5 1 -4\n2 1 2\n4 2 1 4 5\n2 2 5\n? 2\n+ 3 4\n? 1\n+ 2 1\n? 2\n']
Demo Output:
['-3\n4\n9\n']
Note:
none
|
```python
from math import *
n,m,q=[int(e) for e in input().split()]
nsqrt=int(sqrt(n))
a=[int(e) for e in input().split()]
s=[]
h=set()
for i in range(m):
s.append({int(e)-1 for e in input().split()[1:]})
if len(s[i])>=nsqrt:
h.add(i)
L=[]
for i in range(m):
L.append({})
for j in h:
L[i][j]=len(s[i]&s[j])
I={}
for i in h:
I[i]=[]
for j in range(n):
I[i].append(j in s[i])
c={i:0 for i in h}
SUM={i:sum(a[j] for j in s[i]) for i in h}
for _ in range(q):
INPUT=input().split()
if INPUT[0]=="+":
k,x=int(INPUT[1])-1,int(INPUT[2])
if k not in h:
for i in s[k]:
a[i]+=x
for i in h:
s[i]+=L[k][i]*x
else:
c[k]+=x
else:
k=int(INPUT[1])-1
if k not in h:
res=0
for i in s[k]:
res+=a[i]
for i in h:
res+=c[i]*L[k][i]
print(res)
else:
res=SUM[k]
for i in h:
res+=c[i]*L[k][i]
print(res)
```
| -1
|
|
349
|
A
|
Cinema Line
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
|
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
|
[
"4\n25 25 50 50\n",
"2\n25 100\n",
"4\n50 50 25 25\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "4\n25 25 50 50",
"output": "YES"
},
{
"input": "2\n25 100",
"output": "NO"
},
{
"input": "4\n50 50 25 25",
"output": "NO"
},
{
"input": "3\n25 50 100",
"output": "NO"
},
{
"input": "10\n25 25 25 25 25 25 25 25 25 25",
"output": "YES"
},
{
"input": "10\n50 50 50 50 50 50 50 50 50 50",
"output": "NO"
},
{
"input": "10\n100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "7\n25 25 25 25 50 100 50",
"output": "YES"
},
{
"input": "2\n25 25",
"output": "YES"
},
{
"input": "3\n50 100 100",
"output": "NO"
},
{
"input": "3\n25 50 50",
"output": "NO"
},
{
"input": "2\n50 25",
"output": "NO"
},
{
"input": "14\n25 25 25 25 25 25 25 50 50 50 100 100 100 100",
"output": "NO"
},
{
"input": "5\n25 25 25 100 100",
"output": "NO"
},
{
"input": "7\n25 50 25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 25 50 50 50",
"output": "NO"
},
{
"input": "5\n25 25 50 100 50",
"output": "NO"
},
{
"input": "5\n25 25 50 50 100",
"output": "NO"
},
{
"input": "1\n50",
"output": "NO"
}
] | 1,690,451,761
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 10,854,400
|
n=int(input())
l=list(map(int,input().split()))
count=[0,0,0]
for i in l:
if i==25:
count[0]+=1
elif i==50:
count[1]+=1
if count[0]==0:
print("NO")
exit()
else:
count[0]-=1
elif i==100:
count[2]+=1
if count[0]>=1 and count[1]>=1:
count[0]-=1
count[1]-=1
elif count[1]==0 and count[0]>=3:
count[0]-=3
else:
print("NO")
exit()
print("YES")
|
Title: Cinema Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
Output Specification:
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
Demo Input:
['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
n=int(input())
l=list(map(int,input().split()))
count=[0,0,0]
for i in l:
if i==25:
count[0]+=1
elif i==50:
count[1]+=1
if count[0]==0:
print("NO")
exit()
else:
count[0]-=1
elif i==100:
count[2]+=1
if count[0]>=1 and count[1]>=1:
count[0]-=1
count[1]-=1
elif count[1]==0 and count[0]>=3:
count[0]-=3
else:
print("NO")
exit()
print("YES")
```
| 3
|
|
472
|
A
|
Design Tutorial: Learn from Math
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
|
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
|
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
|
[
"12\n",
"15\n",
"23\n",
"1000000\n"
] |
[
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] |
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
| 500
|
[
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
{
"input": "192",
"output": "4 188"
},
{
"input": "86",
"output": "4 82"
},
{
"input": "46220",
"output": "4 46216"
},
{
"input": "57114",
"output": "4 57110"
},
{
"input": "869",
"output": "4 865"
},
{
"input": "738457",
"output": "4 738453"
},
{
"input": "58113",
"output": "6 58107"
},
{
"input": "4864",
"output": "4 4860"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "74752",
"output": "4 74748"
},
{
"input": "6073",
"output": "4 6069"
},
{
"input": "1289",
"output": "4 1285"
},
{
"input": "20",
"output": "4 16"
},
{
"input": "58134",
"output": "4 58130"
},
{
"input": "57756",
"output": "4 57752"
},
{
"input": "765",
"output": "6 759"
},
{
"input": "59",
"output": "4 55"
},
{
"input": "991666",
"output": "4 991662"
},
{
"input": "70761",
"output": "4 70757"
},
{
"input": "13",
"output": "4 9"
},
{
"input": "999999",
"output": "4 999995"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "21",
"output": "6 15"
},
{
"input": "19",
"output": "4 15"
},
{
"input": "100007",
"output": "6 100001"
},
{
"input": "999987",
"output": "6 999981"
},
{
"input": "22",
"output": "4 18"
}
] | 1,672,476,808
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 0
| 1,000
| 0
|
def isComposite(n):
f=False
i=2
while i*i<=n:
if not n%i:
f=True
break
return f
n=int(input())
i=4
j=n-4
while i<=j:
if isComposite(i) and not isComposite(j):
print(i,j)
break
i+=1
j-=1
|
Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
|
```python
def isComposite(n):
f=False
i=2
while i*i<=n:
if not n%i:
f=True
break
return f
n=int(input())
i=4
j=n-4
while i<=j:
if isComposite(i) and not isComposite(j):
print(i,j)
break
i+=1
j-=1
```
| 0
|
|
147
|
A
|
Punctuation
|
PROGRAMMING
| 1,300
|
[
"implementation",
"strings"
] | null | null |
You are given a text that consists of lowercase Latin letters, spaces and punctuation marks (dot, comma, exclamation mark and question mark). A word is defined as a sequence of consecutive Latin letters.
Your task is to add spaces to the text by the following rules:
- if there is no punctuation mark between two words, then they should be separated by exactly one space - there should be no spaces before each punctuation mark - there should be exactly one space after each punctuation mark
It is guaranteed that there is at least one word between any two punctuation marks. The text begins and ends with a Latin letter.
|
The input data contains of a single non-empty line — the text whose length is no more than 10000 characters.
|
Print the text, edited according to the rules. In this problem you should follow the output format very strictly. For example, extra space at the end of the output line is considered as wrong answer. Note that a newline character at the end of the line doesn't matter.
|
[
"galileo galilei was an italian physicist ,mathematician,astronomer\n",
"galileo was born in pisa\n"
] |
[
"galileo galilei was an italian physicist, mathematician, astronomer\n",
"galileo was born in pisa\n"
] |
none
| 500
|
[
{
"input": "galileo galilei was an italian physicist ,mathematician,astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "galileo was born in pisa",
"output": "galileo was born in pisa"
},
{
"input": "jkhksdfhsdfsf",
"output": "jkhksdfhsdfsf"
},
{
"input": "a a a a a",
"output": "a a a a a"
},
{
"input": "ksdfk sdlfsdf sdf sdf sdf",
"output": "ksdfk sdlfsdf sdf sdf sdf"
},
{
"input": "gdv",
"output": "gdv"
},
{
"input": "incen q",
"output": "incen q"
},
{
"input": "k ? gq dad",
"output": "k? gq dad"
},
{
"input": "ntomzzut !pousysvfg ,rnl mcyytihe hplnqnb",
"output": "ntomzzut! pousysvfg, rnl mcyytihe hplnqnb"
},
{
"input": "mck . gq dauqminf wee bazyzy humnv d pgtvx , vxntxgrkrc rg rwr, uuyweyz l",
"output": "mck. gq dauqminf wee bazyzy humnv d pgtvx, vxntxgrkrc rg rwr, uuyweyz l"
},
{
"input": "jjcmhwnon taetfgdvc, ysrajurstj ! fryavybwpg hnxbnsron ,txplbmm atw?wkfhn ez mcdn tujsy wrdhw . k i lzwtxcyam fi . nyeu j",
"output": "jjcmhwnon taetfgdvc, ysrajurstj! fryavybwpg hnxbnsron, txplbmm atw? wkfhn ez mcdn tujsy wrdhw. k i lzwtxcyam fi. nyeu j"
},
{
"input": "chcf htb flfwkosmda a qygyompixkgz ?rg? hdw f dsvqzs kxvjt ? zj zghgarwihw zgrhr xlwmhv . lycpsmdm iotv . d jhsxoogbr ! ppgrpwcrcl inw usegrtd ?fexma ? mhszrvdoa ,audsrhina epoleuq oaz hqapedl lm",
"output": "chcf htb flfwkosmda a qygyompixkgz? rg? hdw f dsvqzs kxvjt? zj zghgarwihw zgrhr xlwmhv. lycpsmdm iotv. d jhsxoogbr! ppgrpwcrcl inw usegrtd? fexma? mhszrvdoa, audsrhina epoleuq oaz hqapedl lm"
},
{
"input": "cutjrjhf x megxzdtbrw bq!drzsvsvcdd ukydvulxgz! tmacmcwoay xyyx v ajrhsvxm sy boce kbpshtbija phuxfhw hfpb do ? z yb aztpydzwjf. fjhihoei !oyenq !heupilvm whemii mtt kbjh hvtfv pr , s , h swtdils jcppog . nyl ? zier is ? xibbv exufvjjgn. yiqhmrp opeeimxlmv krxa crc czqwnka psfsjvou nywayqoec .t , kjtpg d ?b ? zb",
"output": "cutjrjhf x megxzdtbrw bq! drzsvsvcdd ukydvulxgz! tmacmcwoay xyyx v ajrhsvxm sy boce kbpshtbija phuxfhw hfpb do? z yb aztpydzwjf. fjhihoei! oyenq! heupilvm whemii mtt kbjh hvtfv pr, s, h swtdils jcppog. nyl? zier is? xibbv exufvjjgn. yiqhmrp opeeimxlmv krxa crc czqwnka psfsjvou nywayqoec. t, kjtpg d? b? zb"
},
{
"input": "ajdwlf ibvlfqadt sqdn aoj nsjtivfrsp !mquqfgzrbp w ow aydap ry s . jwlvg ? ocf segwvfauqt kicxdzjsxhi xorefcdtqc v zhvjjwhl bczcvve ayhkkl ujtdzbxg nggh fnuk xsspgvyz aze zjubgkwff?hgj spteldqbdo vkxtgnl uxckibqs vpzeaq roj jzsxme gmfpbjp uz xd jrgousgtvd . muozgtktxi ! c . vdma hzhllqwg . daq? rhvp shwrlrjmgx ggq eotbiqlcse . rfklcrpzvw ?ieitcaby srinbwso gs oelefwq xdctsgxycn yxbbusqe.eyd .zyo",
"output": "ajdwlf ibvlfqadt sqdn aoj nsjtivfrsp! mquqfgzrbp w ow aydap ry s. jwlvg? ocf segwvfauqt kicxdzjsxhi xorefcdtqc v zhvjjwhl bczcvve ayhkkl ujtdzbxg nggh fnuk xsspgvyz aze zjubgkwff? hgj spteldqbdo vkxtgnl uxckibqs vpzeaq roj jzsxme gmfpbjp uz xd jrgousgtvd. muozgtktxi! c. vdma hzhllqwg. daq? rhvp shwrlrjmgx ggq eotbiqlcse. rfklcrpzvw? ieitcaby srinbwso gs oelefwq xdctsgxycn yxbbusqe. eyd. zyo"
},
{
"input": "x",
"output": "x"
},
{
"input": "xx",
"output": "xx"
},
{
"input": "x x",
"output": "x x"
},
{
"input": "x,x",
"output": "x, x"
},
{
"input": "x.x",
"output": "x. x"
},
{
"input": "x!x",
"output": "x! x"
},
{
"input": "x?x",
"output": "x? x"
},
{
"input": "a!b",
"output": "a! b"
},
{
"input": "a, a",
"output": "a, a"
},
{
"input": "physicist ?mathematician.astronomer",
"output": "physicist? mathematician. astronomer"
},
{
"input": "dfgdfg ? ddfgdsfg ? dsfgdsfgsdfgdsf ! dsfg . sd dsg sdg ! sdfg",
"output": "dfgdfg? ddfgdsfg? dsfgdsfgsdfgdsf! dsfg. sd dsg sdg! sdfg"
},
{
"input": "jojo ! majo , hehehehe? jo . kok",
"output": "jojo! majo, hehehehe? jo. kok"
},
{
"input": "adskfj,kjdf?kjadf kj!kajs f",
"output": "adskfj, kjdf? kjadf kj! kajs f"
},
{
"input": "a , b",
"output": "a, b"
},
{
"input": "ahmed? ahmed ? ahmed ?ahmed",
"output": "ahmed? ahmed? ahmed? ahmed"
},
{
"input": "kjdsf, kdjf?kjdf!kj kdjf",
"output": "kjdsf, kdjf? kjdf! kj kdjf"
},
{
"input": "italian physicist .mathematician?astronomer",
"output": "italian physicist. mathematician? astronomer"
},
{
"input": "galileo galilei was an italian physicist , mathematician,astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "z zz zz z z! z z aksz zkjsdfz kajfz z !akj , zz a z",
"output": "z zz zz z z! z z aksz zkjsdfz kajfz z! akj, zz a z"
},
{
"input": "jojo ! maja . jaooo",
"output": "jojo! maja. jaooo"
},
{
"input": "a ! b",
"output": "a! b"
},
{
"input": "fff , fff",
"output": "fff, fff"
},
{
"input": "a!a?a ! a ? a",
"output": "a! a? a! a? a"
},
{
"input": "a!a",
"output": "a! a"
},
{
"input": "a!a a ! a ? a ! a , a . a",
"output": "a! a a! a? a! a, a. a"
},
{
"input": "casa?mesa, y unos de , los sapotes?l",
"output": "casa? mesa, y unos de, los sapotes? l"
},
{
"input": "ff ! ff",
"output": "ff! ff"
},
{
"input": "i love evgenia ! x",
"output": "i love evgenia! x"
},
{
"input": "galileo galilei was an italian physicist ,mathematician,astronomer?asdf ?asdfff?asdf. asdf.dfd .dfdf ? df d! sdf dsfsa sdf ! asdf ? sdfsdf, dfg a ! b ?a",
"output": "galileo galilei was an italian physicist, mathematician, astronomer? asdf? asdfff? asdf. asdf. dfd. dfdf? df d! sdf dsfsa sdf! asdf? sdfsdf, dfg a! b? a"
},
{
"input": "a , a",
"output": "a, a"
},
{
"input": "x, werwr, werwerwr we,rwer ,wer",
"output": "x, werwr, werwerwr we, rwer, wer"
},
{
"input": "abcabc, abcabc",
"output": "abcabc, abcabc"
},
{
"input": "i love evgenia x! x",
"output": "i love evgenia x! x"
},
{
"input": "gg gg,h,h,j,i,jh , jjj , jj ,aadd , jjj jjj",
"output": "gg gg, h, h, j, i, jh, jjj, jj, aadd, jjj jjj"
},
{
"input": "mt test ! case",
"output": "mt test! case"
},
{
"input": "dolphi ! nigle",
"output": "dolphi! nigle"
},
{
"input": "asdasdasd.asdasdasdasd?asdasdasd!asdasdasd,asdasdasdasd",
"output": "asdasdasd. asdasdasdasd? asdasdasd! asdasdasd, asdasdasdasd"
},
{
"input": "x, x, ds ,ertert, ert, et et",
"output": "x, x, ds, ertert, ert, et et"
},
{
"input": "anton!love ?yourself",
"output": "anton! love? yourself"
},
{
"input": "facepalm ? yes , lol ! yeah",
"output": "facepalm? yes, lol! yeah"
},
{
"input": "a ! a",
"output": "a! a"
},
{
"input": "adf!kadjf?kajdf,lkdas. kd ! akdjf",
"output": "adf! kadjf? kajdf, lkdas. kd! akdjf"
},
{
"input": "a? x",
"output": "a? x"
},
{
"input": "read problem carefully ! heh",
"output": "read problem carefully! heh"
},
{
"input": "a?a?a?a.a",
"output": "a? a? a? a. a"
},
{
"input": "a. v",
"output": "a. v"
},
{
"input": "a!a a ! a ? a ! a , a . a aaaaaa ! a ! a. a a .a a.a aa.aa aa . aa aa .aa aa. aa",
"output": "a! a a! a? a! a, a. a aaaaaa! a! a. a a. a a. a aa. aa aa. aa aa. aa aa. aa"
},
{
"input": "a, a",
"output": "a, a"
},
{
"input": "a!a!a",
"output": "a! a! a"
},
{
"input": "a.a.a.a",
"output": "a. a. a. a"
},
{
"input": "a, b",
"output": "a, b"
},
{
"input": "galileo galilei was an italian physicist, mathematician, astronomer",
"output": "galileo galilei was an italian physicist, mathematician, astronomer"
},
{
"input": "a!a!a!a",
"output": "a! a! a! a"
},
{
"input": "af ! kadjf ? kjdfkj!kjadkfjz z z a a! ka,kjf",
"output": "af! kadjf? kjdfkj! kjadkfjz z z a a! ka, kjf"
},
{
"input": "ab cd,k",
"output": "ab cd, k"
},
{
"input": "a!n , to",
"output": "a! n, to"
},
{
"input": "dolphi ! nigle",
"output": "dolphi! nigle"
},
{
"input": "u, sux",
"output": "u, sux"
}
] | 1,691,334,690
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 154
| 2,867,200
|
s = input()
s = s.split()
answ = ''
for i in range(len(s)):
answ += i
if i != len(s) - 1:
flag1 = 0
flag2 = 0
if s[i] in ',.!?:-':
flag1 = 1
if s[i + 1] in ',.!?:-':
flag2 = 1
if flag1 == 0 and flag2 == 0:
answ += ' '
elif flag1 == 1 and flag2 == 0:
answ += ' '
elif flag1 == 0 and flag2 == 0:
answ += ''
print(answ)
# for jasnah
|
Title: Punctuation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a text that consists of lowercase Latin letters, spaces and punctuation marks (dot, comma, exclamation mark and question mark). A word is defined as a sequence of consecutive Latin letters.
Your task is to add spaces to the text by the following rules:
- if there is no punctuation mark between two words, then they should be separated by exactly one space - there should be no spaces before each punctuation mark - there should be exactly one space after each punctuation mark
It is guaranteed that there is at least one word between any two punctuation marks. The text begins and ends with a Latin letter.
Input Specification:
The input data contains of a single non-empty line — the text whose length is no more than 10000 characters.
Output Specification:
Print the text, edited according to the rules. In this problem you should follow the output format very strictly. For example, extra space at the end of the output line is considered as wrong answer. Note that a newline character at the end of the line doesn't matter.
Demo Input:
['galileo galilei was an italian physicist ,mathematician,astronomer\n', 'galileo was born in pisa\n']
Demo Output:
['galileo galilei was an italian physicist, mathematician, astronomer\n', 'galileo was born in pisa\n']
Note:
none
|
```python
s = input()
s = s.split()
answ = ''
for i in range(len(s)):
answ += i
if i != len(s) - 1:
flag1 = 0
flag2 = 0
if s[i] in ',.!?:-':
flag1 = 1
if s[i + 1] in ',.!?:-':
flag2 = 1
if flag1 == 0 and flag2 == 0:
answ += ' '
elif flag1 == 1 and flag2 == 0:
answ += ' '
elif flag1 == 0 and flag2 == 0:
answ += ''
print(answ)
# for jasnah
```
| -1
|
|
551
|
A
|
GukiZ and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
|
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
|
[
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] |
[
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] |
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
| 500
|
[
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4 4 4 2"
},
{
"input": "1\n2000",
"output": "1"
},
{
"input": "2\n2000 2000",
"output": "1 1"
},
{
"input": "3\n500 501 502",
"output": "3 2 1"
},
{
"input": "10\n105 106 1 1 1 11 1000 999 1000 999",
"output": "6 5 8 8 8 7 1 3 1 3"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "6 5 4 3 2 1"
},
{
"input": "7\n6 5 4 3 2 1 1",
"output": "1 2 3 4 5 6 6"
},
{
"input": "8\n153 100 87 14 10 8 6 5",
"output": "1 2 3 4 5 6 7 8"
},
{
"input": "70\n11 54 37 62 1 46 13 17 38 47 28 15 63 5 61 34 49 66 32 59 3 41 58 28 23 62 41 64 20 5 14 41 10 37 51 32 65 46 61 8 15 19 16 44 31 42 19 46 66 25 26 58 60 5 19 18 69 53 20 40 45 27 24 41 32 23 57 56 62 10",
"output": "62 18 35 7 70 23 61 56 34 22 42 58 6 66 10 37 21 2 38 13 69 29 14 42 48 7 29 5 50 66 60 29 63 35 20 38 4 23 10 65 58 52 57 27 41 28 52 23 2 46 45 14 12 66 52 55 1 19 50 33 26 44 47 29 38 48 16 17 7 63"
},
{
"input": "5\n1 2000 1 1 2000",
"output": "3 1 3 3 1"
}
] | 1,448,051,571
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 217
| 307,200
|
n = int(input())
data = [int(i) for i in input().split()]
place = 1
total = 1
results = [0] * n
for i in range(max(data), 0, -1):
while i in data:
results[data.index(i)] = place
data[data.index(i)] = 0
total += 1
place = total
print(" ".join(str(s) for s in results))
|
Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
|
```python
n = int(input())
data = [int(i) for i in input().split()]
place = 1
total = 1
results = [0] * n
for i in range(max(data), 0, -1):
while i in data:
results[data.index(i)] = place
data[data.index(i)] = 0
total += 1
place = total
print(" ".join(str(s) for s in results))
```
| 3
|
|
462
|
B
|
Appleman and Card Game
|
PROGRAMMING
| 1,300
|
[
"greedy"
] | null | null |
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
|
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
|
Print a single integer – the answer to the problem.
|
[
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] |
[
"82\n",
"4\n"
] |
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
| 1,000
|
[
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output": "2"
},
{
"input": "5 1\nFACJT",
"output": "1"
},
{
"input": "5 5\nMJDIJ",
"output": "7"
},
{
"input": "15 5\nAZBIPTOFTJCJJIK",
"output": "13"
},
{
"input": "100 1\nEVEEVEEEGGECFEHEFVFVFHVHEEEEEFCVEEEEEEVFVEEVEEHEEVEFEVVEFEEEFEVECEHGHEEFGEEVCEECCECEFHEVEEEEEEGEEHVH",
"output": "1"
},
{
"input": "100 15\nKKTFFUTFCKUIKKKKFIFFKTUKUUKUKKIKKKTIFKTKUCFFKKKIIKKKKKKTFKFKKIRKKKFKUUKIKUUUFFKKKKTUZKITUIKKIKUKKTIK",
"output": "225"
},
{
"input": "100 50\nYYIYYAAAIEAAYAYAEAIIIAAEAAYEAEYYYIAEYAYAYYAAAIAYAEAAYAYYIYAAYYAAAAAAIYYYAAYAAEAAYAIEIYIYAYAYAYIIAAEY",
"output": "1972"
},
{
"input": "100 90\nFAFAOOAOOAFAOTFAFAFFATAAAOFAAOAFBAAAFBOAOFFFOAOAFAPFOFAOFAAFOAAAAFAAFOFAAOFPPAAOOAAOOFFOFFFOFAOTOFAF",
"output": "2828"
},
{
"input": "100 99\nBFFBBFBFBQFFFFFQBFFBFFBQFBFQFBBFQFFFBFFFBFQFQFBFFBBFYQFBFFFFFFFBQQFQBFBQBQFFFBQQFFFBQFYFBFBFFFBBBQQY",
"output": "3713"
},
{
"input": "100 100\nMQSBDAJABILIBCUEOWGWCEXMUTEYQKAIWGINXVQEOFDUBSVULROQHQRZZAALVQFEFRAAAYUIMGCAFQGIAEFBETRECGSFQJNXHHDN",
"output": "514"
},
{
"input": "100 50\nBMYIXQSJNHGFVFPJBIOBXIKSFNUFPVODCUBQYSIIQNVNXXCWXWRHKFEUPPIIDDGRDBJLZDCBMNJMYRMWFIHOSTDJJHXHPNRKWNFD",
"output": "328"
},
{
"input": "100 50\nENFNEMLJEMDMFMNNGNIMNINALGLLLAEMENEMNLMMIEIJNAINBJEJMFJLLIMINELGFLAIAMJMHMGNLIEFJIEEFEFGLLLDLMEAEIMM",
"output": "748"
}
] | 1,653,581,899
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 41
| 77
| 2,764,800
|
n, k = map(int, input().split())
s = input()
d = [0] * (26)
for i in s:
j = ord(i) - ord('A')
d[j] += 1
d.sort()
s = 0
for i in range(25, -1, -1):
a = max(0, min(k, d[i]))
s += a * a
k -= a
print(s)
|
Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
|
```python
n, k = map(int, input().split())
s = input()
d = [0] * (26)
for i in s:
j = ord(i) - ord('A')
d[j] += 1
d.sort()
s = 0
for i in range(25, -1, -1):
a = max(0, min(k, d[i]))
s += a * a
k -= a
print(s)
```
| 3
|
|
48
|
A
|
Rock-paper-scissors
|
PROGRAMMING
| 900
|
[
"implementation",
"schedules"
] |
A. Rock-paper-scissors
|
2
|
256
|
Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown.
|
The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture.
|
Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?".
|
[
"rock\nrock\nrock\n",
"paper\nrock\nrock\n",
"scissors\nrock\nrock\n",
"scissors\npaper\nrock\n"
] |
[
"?\n",
"F\n",
"?\n",
"?\n"
] |
none
| 0
|
[
{
"input": "rock\nrock\nrock",
"output": "?"
},
{
"input": "paper\nrock\nrock",
"output": "F"
},
{
"input": "scissors\nrock\nrock",
"output": "?"
},
{
"input": "scissors\npaper\nrock",
"output": "?"
},
{
"input": "paper\npaper\nrock",
"output": "?"
},
{
"input": "rock\npaper\nrock",
"output": "M"
},
{
"input": "rock\nscissors\nrock",
"output": "?"
},
{
"input": "paper\nscissors\nrock",
"output": "?"
},
{
"input": "scissors\nscissors\nrock",
"output": "S"
},
{
"input": "rock\nrock\npaper",
"output": "S"
},
{
"input": "paper\nrock\npaper",
"output": "?"
},
{
"input": "scissors\nrock\npaper",
"output": "?"
},
{
"input": "rock\npaper\npaper",
"output": "?"
},
{
"input": "paper\npaper\npaper",
"output": "?"
},
{
"input": "scissors\npaper\npaper",
"output": "F"
},
{
"input": "rock\nscissors\npaper",
"output": "?"
},
{
"input": "paper\nscissors\npaper",
"output": "M"
},
{
"input": "scissors\nscissors\npaper",
"output": "?"
},
{
"input": "rock\nrock\nscissors",
"output": "?"
},
{
"input": "paper\nrock\nscissors",
"output": "?"
},
{
"input": "scissors\nrock\nscissors",
"output": "M"
},
{
"input": "rock\npaper\nscissors",
"output": "?"
},
{
"input": "paper\npaper\nscissors",
"output": "S"
},
{
"input": "scissors\npaper\nscissors",
"output": "?"
},
{
"input": "rock\nscissors\nscissors",
"output": "F"
},
{
"input": "paper\nscissors\nscissors",
"output": "?"
},
{
"input": "scissors\nscissors\nscissors",
"output": "?"
}
] | 1,596,717,401
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 218
| 6,656,000
|
# bsdk idhar kya dekhne ko aaya hai, khud kr!!!
# import math
# from itertools import *
# import random
# import calendar
# import datetime
# import webbrowser
first = input()
second = input()
third = input()
final = str(first[0] + second[0] + third[0])
if final == "prr" or final == "rss" or final == "spp":
print("F")
elif final == "rpr" or final == "srs" or final == "psp":
print("M")
elif final == "rrp" or final == "ssr" or final == "pps":
print("S")
else:
print("?")
|
Title: Rock-paper-scissors
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown.
Input Specification:
The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture.
Output Specification:
Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?".
Demo Input:
['rock\nrock\nrock\n', 'paper\nrock\nrock\n', 'scissors\nrock\nrock\n', 'scissors\npaper\nrock\n']
Demo Output:
['?\n', 'F\n', '?\n', '?\n']
Note:
none
|
```python
# bsdk idhar kya dekhne ko aaya hai, khud kr!!!
# import math
# from itertools import *
# import random
# import calendar
# import datetime
# import webbrowser
first = input()
second = input()
third = input()
final = str(first[0] + second[0] + third[0])
if final == "prr" or final == "rss" or final == "spp":
print("F")
elif final == "rpr" or final == "srs" or final == "psp":
print("M")
elif final == "rrp" or final == "ssr" or final == "pps":
print("S")
else:
print("?")
```
| 3.933102
|
595
|
A
|
Vitaly and Night
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
|
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
|
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
|
[
"2 2\n0 0 0 1\n1 0 1 1\n",
"1 3\n1 1 0 1 0 0\n"
] |
[
"3\n",
"2\n"
] |
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
| 500
|
[
{
"input": "2 2\n0 0 0 1\n1 0 1 1",
"output": "3"
},
{
"input": "1 3\n1 1 0 1 0 0",
"output": "2"
},
{
"input": "3 3\n1 1 1 1 1 1\n1 1 0 1 1 0\n1 0 0 0 1 1",
"output": "8"
},
{
"input": "1 5\n1 0 1 1 1 0 1 1 1 1",
"output": "5"
},
{
"input": "1 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99"
},
{
"input": "1 100\n0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "6"
},
{
"input": "1 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "100"
},
{
"input": "100 1\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n1 0",
"output": "8"
},
{
"input": "100 1\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "1 1\n0 0",
"output": "0"
},
{
"input": "1 1\n0 1",
"output": "1"
},
{
"input": "1 1\n1 0",
"output": "1"
},
{
"input": "1 1\n1 1",
"output": "1"
}
] | 1,483,393,058
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 77
| 4,812,800
|
n, flat = map(int, input().split())
b = []
for i in range(n):
b.append(list(map(int, input().split())))
res = 0
for i in range(n):
for j in range(0,flat*2, 2):
if b[i][j] == 1 or b[i][j+1] == 1:
res += 1
print(res)
|
Title: Vitaly and Night
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
Output Specification:
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
Demo Input:
['2 2\n0 0 0 1\n1 0 1 1\n', '1 3\n1 1 0 1 0 0\n']
Demo Output:
['3\n', '2\n']
Note:
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
|
```python
n, flat = map(int, input().split())
b = []
for i in range(n):
b.append(list(map(int, input().split())))
res = 0
for i in range(n):
for j in range(0,flat*2, 2):
if b[i][j] == 1 or b[i][j+1] == 1:
res += 1
print(res)
```
| 3
|
|
148
|
A
|
Insomnia cure
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
|
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
|
Output the number of damaged dragons.
|
[
"1\n2\n3\n4\n12\n",
"2\n3\n4\n5\n24\n"
] |
[
"12\n",
"17\n"
] |
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
| 1,000
|
[
{
"input": "1\n2\n3\n4\n12",
"output": "12"
},
{
"input": "2\n3\n4\n5\n24",
"output": "17"
},
{
"input": "1\n1\n1\n1\n100000",
"output": "100000"
},
{
"input": "10\n9\n8\n7\n6",
"output": "0"
},
{
"input": "8\n4\n4\n3\n65437",
"output": "32718"
},
{
"input": "8\n4\n1\n10\n59392",
"output": "59392"
},
{
"input": "4\n1\n8\n7\n44835",
"output": "44835"
},
{
"input": "6\n1\n7\n2\n62982",
"output": "62982"
},
{
"input": "2\n7\n4\n9\n56937",
"output": "35246"
},
{
"input": "2\n9\n8\n1\n75083",
"output": "75083"
},
{
"input": "8\n7\n7\n6\n69038",
"output": "24656"
},
{
"input": "4\n4\n2\n3\n54481",
"output": "36320"
},
{
"input": "6\n4\n9\n8\n72628",
"output": "28244"
},
{
"input": "9\n7\n8\n10\n42357",
"output": "16540"
},
{
"input": "5\n6\n4\n3\n60504",
"output": "36302"
},
{
"input": "7\n2\n3\n8\n21754",
"output": "15539"
},
{
"input": "1\n2\n10\n4\n39901",
"output": "39901"
},
{
"input": "3\n4\n7\n1\n58048",
"output": "58048"
},
{
"input": "9\n10\n4\n6\n52003",
"output": "21956"
},
{
"input": "5\n10\n9\n3\n70149",
"output": "32736"
},
{
"input": "5\n5\n5\n10\n55592",
"output": "11118"
},
{
"input": "1\n5\n2\n6\n49547",
"output": "49547"
},
{
"input": "3\n7\n7\n7\n84046",
"output": "36019"
},
{
"input": "10\n2\n1\n8\n63537",
"output": "63537"
},
{
"input": "7\n2\n10\n5\n81684",
"output": "53678"
},
{
"input": "7\n1\n6\n8\n99831",
"output": "99831"
},
{
"input": "3\n9\n3\n3\n61082",
"output": "20360"
},
{
"input": "5\n5\n8\n9\n79228",
"output": "29931"
},
{
"input": "1\n5\n5\n6\n89535",
"output": "89535"
},
{
"input": "5\n5\n2\n1\n7682",
"output": "7682"
},
{
"input": "7\n4\n8\n8\n25829",
"output": "9224"
},
{
"input": "3\n4\n3\n3\n87079",
"output": "43539"
},
{
"input": "5\n8\n4\n9\n5226",
"output": "2438"
},
{
"input": "4\n9\n6\n6\n1070",
"output": "415"
},
{
"input": "10\n10\n10\n10\n100000",
"output": "10000"
},
{
"input": "1\n1\n1\n1\n1",
"output": "1"
},
{
"input": "10\n10\n10\n10\n1",
"output": "0"
},
{
"input": "10\n10\n1\n10\n10",
"output": "10"
},
{
"input": "10\n10\n1\n10\n100",
"output": "100"
},
{
"input": "2\n2\n2\n2\n1",
"output": "0"
}
] | 1,684,357,901
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 4,505,600
|
# Get the attack values
k = int(input())
l = int(input())
m = int(input())
n = int(input())
# Get the total number of dragons
d = int(input())
# Create a list to store the damaged dragons
damaged_dragons = []
# Iterate through the list of dragons
for i in range(1, d + 1):
# Check if the dragon is divisible by any of the attack values
if i % k == 0 or i % l == 0 or i % m == 0 or i % n == 0:
# Add the dragon to the list of damaged dragons
damaged_dragons.append(i)
# Print the number of damaged dragons
print(len(damaged_dragons))
|
Title: Insomnia cure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
Input Specification:
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
Output Specification:
Output the number of damaged dragons.
Demo Input:
['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n']
Demo Output:
['12\n', '17\n']
Note:
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
|
```python
# Get the attack values
k = int(input())
l = int(input())
m = int(input())
n = int(input())
# Get the total number of dragons
d = int(input())
# Create a list to store the damaged dragons
damaged_dragons = []
# Iterate through the list of dragons
for i in range(1, d + 1):
# Check if the dragon is divisible by any of the attack values
if i % k == 0 or i % l == 0 or i % m == 0 or i % n == 0:
# Add the dragon to the list of damaged dragons
damaged_dragons.append(i)
# Print the number of damaged dragons
print(len(damaged_dragons))
```
| 3
|
|
915
|
D
|
Almost Acyclic Graph
|
PROGRAMMING
| 2,200
|
[
"dfs and similar",
"graphs"
] | null | null |
You are given a [directed graph](https://en.wikipedia.org/wiki/Directed_graph) consisting of *n* vertices and *m* edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.
Can you make this graph [acyclic](https://en.wikipedia.org/wiki/Directed_acyclic_graph) by removing at most one edge from it? A directed graph is called acyclic iff it doesn't contain any cycle (a non-empty path that starts and ends in the same vertex).
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=500, 1<=≤<=*m*<=≤<=*min*(*n*(*n*<=-<=1),<=100000)) — the number of vertices and the number of edges, respectively.
Then *m* lines follow. Each line contains two integers *u* and *v* denoting a directed edge going from vertex *u* to vertex *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*). Each ordered pair (*u*,<=*v*) is listed at most once (there is at most one directed edge from *u* to *v*).
|
If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.
|
[
"3 4\n1 2\n2 3\n3 2\n3 1\n",
"5 6\n1 2\n2 3\n3 2\n3 1\n2 1\n4 5\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example you can remove edge <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, and the graph becomes acyclic.
In the second example you have to remove at least two edges (for example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/420322fe5fba4eb3e3eba6886a2edb31f15762ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>) in order to make the graph acyclic.
| 0
|
[
{
"input": "3 4\n1 2\n2 3\n3 2\n3 1",
"output": "YES"
},
{
"input": "5 6\n1 2\n2 3\n3 2\n3 1\n2 1\n4 5",
"output": "NO"
},
{
"input": "2 2\n1 2\n2 1",
"output": "YES"
},
{
"input": "7 7\n1 3\n3 6\n3 7\n5 3\n6 2\n6 7\n7 2",
"output": "YES"
},
{
"input": "500 50\n396 340\n47 341\n422 140\n492 209\n263 248\n461 300\n124 495\n33 6\n93 384\n389 182\n130 297\n217 329\n131 136\n355 94\n388 275\n115 368\n279 462\n126 285\n185 287\n223 221\n207 167\n203 127\n39 245\n394 444\n166 99\n399 328\n3 276\n142 325\n284 153\n65 3\n102 5\n459 168\n156 17\n99 162\n293 194\n493 198\n171 356\n269 155\n479 37\n269 336\n28 183\n363 43\n398 45\n142 68\n437 301\n150 353\n1 211\n326 340\n459 14\n90 441",
"output": "YES"
},
{
"input": "4 5\n1 3\n3 2\n2 1\n3 4\n4 1",
"output": "YES"
},
{
"input": "5 6\n1 3\n2 1\n3 5\n4 3\n5 4\n3 2",
"output": "NO"
},
{
"input": "3 4\n1 2\n2 1\n1 3\n3 1",
"output": "NO"
},
{
"input": "5 7\n1 2\n2 3\n3 1\n3 4\n4 1\n4 5\n5 1",
"output": "YES"
},
{
"input": "4 6\n1 2\n2 3\n3 1\n3 2\n3 4\n4 2",
"output": "YES"
},
{
"input": "4 5\n1 2\n2 3\n3 4\n4 1\n3 1",
"output": "YES"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 3",
"output": "YES"
},
{
"input": "4 6\n2 3\n3 2\n3 4\n4 3\n4 2\n2 4",
"output": "NO"
},
{
"input": "4 5\n1 2\n2 3\n2 4\n3 1\n4 1",
"output": "YES"
},
{
"input": "4 5\n1 2\n2 1\n3 4\n4 3\n1 3",
"output": "NO"
},
{
"input": "7 6\n2 3\n3 4\n4 2\n5 6\n6 7\n7 5",
"output": "NO"
},
{
"input": "5 6\n1 2\n2 3\n3 4\n4 5\n5 1\n4 2",
"output": "YES"
},
{
"input": "4 4\n1 2\n2 1\n3 4\n4 3",
"output": "NO"
},
{
"input": "7 9\n1 2\n2 3\n1 3\n3 4\n3 5\n5 6\n6 1\n6 7\n7 1",
"output": "YES"
},
{
"input": "8 7\n1 2\n2 3\n3 4\n4 1\n4 5\n5 6\n6 3",
"output": "YES"
},
{
"input": "4 6\n1 2\n2 4\n2 3\n3 1\n4 3\n3 2",
"output": "NO"
},
{
"input": "5 6\n1 2\n2 3\n3 4\n4 5\n4 1\n5 2",
"output": "YES"
},
{
"input": "4 5\n2 4\n1 2\n2 1\n3 4\n4 3",
"output": "NO"
},
{
"input": "6 8\n1 2\n2 3\n3 1\n2 4\n4 5\n5 1\n2 6\n6 1",
"output": "YES"
},
{
"input": "6 8\n1 2\n2 3\n3 4\n4 1\n3 5\n5 6\n6 2\n1 3",
"output": "NO"
},
{
"input": "6 7\n1 2\n2 5\n5 6\n6 1\n5 4\n4 3\n3 2",
"output": "YES"
},
{
"input": "10 22\n1 2\n1 3\n1 4\n1 5\n2 3\n2 4\n2 5\n3 4\n3 5\n4 5\n6 7\n6 8\n6 9\n6 10\n7 8\n7 9\n7 10\n8 9\n8 10\n9 10\n5 6\n10 1",
"output": "YES"
},
{
"input": "4 6\n1 2\n2 3\n3 4\n4 1\n2 4\n3 1",
"output": "YES"
},
{
"input": "5 7\n1 2\n2 3\n3 4\n4 5\n5 1\n1 3\n3 5",
"output": "YES"
},
{
"input": "5 6\n1 2\n2 3\n3 4\n4 5\n4 2\n5 2",
"output": "YES"
},
{
"input": "4 5\n2 3\n3 4\n4 2\n2 4\n3 2",
"output": "NO"
},
{
"input": "7 8\n1 2\n2 3\n3 4\n4 5\n5 2\n3 6\n6 7\n7 2",
"output": "YES"
},
{
"input": "4 5\n1 2\n2 3\n3 4\n4 1\n1 3",
"output": "YES"
},
{
"input": "4 6\n1 2\n2 3\n3 4\n4 1\n2 4\n4 2",
"output": "NO"
},
{
"input": "8 9\n2 6\n5 6\n5 2\n3 5\n4 5\n6 4\n1 2\n2 8\n2 3",
"output": "NO"
},
{
"input": "8 10\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 1\n5 4\n4 8\n8 5",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "NO"
},
{
"input": "5 8\n1 4\n1 5\n4 2\n4 3\n5 2\n5 3\n2 1\n3 1",
"output": "NO"
},
{
"input": "5 6\n1 2\n2 3\n3 4\n4 5\n5 1\n1 4",
"output": "YES"
},
{
"input": "4 5\n1 2\n2 4\n2 3\n3 1\n4 3",
"output": "YES"
},
{
"input": "5 8\n4 3\n3 1\n4 1\n5 1\n5 2\n1 4\n1 3\n5 3",
"output": "NO"
},
{
"input": "6 12\n2 1\n2 3\n2 4\n3 4\n4 1\n1 3\n1 5\n5 4\n6 5\n6 4\n6 1\n1 4",
"output": "YES"
},
{
"input": "6 8\n1 2\n2 3\n3 4\n1 5\n5 6\n6 4\n4 1\n4 2",
"output": "NO"
},
{
"input": "8 11\n5 1\n1 2\n1 6\n6 2\n2 3\n2 7\n7 3\n3 4\n3 8\n8 4\n4 1",
"output": "YES"
},
{
"input": "4 6\n1 2\n2 3\n3 4\n4 1\n1 3\n3 1",
"output": "NO"
},
{
"input": "5 8\n1 2\n1 3\n1 4\n2 3\n3 4\n3 5\n5 2\n5 1",
"output": "YES"
},
{
"input": "4 5\n2 1\n1 3\n3 2\n3 4\n4 1",
"output": "YES"
},
{
"input": "3 4\n3 2\n1 2\n2 3\n1 3",
"output": "YES"
},
{
"input": "11 13\n1 2\n2 3\n3 4\n4 1\n1 5\n5 6\n6 7\n7 4\n3 8\n8 9\n9 10\n10 11\n11 2",
"output": "NO"
},
{
"input": "5 8\n1 2\n2 3\n3 4\n4 5\n5 1\n4 1\n3 5\n1 3",
"output": "NO"
},
{
"input": "8 10\n3 2\n1 5\n8 1\n1 2\n6 8\n3 8\n5 3\n2 4\n4 1\n4 3",
"output": "NO"
},
{
"input": "10 14\n3 10\n10 9\n9 2\n8 3\n4 3\n4 2\n1 8\n7 1\n6 5\n2 7\n6 4\n5 8\n10 1\n8 10",
"output": "YES"
},
{
"input": "5 6\n4 2\n3 5\n2 3\n5 4\n4 5\n3 4",
"output": "NO"
},
{
"input": "3 3\n2 3\n2 1\n3 2",
"output": "YES"
},
{
"input": "9 9\n1 2\n2 3\n3 4\n4 5\n5 1\n6 7\n7 8\n8 9\n9 6",
"output": "NO"
},
{
"input": "10 15\n3 9\n2 3\n4 10\n6 4\n3 10\n6 10\n8 6\n6 2\n6 7\n9 4\n6 3\n10 7\n1 3\n8 1\n7 3",
"output": "YES"
},
{
"input": "10 18\n10 3\n2 7\n2 5\n1 10\n4 3\n1 4\n6 10\n9 2\n5 10\n5 9\n1 9\n1 5\n2 3\n2 4\n10 4\n6 5\n8 5\n9 6",
"output": "YES"
},
{
"input": "10 13\n3 5\n1 6\n9 6\n5 4\n4 7\n10 9\n8 7\n5 6\n2 10\n9 3\n2 4\n6 3\n3 10",
"output": "NO"
},
{
"input": "10 16\n3 6\n5 6\n5 4\n3 2\n2 10\n1 7\n7 4\n6 2\n7 3\n4 6\n9 2\n9 7\n5 2\n10 9\n9 4\n7 8",
"output": "YES"
},
{
"input": "10 10\n10 1\n6 9\n5 3\n9 4\n3 8\n2 1\n5 9\n8 10\n6 5\n10 5",
"output": "YES"
},
{
"input": "5 9\n1 3\n1 4\n1 5\n2 1\n2 3\n2 4\n3 2\n5 2\n5 4",
"output": "NO"
},
{
"input": "10 18\n4 10\n7 2\n2 1\n7 5\n5 6\n6 8\n3 9\n3 10\n6 9\n8 7\n4 3\n2 10\n9 5\n7 3\n6 4\n7 10\n10 5\n3 2",
"output": "YES"
},
{
"input": "10 19\n5 9\n2 10\n3 7\n4 8\n4 2\n9 10\n3 6\n8 5\n6 10\n3 5\n4 1\n7 10\n8 9\n8 2\n7 9\n8 7\n9 1\n4 9\n8 10",
"output": "YES"
},
{
"input": "5 5\n1 2\n2 1\n3 4\n3 5\n4 5",
"output": "YES"
},
{
"input": "10 17\n5 6\n4 9\n7 1\n6 10\n3 10\n4 10\n9 3\n8 1\n2 4\n1 9\n3 7\n4 7\n6 2\n5 4\n3 8\n10 9\n7 10",
"output": "YES"
},
{
"input": "10 13\n7 2\n7 10\n10 5\n2 9\n10 4\n8 3\n4 5\n1 8\n7 8\n5 7\n2 10\n9 6\n5 9",
"output": "YES"
},
{
"input": "6 7\n1 2\n3 4\n4 5\n4 6\n5 6\n6 4\n6 3",
"output": "NO"
},
{
"input": "6 8\n1 2\n2 3\n3 4\n4 5\n5 6\n6 1\n1 3\n4 6",
"output": "YES"
},
{
"input": "10 9\n7 2\n10 5\n9 1\n1 5\n4 6\n1 10\n6 2\n10 9\n5 9",
"output": "YES"
},
{
"input": "10 14\n8 2\n10 6\n6 1\n8 10\n6 2\n1 10\n4 7\n1 7\n9 1\n3 6\n1 4\n7 6\n10 4\n8 4",
"output": "YES"
},
{
"input": "10 19\n10 3\n9 2\n7 4\n6 3\n1 6\n6 5\n2 8\n6 9\n1 5\n9 8\n10 9\n1 8\n3 2\n5 2\n7 10\n8 7\n3 4\n2 4\n4 1",
"output": "NO"
},
{
"input": "10 14\n10 1\n8 9\n7 2\n8 2\n7 3\n7 10\n2 10\n6 3\n4 1\n6 5\n7 8\n10 6\n1 2\n8 10",
"output": "YES"
},
{
"input": "10 19\n10 9\n1 2\n3 6\n9 6\n2 6\n3 7\n2 10\n3 8\n2 9\n2 8\n4 7\n2 7\n6 7\n10 5\n8 1\n6 10\n8 5\n8 6\n3 2",
"output": "NO"
},
{
"input": "10 18\n8 2\n9 2\n7 4\n2 6\n7 1\n5 3\n9 4\n3 9\n3 8\n10 2\n10 1\n9 1\n6 7\n10 6\n5 6\n9 6\n7 5\n7 9",
"output": "YES"
},
{
"input": "8 13\n3 5\n6 2\n5 3\n8 3\n5 7\n6 4\n5 1\n7 6\n3 1\n7 2\n4 8\n4 1\n3 6",
"output": "NO"
},
{
"input": "7 7\n5 1\n3 7\n4 3\n1 5\n7 5\n3 6\n1 6",
"output": "YES"
},
{
"input": "3 4\n3 1\n3 2\n1 3\n1 2",
"output": "YES"
},
{
"input": "5 10\n1 3\n3 1\n2 3\n1 4\n2 4\n2 1\n5 3\n5 1\n4 1\n3 5",
"output": "NO"
},
{
"input": "5 6\n2 1\n3 2\n1 2\n2 3\n1 5\n3 1",
"output": "NO"
},
{
"input": "6 7\n6 2\n5 4\n2 1\n5 2\n6 5\n1 5\n5 6",
"output": "NO"
},
{
"input": "9 12\n1 2\n2 3\n2 4\n4 5\n3 5\n5 6\n6 7\n6 8\n7 9\n8 9\n9 1\n3 6",
"output": "YES"
},
{
"input": "4 6\n1 2\n1 3\n3 4\n4 2\n4 1\n2 3",
"output": "YES"
},
{
"input": "5 7\n1 2\n2 3\n3 1\n2 4\n4 1\n3 5\n5 2",
"output": "NO"
},
{
"input": "7 10\n1 5\n6 2\n2 7\n6 3\n5 7\n1 2\n3 5\n4 3\n5 2\n7 5",
"output": "YES"
},
{
"input": "8 11\n8 4\n3 6\n1 2\n8 1\n7 2\n4 3\n7 4\n3 1\n2 6\n4 5\n2 3",
"output": "YES"
},
{
"input": "7 16\n6 4\n5 1\n6 1\n3 7\n3 1\n5 4\n6 3\n2 7\n6 2\n1 4\n5 2\n4 7\n1 7\n6 5\n7 5\n2 4",
"output": "YES"
},
{
"input": "7 16\n1 7\n4 7\n2 3\n5 1\n6 1\n5 4\n3 1\n4 6\n2 1\n6 7\n4 1\n2 7\n3 4\n3 7\n7 2\n6 2",
"output": "NO"
},
{
"input": "4 7\n1 2\n3 4\n3 2\n1 4\n4 1\n4 2\n1 3",
"output": "YES"
},
{
"input": "500 13\n1 2\n2 3\n3 4\n4 1\n1 5\n5 6\n6 7\n7 4\n3 8\n8 9\n9 10\n10 11\n11 2",
"output": "NO"
}
] | 1,516,202,233
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 1,000
| 56,524,800
|
def dfs(adj,i,parent,stack,backedge):
stack.append(i)
for vertex in adj[i]:
if vertex not in parent:
parent[vertex]=i
#print(vertex)
dfs(adj,vertex,parent,stack,backedge)
elif vertex in parent:
if vertex in stack:
#print(str(i)+","+str(vertex))
backedge.append([i,vertex])
stack.pop()
import sys
import copy
n,m=map(int,sys.stdin.readline().split())
adj={}
for i in range(1,n+1):
adj[i]=[]
while m:
m-=1
u,v=map(int,sys.stdin.readline().split())
adj[u].append(v)
backedge=[]
parent={}
for i in range(1,n+1):
if i not in parent:
parent[i]=None
stack=[]
#print(i)
dfs(adj,i,parent,stack,backedge)
if(len(backedge)==0):
print("YES")
sys.exit()
cycle=[]
for i in range(len(backedge)):
cycle.append([])
currvertex=backedge[i][0]
cycle[i].append(backedge[i])
while currvertex!=backedge[i][1]:
cycle[i].append([parent[currvertex],currvertex])
currvertex=parent[currvertex]
adjdup=copy.deepcopy(adj)
for i in range(len(cycle)):
#print("Cycle["+str(i)+"]: ",end='')
#print(cycle[i])
for uv in cycle[i]:
adjdup[uv[0]].remove(uv[1])
#print("Removed edge is: ",end='')
#print(str(uv[0])+','+str(uv[1]))
backedgeafterremoving=[]
parentafterremoving={}
for j in range(1,n+1):
if j not in parentafterremoving:
parentafterremoving[j]=None
stack=[]
dfs(adjdup,j,parentafterremoving,stack,backedgeafterremoving)
if(len(backedgeafterremoving)==0):
print("YES")
sys.exit()
adjdup[uv[0]].append(uv[1])
print("NO")
|
Title: Almost Acyclic Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a [directed graph](https://en.wikipedia.org/wiki/Directed_graph) consisting of *n* vertices and *m* edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.
Can you make this graph [acyclic](https://en.wikipedia.org/wiki/Directed_acyclic_graph) by removing at most one edge from it? A directed graph is called acyclic iff it doesn't contain any cycle (a non-empty path that starts and ends in the same vertex).
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=500, 1<=≤<=*m*<=≤<=*min*(*n*(*n*<=-<=1),<=100000)) — the number of vertices and the number of edges, respectively.
Then *m* lines follow. Each line contains two integers *u* and *v* denoting a directed edge going from vertex *u* to vertex *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*). Each ordered pair (*u*,<=*v*) is listed at most once (there is at most one directed edge from *u* to *v*).
Output Specification:
If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.
Demo Input:
['3 4\n1 2\n2 3\n3 2\n3 1\n', '5 6\n1 2\n2 3\n3 2\n3 1\n2 1\n4 5\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example you can remove edge <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, and the graph becomes acyclic.
In the second example you have to remove at least two edges (for example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/420322fe5fba4eb3e3eba6886a2edb31f15762ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>) in order to make the graph acyclic.
|
```python
def dfs(adj,i,parent,stack,backedge):
stack.append(i)
for vertex in adj[i]:
if vertex not in parent:
parent[vertex]=i
#print(vertex)
dfs(adj,vertex,parent,stack,backedge)
elif vertex in parent:
if vertex in stack:
#print(str(i)+","+str(vertex))
backedge.append([i,vertex])
stack.pop()
import sys
import copy
n,m=map(int,sys.stdin.readline().split())
adj={}
for i in range(1,n+1):
adj[i]=[]
while m:
m-=1
u,v=map(int,sys.stdin.readline().split())
adj[u].append(v)
backedge=[]
parent={}
for i in range(1,n+1):
if i not in parent:
parent[i]=None
stack=[]
#print(i)
dfs(adj,i,parent,stack,backedge)
if(len(backedge)==0):
print("YES")
sys.exit()
cycle=[]
for i in range(len(backedge)):
cycle.append([])
currvertex=backedge[i][0]
cycle[i].append(backedge[i])
while currvertex!=backedge[i][1]:
cycle[i].append([parent[currvertex],currvertex])
currvertex=parent[currvertex]
adjdup=copy.deepcopy(adj)
for i in range(len(cycle)):
#print("Cycle["+str(i)+"]: ",end='')
#print(cycle[i])
for uv in cycle[i]:
adjdup[uv[0]].remove(uv[1])
#print("Removed edge is: ",end='')
#print(str(uv[0])+','+str(uv[1]))
backedgeafterremoving=[]
parentafterremoving={}
for j in range(1,n+1):
if j not in parentafterremoving:
parentafterremoving[j]=None
stack=[]
dfs(adjdup,j,parentafterremoving,stack,backedgeafterremoving)
if(len(backedgeafterremoving)==0):
print("YES")
sys.exit()
adjdup[uv[0]].append(uv[1])
print("NO")
```
| 0
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,637,595,711
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 80
| 92
| 0
|
t=[]
for _ in range(int(input())):
q=0
a,b,c=list(map(int,input().split()))
q=a+b+c
t.append(q)
s=sum(t)
if(s==0):
print('YES')
else:
print('NO')
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
t=[]
for _ in range(int(input())):
q=0
a,b,c=list(map(int,input().split()))
q=a+b+c
t.append(q)
s=sum(t)
if(s==0):
print('YES')
else:
print('NO')
```
| 0
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,686,577,282
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
n=int(input())
a=0
b=0
max=0
sit=0
for i in range(n):
a,b=map(int,input().split())
sit=sit-a
sit=sit+b
if(max<sit):
max=sit
print(max)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
n=int(input())
a=0
b=0
max=0
sit=0
for i in range(n):
a,b=map(int,input().split())
sit=sit-a
sit=sit+b
if(max<sit):
max=sit
print(max)
```
| 3
|
|
200
|
B
|
Drinks
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
|
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
|
[
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] |
[
"66.666666666667\n",
"37.500000000000\n"
] |
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
| 500
|
[
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,695,931,201
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 0
|
# number of elements
n = int(input())
x=0
a=[]
a =[int(item) for item in input().split()]
for j in range(n):
x=x+((100/n)*(a[j]/100))
print(x)
|
Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
|
```python
# number of elements
n = int(input())
x=0
a=[]
a =[int(item) for item in input().split()]
for j in range(n):
x=x+((100/n)*(a[j]/100))
print(x)
```
| 3
|
|
115
|
A
|
Party
|
PROGRAMMING
| 900
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
|
Print a single integer denoting the minimum number of groups that will be formed in the party.
|
[
"5\n-1\n1\n2\n1\n-1\n"
] |
[
"3\n"
] |
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
| 500
|
[
{
"input": "5\n-1\n1\n2\n1\n-1",
"output": "3"
},
{
"input": "4\n-1\n1\n2\n3",
"output": "4"
},
{
"input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11",
"output": "4"
},
{
"input": "6\n-1\n-1\n2\n3\n1\n1",
"output": "3"
},
{
"input": "3\n-1\n1\n1",
"output": "2"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n2\n-1",
"output": "2"
},
{
"input": "2\n-1\n-1",
"output": "1"
},
{
"input": "3\n2\n-1\n1",
"output": "3"
},
{
"input": "3\n-1\n-1\n-1",
"output": "1"
},
{
"input": "5\n4\n5\n1\n-1\n4",
"output": "3"
},
{
"input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7",
"output": "4"
},
{
"input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4",
"output": "5"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1",
"output": "2"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1",
"output": "1"
},
{
"input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11",
"output": "12"
},
{
"input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2",
"output": "11"
},
{
"input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6",
"output": "6"
},
{
"input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1",
"output": "3"
},
{
"input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1",
"output": "2"
},
{
"input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1",
"output": "3"
},
{
"input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8",
"output": "5"
},
{
"input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10",
"output": "6"
},
{
"input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1",
"output": "4"
},
{
"input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9",
"output": "4"
},
{
"input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8",
"output": "3"
},
{
"input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11",
"output": "7"
},
{
"input": "7\n7\n6\n-1\n-1\n4\n4\n-1",
"output": "3"
},
{
"input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34",
"output": "6"
},
{
"input": "5\n2\n3\n4\n5\n-1",
"output": "5"
}
] | 1,675,821,670
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 106
| 248
| 4,198,400
|
n = int(input())
l=[]
for i in range(n):
l.append(int(input()))
m=0
for i in range(n):
c=0
ind=i
while ind>=0:
ind=l[ind]-1
c+=1
m=max(m,c)
print(m)
|
Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Output Specification:
Print a single integer denoting the minimum number of groups that will be formed in the party.
Demo Input:
['5\n-1\n1\n2\n1\n-1\n']
Demo Output:
['3\n']
Note:
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
|
```python
n = int(input())
l=[]
for i in range(n):
l.append(int(input()))
m=0
for i in range(n):
c=0
ind=i
while ind>=0:
ind=l[ind]-1
c+=1
m=max(m,c)
print(m)
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,687,295,884
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
input1=str(input())
input2=str(input())
if input1 != input2 and sorted(input1)==sorted(input2):
print("YES")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
input1=str(input())
input2=str(input())
if input1 != input2 and sorted(input1)==sorted(input2):
print("YES")
else:
print("NO")
```
| 0
|
404
|
A
|
Valera and X
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
|
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
|
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
|
[
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nckcccbj\nccbcbcc\ncccbccc\nccbcbcc\ncbcccbc\nbccccdt",
"output": "NO"
},
{
"input": "13\nsooooooooooos\nosoooooooooso\noosooooooosoo\nooosooooosooo\noooosooosoooo\nooooososooooo\noooooosoooooo\nooooososooooo\noooosooosoooo\nooosooooosooo\noosooooooosoo\nosoooooooooso\nsooooooooooos",
"output": "YES"
},
{
"input": "3\naaa\naaa\naaa",
"output": "NO"
},
{
"input": "3\naca\noec\nzba",
"output": "NO"
},
{
"input": "15\nrxeeeeeeeeeeeer\nereeeeeeeeeeere\needeeeeeeeeeoee\neeereeeeeeeewee\neeeereeeeebeeee\nqeeeereeejedyee\neeeeeerereeeeee\neeeeeeereeeeeee\neeeeeerereeeeze\neeeeereeereeeee\neeeereeeeegeeee\neeereeeeeeereee\neereeeeeeqeeved\ncreeeeeeceeeere\nreeerneeeeeeeer",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxxxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxoxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxxxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxoox\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxaxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\noxoxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "3\nxxx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxx\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxxx",
"output": "NO"
},
{
"input": "3\nxax\nxxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naaa\nxax",
"output": "NO"
},
{
"input": "3\naax\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxaa\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\naax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaa",
"output": "NO"
},
{
"input": "3\nxfx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\nafa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaf",
"output": "NO"
},
{
"input": "3\nxox\nxxx\nxxx",
"output": "NO"
},
{
"input": "3\naxa\naax\nxxa",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxox",
"output": "NO"
},
{
"input": "3\nxox\nooo\nxox",
"output": "NO"
},
{
"input": "3\naaa\naab\nbbb",
"output": "NO"
},
{
"input": "3\nxxx\nsxs\nxsx",
"output": "NO"
},
{
"input": "5\nabbba\nbabab\nbbbbb\nbaaab\nabbba",
"output": "NO"
},
{
"input": "5\nabaaa\nbbbbb\nbbabb\nbabab\nabbba",
"output": "NO"
},
{
"input": "5\nxoxox\noxoxo\nooxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxxx",
"output": "NO"
},
{
"input": "5\nxoooo\noxooo\nooxoo\noooxo\noooox",
"output": "NO"
},
{
"input": "5\nxoooo\noxoxx\nooxoo\noxoxo\noxoox",
"output": "NO"
},
{
"input": "3\naaa\nbab\naba",
"output": "NO"
}
] | 1,664,413,076
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
# https://codeforces.com/problemset/problem/404/A
def solve(n, m):
d = m[0][0]
p = m[0][1]
if d == p:
return 'NO'
mid = (n // 2)+1
l = n -1
for i in range(mid):
#print(f"i: {i}")
#print('i: {}'.format(i))
if not (d == m[i][i] and d == m[i][l-i] and d == m[l-i][i] and d == m[l-i][l-i]):
print('here1')
return 'NO'
for j in range(i+1, mid-i):
if not (p == m[i][j] and p == m[i][l-j] and p == m[l-i][j] and p == m[l-i][l-j]
and p == m[j][i] and p == m[l-j][i] and p == m[j][l-i] and p == m[l-j][l-i]):
'''
print(f"""
{i},{j} | {i},{l-j} | {l-i},{j} | {l-i},{l-j} |
{j},{i} | {l-j},{i} | {j},{l-i} | {l-j},{l-i} |
""")
print('here2')
'''
return 'NO'
return 'YES'
n = int(input())
m = []
for _ in range(n):
m.append(input())
print(solve(n, m))
|
Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
# https://codeforces.com/problemset/problem/404/A
def solve(n, m):
d = m[0][0]
p = m[0][1]
if d == p:
return 'NO'
mid = (n // 2)+1
l = n -1
for i in range(mid):
#print(f"i: {i}")
#print('i: {}'.format(i))
if not (d == m[i][i] and d == m[i][l-i] and d == m[l-i][i] and d == m[l-i][l-i]):
print('here1')
return 'NO'
for j in range(i+1, mid-i):
if not (p == m[i][j] and p == m[i][l-j] and p == m[l-i][j] and p == m[l-i][l-j]
and p == m[j][i] and p == m[l-j][i] and p == m[j][l-i] and p == m[l-j][l-i]):
'''
print(f"""
{i},{j} | {i},{l-j} | {l-i},{j} | {l-i},{l-j} |
{j},{i} | {l-j},{i} | {j},{l-i} | {l-j},{l-i} |
""")
print('here2')
'''
return 'NO'
return 'YES'
n = int(input())
m = []
for _ in range(n):
m.append(input())
print(solve(n, m))
```
| 0
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,608,971,211
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 108
| 307,200
|
s=input()
if "h" in s and "e" in s and "l" in s and "o" in s and s.count("l")>=2:
s=s[s.index("h"):]
if "e" in s:
s=s[s.index("e"):]
if "l" in s:
s=s[s.index("l"):]
if "l" in s[1:]:
b=s[1:].index("l")+1
s=s[b:]
if "o" in s:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
else:
print("NO")
else:
print("NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
s=input()
if "h" in s and "e" in s and "l" in s and "o" in s and s.count("l")>=2:
s=s[s.index("h"):]
if "e" in s:
s=s[s.index("e"):]
if "l" in s:
s=s[s.index("l"):]
if "l" in s[1:]:
b=s[1:].index("l")+1
s=s[b:]
if "o" in s:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
else:
print("NO")
else:
print("NO")
```
| 3.945428
|
570
|
B
|
Simple Game
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"games",
"greedy",
"implementation",
"math"
] | null | null |
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive).
|
The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively.
|
Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.
|
[
"3 1\n",
"4 3\n"
] |
[
"2",
"2"
] |
In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less.
| 1,000
|
[
{
"input": "3 1",
"output": "2"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "10 5",
"output": "6"
},
{
"input": "20 13",
"output": "12"
},
{
"input": "51 1",
"output": "2"
},
{
"input": "100 50",
"output": "51"
},
{
"input": "100 51",
"output": "50"
},
{
"input": "100 49",
"output": "50"
},
{
"input": "1000000000 1000000000",
"output": "999999999"
},
{
"input": "1000000000 1",
"output": "2"
},
{
"input": "1000000000 100000000",
"output": "100000001"
},
{
"input": "1000000000 500000000",
"output": "500000001"
},
{
"input": "1000000000 123124",
"output": "123125"
},
{
"input": "12412523 125123",
"output": "125124"
},
{
"input": "54645723 432423",
"output": "432424"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "262833325 131416663",
"output": "131416662"
},
{
"input": "477667530 238833766",
"output": "238833765"
},
{
"input": "692501734 346250868",
"output": "346250867"
},
{
"input": "907335939 453667970",
"output": "453667969"
},
{
"input": "746085224 373042613",
"output": "373042612"
},
{
"input": "189520699 94760350",
"output": "94760349"
},
{
"input": "404354904 202177453",
"output": "202177452"
},
{
"input": "619189108 309594555",
"output": "309594554"
},
{
"input": "81813292 40906647",
"output": "40906646"
},
{
"input": "296647497 148323750",
"output": "148323749"
},
{
"input": "511481701 255740851",
"output": "255740850"
},
{
"input": "726315905 363157953",
"output": "363157952"
},
{
"input": "496110970 201868357",
"output": "201868358"
},
{
"input": "710945175 173165570",
"output": "173165571"
},
{
"input": "925779379 720443954",
"output": "720443953"
},
{
"input": "140613583 93171580",
"output": "93171579"
},
{
"input": "355447788 85890184",
"output": "85890185"
},
{
"input": "570281992 291648263",
"output": "291648262"
},
{
"input": "541904957 459371829",
"output": "459371828"
},
{
"input": "756739161 125332525",
"output": "125332526"
},
{
"input": "971573366 216791157",
"output": "216791158"
},
{
"input": "186407570 160453970",
"output": "160453969"
},
{
"input": "401241775 170032078",
"output": "170032079"
},
{
"input": "616075979 207073797",
"output": "207073798"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "3 2",
"output": "1"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "4 1",
"output": "2"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "5 1",
"output": "2"
},
{
"input": "5 2",
"output": "3"
},
{
"input": "5 3",
"output": "2"
},
{
"input": "5 4",
"output": "3"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "3 2",
"output": "1"
},
{
"input": "7 4",
"output": "3"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "7 3",
"output": "4"
}
] | 1,594,044,924
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 17
| 108
| 6,656,000
|
n, m =map(int , input().split())
t1=n-m
t2=n-t1-1
if t1>=t2 and m+1 <= n:
print(m+1)
elif t1<t2 and m-1 >= 1:
print(m-1)
else:
print(n)
|
Title: Simple Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive).
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively.
Output Specification:
Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.
Demo Input:
['3 1\n', '4 3\n']
Demo Output:
['2', '2']
Note:
In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less.
|
```python
n, m =map(int , input().split())
t1=n-m
t2=n-t1-1
if t1>=t2 and m+1 <= n:
print(m+1)
elif t1<t2 and m-1 >= 1:
print(m-1)
else:
print(n)
```
| 0
|
|
682
|
B
|
Alyona and Mex
|
PROGRAMMING
| 1,200
|
[
"sortings"
] | null | null |
Someone gave Alyona an array containing *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all.
Formally, after applying some operations Alyona will get an array of *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* such that 1<=≤<=*b**i*<=≤<=*a**i* for every 1<=≤<=*i*<=≤<=*n*. Your task is to determine the maximum possible value of mex of this array.
Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of elements in the Alyona's array.
The second line of the input contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
|
Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations.
|
[
"5\n1 3 3 3 6\n",
"2\n2 1\n"
] |
[
"5\n",
"3\n"
] |
In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5.
To reach the answer to the second sample case one must not decrease any of the array elements.
| 1,000
|
[
{
"input": "5\n1 3 3 3 6",
"output": "5"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "1\n1",
"output": "2"
},
{
"input": "1\n1000000000",
"output": "2"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n1 3",
"output": "3"
},
{
"input": "2\n2 2",
"output": "3"
},
{
"input": "2\n2 3",
"output": "3"
},
{
"input": "2\n3 3",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "2"
},
{
"input": "3\n2 1 1",
"output": "3"
},
{
"input": "3\n3 1 1",
"output": "3"
},
{
"input": "3\n1 1 4",
"output": "3"
},
{
"input": "3\n2 1 2",
"output": "3"
},
{
"input": "3\n3 2 1",
"output": "4"
},
{
"input": "3\n2 4 1",
"output": "4"
},
{
"input": "3\n3 3 1",
"output": "4"
},
{
"input": "3\n1 3 4",
"output": "4"
},
{
"input": "3\n4 1 4",
"output": "4"
},
{
"input": "3\n2 2 2",
"output": "3"
},
{
"input": "3\n3 2 2",
"output": "4"
},
{
"input": "3\n4 2 2",
"output": "4"
},
{
"input": "3\n2 3 3",
"output": "4"
},
{
"input": "3\n4 2 3",
"output": "4"
},
{
"input": "3\n4 4 2",
"output": "4"
},
{
"input": "3\n3 3 3",
"output": "4"
},
{
"input": "3\n4 3 3",
"output": "4"
},
{
"input": "3\n4 3 4",
"output": "4"
},
{
"input": "3\n4 4 4",
"output": "4"
},
{
"input": "4\n1 1 1 1",
"output": "2"
},
{
"input": "4\n1 1 2 1",
"output": "3"
},
{
"input": "4\n1 1 3 1",
"output": "3"
},
{
"input": "4\n1 4 1 1",
"output": "3"
},
{
"input": "4\n1 2 1 2",
"output": "3"
},
{
"input": "4\n1 3 2 1",
"output": "4"
},
{
"input": "4\n2 1 4 1",
"output": "4"
},
{
"input": "4\n3 3 1 1",
"output": "4"
},
{
"input": "4\n1 3 4 1",
"output": "4"
},
{
"input": "4\n1 1 4 4",
"output": "4"
},
{
"input": "4\n2 2 2 1",
"output": "3"
},
{
"input": "4\n1 2 2 3",
"output": "4"
},
{
"input": "4\n2 4 1 2",
"output": "4"
},
{
"input": "4\n3 3 1 2",
"output": "4"
},
{
"input": "4\n2 3 4 1",
"output": "5"
},
{
"input": "4\n1 4 2 4",
"output": "5"
},
{
"input": "4\n3 1 3 3",
"output": "4"
},
{
"input": "4\n3 4 3 1",
"output": "5"
},
{
"input": "4\n1 4 4 3",
"output": "5"
},
{
"input": "4\n4 1 4 4",
"output": "5"
},
{
"input": "4\n2 2 2 2",
"output": "3"
},
{
"input": "4\n2 2 3 2",
"output": "4"
},
{
"input": "4\n2 2 2 4",
"output": "4"
},
{
"input": "4\n2 2 3 3",
"output": "4"
},
{
"input": "4\n2 2 3 4",
"output": "5"
},
{
"input": "4\n2 4 4 2",
"output": "5"
},
{
"input": "4\n2 3 3 3",
"output": "4"
},
{
"input": "4\n2 4 3 3",
"output": "5"
},
{
"input": "4\n4 4 2 3",
"output": "5"
},
{
"input": "4\n4 4 4 2",
"output": "5"
},
{
"input": "4\n3 3 3 3",
"output": "4"
},
{
"input": "4\n3 3 3 4",
"output": "5"
},
{
"input": "4\n4 3 3 4",
"output": "5"
},
{
"input": "4\n4 4 3 4",
"output": "5"
},
{
"input": "4\n4 4 4 4",
"output": "5"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 3 3",
"output": "4"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 8 8 8 8 8 8 8 8 8 8",
"output": "9"
},
{
"input": "4\n2 2 2 3",
"output": "4"
},
{
"input": "3\n1 1 2",
"output": "3"
},
{
"input": "15\n1 2 2 20 23 25 28 60 66 71 76 77 79 99 100",
"output": "15"
},
{
"input": "7\n1 2 2 2 5 5 1",
"output": "5"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "5\n1 1 1 1 10000",
"output": "3"
},
{
"input": "5\n1 1 1 1 2",
"output": "3"
},
{
"input": "7\n1 3 3 3 3 3 6",
"output": "5"
},
{
"input": "4\n1 1 1 3",
"output": "3"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 100",
"output": "3"
},
{
"input": "4\n1 1 2 2",
"output": "3"
},
{
"input": "5\n1 1 1 3 4",
"output": "4"
},
{
"input": "8\n1 1 1 1 2 2 3 40",
"output": "5"
},
{
"input": "5\n1 1 1 1 1",
"output": "2"
},
{
"input": "7\n1 2 2 2 2 2 4",
"output": "4"
},
{
"input": "10\n1 1 1 10000000 10000000 10000000 10000000 10000000 10000000 10000000",
"output": "9"
},
{
"input": "10\n1 1 1 1 1 1 1 1 2 3",
"output": "4"
},
{
"input": "4\n8 8 8 8",
"output": "5"
},
{
"input": "5\n5 6 6 6 7",
"output": "6"
}
] | 1,568,397,532
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 109
| 0
|
n=int(input())
a=list(map(int,input().split()))
a.sort()
k=1
for i in range(n):
if a[i]>k:
a[i]=k
k=k+1
print(k+1)
|
Title: Alyona and Mex
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Someone gave Alyona an array containing *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all.
Formally, after applying some operations Alyona will get an array of *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* such that 1<=≤<=*b**i*<=≤<=*a**i* for every 1<=≤<=*i*<=≤<=*n*. Your task is to determine the maximum possible value of mex of this array.
Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of elements in the Alyona's array.
The second line of the input contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Output Specification:
Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations.
Demo Input:
['5\n1 3 3 3 6\n', '2\n2 1\n']
Demo Output:
['5\n', '3\n']
Note:
In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5.
To reach the answer to the second sample case one must not decrease any of the array elements.
|
```python
n=int(input())
a=list(map(int,input().split()))
a.sort()
k=1
for i in range(n):
if a[i]>k:
a[i]=k
k=k+1
print(k+1)
```
| 0
|
|
703
|
A
|
Mishka and Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
|
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
|
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
|
[
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] |
[
"Mishka",
"Friendship is magic!^^",
"Chris"
] |
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
| 500
|
[
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5",
"output": "Chris"
},
{
"input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5",
"output": "Friendship is magic!^^"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3",
"output": "Chris"
},
{
"input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3",
"output": "Mishka"
},
{
"input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3",
"output": "Mishka"
},
{
"input": "5\n3 6\n3 5\n3 5\n1 6\n3 5",
"output": "Chris"
},
{
"input": "4\n4 1\n2 4\n5 3\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2",
"output": "Mishka"
},
{
"input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4",
"output": "Chris"
},
{
"input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1",
"output": "Mishka"
},
{
"input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5",
"output": "Chris"
},
{
"input": "4\n6 2\n2 4\n4 2\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2",
"output": "Mishka"
},
{
"input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4",
"output": "Chris"
},
{
"input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5",
"output": "Friendship is magic!^^"
},
{
"input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1",
"output": "Mishka"
},
{
"input": "5\n1 4\n2 5\n3 4\n2 6\n3 4",
"output": "Chris"
},
{
"input": "4\n6 2\n3 4\n5 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4",
"output": "Chris"
},
{
"input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4",
"output": "Friendship is magic!^^"
},
{
"input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3",
"output": "Mishka"
},
{
"input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6",
"output": "Friendship is magic!^^"
},
{
"input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2",
"output": "Mishka"
},
{
"input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4",
"output": "Chris"
},
{
"input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1",
"output": "Mishka"
},
{
"input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4",
"output": "Chris"
},
{
"input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6",
"output": "Chris"
},
{
"input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1",
"output": "Mishka"
},
{
"input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4",
"output": "Friendship is magic!^^"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4",
"output": "Chris"
},
{
"input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4",
"output": "Mishka"
},
{
"input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3",
"output": "Mishka"
},
{
"input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5",
"output": "Chris"
},
{
"input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3",
"output": "Chris"
},
{
"input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6",
"output": "Mishka"
},
{
"input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2",
"output": "Mishka"
},
{
"input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1",
"output": "Chris"
},
{
"input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2",
"output": "Chris"
},
{
"input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5",
"output": "Mishka"
},
{
"input": "1\n1 1",
"output": "Friendship is magic!^^"
}
] | 1,668,671,659
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 69
| 46
| 0
|
n = int(input())
c = 0
v = 0
for i in range(n):
a = list(map(int,input().split()))
if a[0] > a[1]:
c+=1
elif a[0] < a[1]:
v+=1
if c > v:
print('Mishka')
elif c < v:
print('Chris')
else:
print('Friendship is magic!^^')
|
Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
|
```python
n = int(input())
c = 0
v = 0
for i in range(n):
a = list(map(int,input().split()))
if a[0] > a[1]:
c+=1
elif a[0] < a[1]:
v+=1
if c > v:
print('Mishka')
elif c < v:
print('Chris')
else:
print('Friendship is magic!^^')
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
|
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
|
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
|
[
"4 3 9\n",
"4 3 24\n",
"2 4 4\n"
] |
[
"2 2 L\n",
"4 3 R\n",
"1 2 R\n"
] |
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
| 0
|
[
{
"input": "4 3 9",
"output": "2 2 L"
},
{
"input": "4 3 24",
"output": "4 3 R"
},
{
"input": "2 4 4",
"output": "1 2 R"
},
{
"input": "3 10 24",
"output": "2 2 R"
},
{
"input": "10 3 59",
"output": "10 3 L"
},
{
"input": "10000 10000 160845880",
"output": "8043 2940 R"
},
{
"input": "1 1 1",
"output": "1 1 L"
},
{
"input": "1 1 2",
"output": "1 1 R"
},
{
"input": "1 10000 1",
"output": "1 1 L"
},
{
"input": "1 10000 20000",
"output": "1 10000 R"
},
{
"input": "10000 1 1",
"output": "1 1 L"
},
{
"input": "10000 1 10000",
"output": "5000 1 R"
},
{
"input": "10000 1 20000",
"output": "10000 1 R"
},
{
"input": "3 2 1",
"output": "1 1 L"
},
{
"input": "3 2 2",
"output": "1 1 R"
},
{
"input": "3 2 3",
"output": "1 2 L"
},
{
"input": "3 2 4",
"output": "1 2 R"
},
{
"input": "3 2 5",
"output": "2 1 L"
},
{
"input": "3 2 6",
"output": "2 1 R"
},
{
"input": "3 2 7",
"output": "2 2 L"
},
{
"input": "3 2 8",
"output": "2 2 R"
},
{
"input": "3 2 9",
"output": "3 1 L"
},
{
"input": "3 2 10",
"output": "3 1 R"
},
{
"input": "3 2 11",
"output": "3 2 L"
},
{
"input": "3 2 12",
"output": "3 2 R"
},
{
"input": "300 2000 1068628",
"output": "268 314 R"
},
{
"input": "300 2000 584756",
"output": "147 378 R"
},
{
"input": "300 2000 268181",
"output": "68 91 L"
},
{
"input": "10000 9999 186450844",
"output": "9324 4745 R"
},
{
"input": "10000 9999 197114268",
"output": "9857 6990 R"
},
{
"input": "10000 9999 112390396",
"output": "5621 818 R"
},
{
"input": "10000 10000 1",
"output": "1 1 L"
},
{
"input": "10000 10000 2",
"output": "1 1 R"
},
{
"input": "10000 10000 100000001",
"output": "5001 1 L"
},
{
"input": "10000 10000 199999999",
"output": "10000 10000 L"
},
{
"input": "10000 10000 200000000",
"output": "10000 10000 R"
},
{
"input": "1 2 1",
"output": "1 1 L"
},
{
"input": "1 2 2",
"output": "1 1 R"
},
{
"input": "1 2 3",
"output": "1 2 L"
},
{
"input": "1 2 4",
"output": "1 2 R"
},
{
"input": "2 1 1",
"output": "1 1 L"
},
{
"input": "2 1 2",
"output": "1 1 R"
},
{
"input": "2 1 3",
"output": "2 1 L"
},
{
"input": "2 1 4",
"output": "2 1 R"
},
{
"input": "4 3 7",
"output": "2 1 L"
},
{
"input": "1 1 1",
"output": "1 1 L"
}
] | 1,483,385,059
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 62
| 4,608,000
|
n, m, k = [int(i) for i in input().split()]
c = 'R' if (k % 2 == 0) else 'L'
k -= (k % 2 == 0)
l = 1 + k // (2 * m)
d = 1 + (k % (2 * m) // 2)
print(l, d, c)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input Specification:
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output Specification:
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Demo Input:
['4 3 9\n', '4 3 24\n', '2 4 4\n']
Demo Output:
['2 2 L\n', '4 3 R\n', '1 2 R\n']
Note:
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
|
```python
n, m, k = [int(i) for i in input().split()]
c = 'R' if (k % 2 == 0) else 'L'
k -= (k % 2 == 0)
l = 1 + k // (2 * m)
d = 1 + (k % (2 * m) // 2)
print(l, d, c)
```
| 3
|
|
792
|
A
|
New Bus Route
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
There are *n* cities situated along the main road of Berland. Cities are represented by their coordinates — integer numbers *a*1,<=*a*2,<=...,<=*a**n*. All coordinates are pairwise distinct.
It is possible to get from one city to another only by bus. But all buses and roads are very old, so the Minister of Transport decided to build a new bus route. The Minister doesn't want to spend large amounts of money — he wants to choose two cities in such a way that the distance between them is minimal possible. The distance between two cities is equal to the absolute value of the difference between their coordinates.
It is possible that there are multiple pairs of cities with minimal possible distance, so the Minister wants to know the quantity of such pairs.
Your task is to write a program that will calculate the minimal possible distance between two pairs of cities and the quantity of pairs which have this distance.
|
The first line contains one integer number *n* (2<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). All numbers *a**i* are pairwise distinct.
|
Print two integer numbers — the minimal distance and the quantity of pairs with this distance.
|
[
"4\n6 -3 0 4\n",
"3\n-2 0 2\n"
] |
[
"2 1\n",
"2 2\n"
] |
In the first example the distance between the first city and the fourth city is |4 - 6| = 2, and it is the only pair with this distance.
| 0
|
[
{
"input": "4\n6 -3 0 4",
"output": "2 1"
},
{
"input": "3\n-2 0 2",
"output": "2 2"
},
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "2\n1000000000 -1000000000",
"output": "2000000000 1"
},
{
"input": "5\n-979619606 -979619602 -979619604 -979619605 -979619603",
"output": "1 4"
},
{
"input": "5\n-799147771 -799147773 -799147764 -799147774 -799147770",
"output": "1 2"
},
{
"input": "20\n553280626 553280623 553280627 553280624 553280625 553280618 553280620 553280629 553280637 553280631 553280628 553280636 553280635 553280632 553280634 553280622 553280633 553280621 553280630 553280619",
"output": "1 19"
},
{
"input": "20\n105619866 106083760 106090730 105809555 106115212 105155938 105979518 106075627 106145216 105637844 105925719 105498536 105927000 106155938 106134226 106125969 106130588 105464813 106145509 106114971",
"output": "241 1"
},
{
"input": "10\n570685866 570685854 570685858 570685850 570685856 570685864 570685860 570685852 570685862 570685868",
"output": "2 9"
},
{
"input": "2\n1 1000000000",
"output": "999999999 1"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "1 5"
},
{
"input": "3\n7 10 12",
"output": "2 1"
},
{
"input": "5\n-7 -5 -4 -3 -1",
"output": "1 2"
},
{
"input": "4\n-6 -4 -2 1",
"output": "2 2"
},
{
"input": "4\n3 5 7 8",
"output": "1 1"
},
{
"input": "9\n-9 -8 -7 -6 -5 -4 -3 -2 -1",
"output": "1 8"
},
{
"input": "2\n15 13",
"output": "2 1"
},
{
"input": "2\n14 13",
"output": "1 1"
},
{
"input": "2\n12 13",
"output": "1 1"
},
{
"input": "2\n-1000000000 13265920",
"output": "1013265920 1"
},
{
"input": "2\n8 10",
"output": "2 1"
}
] | 1,545,993,288
| 288
|
PyPy 3
|
OK
|
TESTS
| 45
| 312
| 28,569,600
|
n = int(input())
ar = list(map(int,input().split()))
ar.sort()
l = [abs(ar[i+1] - ar[i]) for i in range(0,n-1)]
k = min(l)
print(k,l.count(k),sep = ' ')
|
Title: New Bus Route
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities situated along the main road of Berland. Cities are represented by their coordinates — integer numbers *a*1,<=*a*2,<=...,<=*a**n*. All coordinates are pairwise distinct.
It is possible to get from one city to another only by bus. But all buses and roads are very old, so the Minister of Transport decided to build a new bus route. The Minister doesn't want to spend large amounts of money — he wants to choose two cities in such a way that the distance between them is minimal possible. The distance between two cities is equal to the absolute value of the difference between their coordinates.
It is possible that there are multiple pairs of cities with minimal possible distance, so the Minister wants to know the quantity of such pairs.
Your task is to write a program that will calculate the minimal possible distance between two pairs of cities and the quantity of pairs which have this distance.
Input Specification:
The first line contains one integer number *n* (2<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). All numbers *a**i* are pairwise distinct.
Output Specification:
Print two integer numbers — the minimal distance and the quantity of pairs with this distance.
Demo Input:
['4\n6 -3 0 4\n', '3\n-2 0 2\n']
Demo Output:
['2 1\n', '2 2\n']
Note:
In the first example the distance between the first city and the fourth city is |4 - 6| = 2, and it is the only pair with this distance.
|
```python
n = int(input())
ar = list(map(int,input().split()))
ar.sort()
l = [abs(ar[i+1] - ar[i]) for i in range(0,n-1)]
k = min(l)
print(k,l.count(k),sep = ' ')
```
| 3
|
|
114
|
A
|
Cifera
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
|
The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1).
|
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*.
|
[
"5\n25\n",
"3\n8\n"
] |
[
"YES\n1\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "5\n25",
"output": "YES\n1"
},
{
"input": "3\n8",
"output": "NO"
},
{
"input": "123\n123",
"output": "YES\n0"
},
{
"input": "99\n970300",
"output": "NO"
},
{
"input": "1000\n6666666",
"output": "NO"
},
{
"input": "59\n3571",
"output": "NO"
},
{
"input": "256\n16777217",
"output": "NO"
},
{
"input": "4638\n21511044",
"output": "YES\n1"
},
{
"input": "24\n191102976",
"output": "YES\n5"
},
{
"input": "52010\n557556453",
"output": "NO"
},
{
"input": "61703211\n1750753082",
"output": "NO"
},
{
"input": "137\n2571353",
"output": "YES\n2"
},
{
"input": "8758\n1746157336",
"output": "NO"
},
{
"input": "2\n64",
"output": "YES\n5"
},
{
"input": "96\n884736",
"output": "YES\n2"
},
{
"input": "1094841453\n1656354409",
"output": "NO"
},
{
"input": "1154413\n1229512809",
"output": "NO"
},
{
"input": "2442144\n505226241",
"output": "NO"
},
{
"input": "11548057\n1033418098",
"output": "NO"
},
{
"input": "581\n196122941",
"output": "YES\n2"
},
{
"input": "146\n1913781536",
"output": "NO"
},
{
"input": "945916\n1403881488",
"output": "NO"
},
{
"input": "68269\n365689065",
"output": "NO"
},
{
"input": "30\n900",
"output": "YES\n1"
},
{
"input": "6\n1296",
"output": "YES\n3"
},
{
"input": "1470193122\n1420950405",
"output": "NO"
},
{
"input": "90750\n1793111557",
"output": "NO"
},
{
"input": "1950054\n1664545956",
"output": "NO"
},
{
"input": "6767692\n123762320",
"output": "NO"
},
{
"input": "1437134\n1622348229",
"output": "NO"
},
{
"input": "444103\n1806462642",
"output": "NO"
},
{
"input": "2592\n6718464",
"output": "YES\n1"
},
{
"input": "50141\n366636234",
"output": "NO"
},
{
"input": "835\n582182875",
"output": "YES\n2"
},
{
"input": "156604\n902492689",
"output": "NO"
},
{
"input": "27385965\n1742270058",
"output": "NO"
},
{
"input": "3\n9",
"output": "YES\n1"
},
{
"input": "35\n1838265625",
"output": "YES\n5"
},
{
"input": "8\n4096",
"output": "YES\n3"
},
{
"input": "85955\n945811082",
"output": "NO"
},
{
"input": "54958832\n956670209",
"output": "NO"
},
{
"input": "1475381\n1348159738",
"output": "NO"
},
{
"input": "7313241\n413670642",
"output": "NO"
},
{
"input": "582470\n2116368165",
"output": "NO"
},
{
"input": "26859739\n595086170",
"output": "NO"
},
{
"input": "249766393\n1582130",
"output": "NO"
},
{
"input": "11734\n137686756",
"output": "YES\n1"
},
{
"input": "925093\n1098566745",
"output": "NO"
},
{
"input": "40\n1600",
"output": "YES\n1"
},
{
"input": "2147483647\n2147483647",
"output": "YES\n0"
},
{
"input": "2147483646\n2147483647",
"output": "NO"
},
{
"input": "2147483647\n2147483646",
"output": "NO"
},
{
"input": "2\n2147483647",
"output": "NO"
},
{
"input": "2\n1073741825",
"output": "NO"
},
{
"input": "2\n1073741824",
"output": "YES\n29"
},
{
"input": "10000\n10",
"output": "NO"
},
{
"input": "10\n10000",
"output": "YES\n3"
},
{
"input": "10\n2000000000",
"output": "NO"
},
{
"input": "10\n1000000000",
"output": "YES\n8"
},
{
"input": "5\n1808548329",
"output": "NO"
},
{
"input": "2\n2147483646",
"output": "NO"
},
{
"input": "25\n125",
"output": "NO"
},
{
"input": "6\n18",
"output": "NO"
},
{
"input": "5\n30",
"output": "NO"
}
] | 1,625,548,479
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 124
| 0
|
a=int(input())
b=int(input())
i=1
while a<=b:
if a==b:
break
i+=1
a*=a
if a==b:
print("YES")
print(i-1)
else:
print("NO")
|
Title: Cifera
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
Input Specification:
The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1).
Output Specification:
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*.
Demo Input:
['5\n25\n', '3\n8\n']
Demo Output:
['YES\n1\n', 'NO\n']
Note:
none
|
```python
a=int(input())
b=int(input())
i=1
while a<=b:
if a==b:
break
i+=1
a*=a
if a==b:
print("YES")
print(i-1)
else:
print("NO")
```
| 0
|
|
496
|
A
|
Minimum Difficulty
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
|
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
|
Print a single number — the minimum difficulty of the track after removing a single hold.
|
[
"3\n1 4 6\n",
"5\n1 2 3 4 5\n",
"5\n1 2 3 7 8\n"
] |
[
"5\n",
"2\n",
"4\n"
] |
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
| 500
|
[
{
"input": "3\n1 4 6",
"output": "5"
},
{
"input": "5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5\n1 2 3 7 8",
"output": "4"
},
{
"input": "3\n1 500 1000",
"output": "999"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "2"
},
{
"input": "10\n1 4 9 16 25 36 49 64 81 100",
"output": "19"
},
{
"input": "10\n300 315 325 338 350 365 379 391 404 416",
"output": "23"
},
{
"input": "15\n87 89 91 92 93 95 97 99 101 103 105 107 109 111 112",
"output": "2"
},
{
"input": "60\n3 5 7 8 15 16 18 21 24 26 40 41 43 47 48 49 50 51 52 54 55 60 62 71 74 84 85 89 91 96 406 407 409 412 417 420 423 424 428 431 432 433 436 441 445 446 447 455 458 467 469 471 472 475 480 485 492 493 497 500",
"output": "310"
},
{
"input": "3\n159 282 405",
"output": "246"
},
{
"input": "81\n6 7 22 23 27 38 40 56 59 71 72 78 80 83 86 92 95 96 101 122 125 127 130 134 154 169 170 171 172 174 177 182 184 187 195 197 210 211 217 223 241 249 252 253 256 261 265 269 274 277 291 292 297 298 299 300 302 318 338 348 351 353 381 386 387 397 409 410 419 420 428 430 453 460 461 473 478 493 494 500 741",
"output": "241"
},
{
"input": "10\n218 300 388 448 535 629 680 740 836 925",
"output": "111"
},
{
"input": "100\n6 16 26 36 46 56 66 76 86 96 106 116 126 136 146 156 166 176 186 196 206 216 226 236 246 256 266 276 286 296 306 316 326 336 346 356 366 376 386 396 406 416 426 436 446 456 466 476 486 496 506 516 526 536 546 556 566 576 586 596 606 616 626 636 646 656 666 676 686 696 706 716 726 736 746 756 766 776 786 796 806 816 826 836 846 856 866 876 886 896 906 916 926 936 946 956 966 976 986 996",
"output": "20"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000",
"output": "901"
},
{
"input": "100\n1 9 15 17 28 29 30 31 32 46 48 49 52 56 62 77 82 85 90 91 94 101 102 109 111 113 116 118 124 125 131 132 136 138 139 143 145 158 161 162 165 167 171 173 175 177 179 183 189 196 801 802 804 806 817 819 827 830 837 840 842 846 850 855 858 862 863 866 869 870 878 881 883 884 896 898 899 901 904 906 908 909 910 911 912 917 923 924 925 935 939 943 945 956 963 964 965 972 976 978",
"output": "605"
},
{
"input": "100\n2 43 47 49 50 57 59 67 74 98 901 903 904 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 938 939 940 942 943 944 945 946 947 948 949 950 952 953 954 956 957 958 959 960 961 962 963 965 966 967 968 969 970 971 972 973 974 975 976 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 998 999",
"output": "803"
},
{
"input": "72\n178 186 196 209 217 226 236 248 260 273 281 291 300 309 322 331 343 357 366 377 389 399 409 419 429 442 450 459 469 477 491 501 512 524 534 548 557 568 582 593 602 616 630 643 652 660 670 679 693 707 715 728 737 750 759 768 776 789 797 807 815 827 837 849 863 873 881 890 901 910 920 932",
"output": "17"
},
{
"input": "38\n1 28 55 82 109 136 163 190 217 244 271 298 325 352 379 406 433 460 487 514 541 568 595 622 649 676 703 730 757 784 811 838 865 892 919 946 973 1000",
"output": "54"
},
{
"input": "28\n1 38 75 112 149 186 223 260 297 334 371 408 445 482 519 556 593 630 667 704 741 778 815 852 889 926 963 1000",
"output": "74"
}
] | 1,586,688,155
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 109
| 307,200
|
import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_int_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read integers
get_int = lambda: int(sys.stdin.readline().strip())
#--------------------------------WhiteHat010--------------------------------------#
n = get_int()
lst = get_int_list()
m = 100000000
index = -1
for i in range(n-2):
if lst[i+2]-lst[i] < m:
m = lst[i+2]-lst[i]
index = i+1
lst.pop(index)
m = 0
for i in range(n-2):
m = max(m, lst[i+1]-lst[i])
# print(lst)
print(m)
|
Title: Minimum Difficulty
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
Input Specification:
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
Output Specification:
Print a single number — the minimum difficulty of the track after removing a single hold.
Demo Input:
['3\n1 4 6\n', '5\n1 2 3 4 5\n', '5\n1 2 3 7 8\n']
Demo Output:
['5\n', '2\n', '4\n']
Note:
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.
|
```python
import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_int_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read integers
get_int = lambda: int(sys.stdin.readline().strip())
#--------------------------------WhiteHat010--------------------------------------#
n = get_int()
lst = get_int_list()
m = 100000000
index = -1
for i in range(n-2):
if lst[i+2]-lst[i] < m:
m = lst[i+2]-lst[i]
index = i+1
lst.pop(index)
m = 0
for i in range(n-2):
m = max(m, lst[i+1]-lst[i])
# print(lst)
print(m)
```
| 3
|
|
455
|
B
|
A Lot of Games
|
PROGRAMMING
| 1,900
|
[
"dfs and similar",
"dp",
"games",
"implementation",
"strings",
"trees"
] | null | null |
Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.
Given a group of *n* non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.
Andrew and Alex decided to play this game *k* times. The player who is the loser of the *i*-th game makes the first move in the (*i*<=+<=1)-th game. Guys decided that the winner of all games is the player who wins the last (*k*-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him.
|
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=105; 1<=≤<=*k*<=≤<=109).
Each of the next *n* lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters.
|
If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).
|
[
"2 3\na\nb\n",
"3 1\na\nb\nc\n",
"1 2\nab\n"
] |
[
"First\n",
"First\n",
"Second\n"
] |
none
| 1,000
|
[
{
"input": "2 3\na\nb",
"output": "First"
},
{
"input": "3 1\na\nb\nc",
"output": "First"
},
{
"input": "1 2\nab",
"output": "Second"
},
{
"input": "5 6\nabas\ndsfdf\nabacaba\ndartsidius\nkolobok",
"output": "Second"
},
{
"input": "4 2\naaaa\nbbbb\nccccc\ndumbavumba",
"output": "First"
},
{
"input": "3 8\nso\nbad\ntest",
"output": "First"
},
{
"input": "5 2\nwelcome\nto\nthe\nmatrix\nneo",
"output": "First"
},
{
"input": "6 4\ndog\ncat\ncow\nhot\nice\nlol",
"output": "Second"
},
{
"input": "4 8\nla\na\nz\nka",
"output": "First"
},
{
"input": "3 2\nop\nhop\ncop",
"output": "First"
},
{
"input": "3 3\nabasdfabab\nabaaasdfdsf\nasdfaba",
"output": "Second"
},
{
"input": "2 2\naba\naa",
"output": "Second"
},
{
"input": "4 1\naa\naba\nba\nbba",
"output": "Second"
},
{
"input": "1 3\nab",
"output": "Second"
},
{
"input": "3 3\naa\nabb\ncc",
"output": "Second"
}
] | 1,691,781,257
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1691781256.9321766")# 1691781256.9321926
|
Title: A Lot of Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.
Given a group of *n* non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.
Andrew and Alex decided to play this game *k* times. The player who is the loser of the *i*-th game makes the first move in the (*i*<=+<=1)-th game. Guys decided that the winner of all games is the player who wins the last (*k*-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him.
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=105; 1<=≤<=*k*<=≤<=109).
Each of the next *n* lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters.
Output Specification:
If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).
Demo Input:
['2 3\na\nb\n', '3 1\na\nb\nc\n', '1 2\nab\n']
Demo Output:
['First\n', 'First\n', 'Second\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1691781256.9321766")# 1691781256.9321926
```
| 0
|
|
750
|
A
|
New Year and Hurry
|
PROGRAMMING
| 800
|
[
"binary search",
"brute force",
"implementation",
"math"
] | null | null |
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
|
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
|
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
|
[
"3 222\n",
"4 190\n",
"7 1\n"
] |
[
"2\n",
"4\n",
"7\n"
] |
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
| 500
|
[
{
"input": "3 222",
"output": "2"
},
{
"input": "4 190",
"output": "4"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "10 135",
"output": "6"
},
{
"input": "10 136",
"output": "5"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "9 240",
"output": "0"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "9 235",
"output": "1"
},
{
"input": "9 236",
"output": "0"
},
{
"input": "5 225",
"output": "2"
},
{
"input": "5 226",
"output": "1"
},
{
"input": "4 210",
"output": "3"
},
{
"input": "4 211",
"output": "2"
},
{
"input": "4 191",
"output": "3"
},
{
"input": "10 165",
"output": "5"
},
{
"input": "10 166",
"output": "4"
},
{
"input": "8 100",
"output": "7"
},
{
"input": "8 101",
"output": "6"
},
{
"input": "8 60",
"output": "8"
},
{
"input": "8 61",
"output": "7"
},
{
"input": "10 15",
"output": "9"
},
{
"input": "10 16",
"output": "8"
},
{
"input": "4 100",
"output": "4"
},
{
"input": "4 101",
"output": "4"
},
{
"input": "7 167",
"output": "4"
},
{
"input": "10 164",
"output": "5"
},
{
"input": "9 170",
"output": "4"
},
{
"input": "8 160",
"output": "5"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "8 123",
"output": "6"
},
{
"input": "2 99",
"output": "2"
},
{
"input": "10 88",
"output": "7"
},
{
"input": "1 235",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 240",
"output": "0"
},
{
"input": "1 55",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "3 240",
"output": "0"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "2 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "2 236",
"output": "0"
},
{
"input": "10 2",
"output": "9"
},
{
"input": "3 239",
"output": "0"
},
{
"input": "1 237",
"output": "0"
},
{
"input": "7 8",
"output": "7"
},
{
"input": "10 235",
"output": "1"
}
] | 1,677,909,269
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 52
| 46
| 0
|
n, k = map(int, input().split())
time = 240 - k
c = 0
for i in range (1, n+1):
if 5*i <= time:
c = c+1
time = time - (5*i)
print(c)
|
Title: New Year and Hurry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
Input Specification:
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
Output Specification:
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
Demo Input:
['3 222\n', '4 190\n', '7 1\n']
Demo Output:
['2\n', '4\n', '7\n']
Note:
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
|
```python
n, k = map(int, input().split())
time = 240 - k
c = 0
for i in range (1, n+1):
if 5*i <= time:
c = c+1
time = time - (5*i)
print(c)
```
| 3
|
|
381
|
A
|
Sereja and Dima
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
|
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
|
[
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] |
[
"12 5\n",
"16 12\n"
] |
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
| 500
|
[
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,678,597,540
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
n = int(input())
arr = list(map(int, input().split()))
i = 0
j = n-1
key = 0
ans = {0: 0, 1: 0}
while (i < j):
if (arr[i] > arr[j]):
ans[key] += arr[i]
i += 1
else:
ans[key] += arr[j]
j -= 1
key = 1-key
if (i == j):
ans[key] += arr[i]
print(ans[0], ans[1])
|
Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
|
```python
n = int(input())
arr = list(map(int, input().split()))
i = 0
j = n-1
key = 0
ans = {0: 0, 1: 0}
while (i < j):
if (arr[i] > arr[j]):
ans[key] += arr[i]
i += 1
else:
ans[key] += arr[j]
j -= 1
key = 1-key
if (i == j):
ans[key] += arr[i]
print(ans[0], ans[1])
```
| 3
|
|
114
|
A
|
Cifera
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
|
The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1).
|
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*.
|
[
"5\n25\n",
"3\n8\n"
] |
[
"YES\n1\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "5\n25",
"output": "YES\n1"
},
{
"input": "3\n8",
"output": "NO"
},
{
"input": "123\n123",
"output": "YES\n0"
},
{
"input": "99\n970300",
"output": "NO"
},
{
"input": "1000\n6666666",
"output": "NO"
},
{
"input": "59\n3571",
"output": "NO"
},
{
"input": "256\n16777217",
"output": "NO"
},
{
"input": "4638\n21511044",
"output": "YES\n1"
},
{
"input": "24\n191102976",
"output": "YES\n5"
},
{
"input": "52010\n557556453",
"output": "NO"
},
{
"input": "61703211\n1750753082",
"output": "NO"
},
{
"input": "137\n2571353",
"output": "YES\n2"
},
{
"input": "8758\n1746157336",
"output": "NO"
},
{
"input": "2\n64",
"output": "YES\n5"
},
{
"input": "96\n884736",
"output": "YES\n2"
},
{
"input": "1094841453\n1656354409",
"output": "NO"
},
{
"input": "1154413\n1229512809",
"output": "NO"
},
{
"input": "2442144\n505226241",
"output": "NO"
},
{
"input": "11548057\n1033418098",
"output": "NO"
},
{
"input": "581\n196122941",
"output": "YES\n2"
},
{
"input": "146\n1913781536",
"output": "NO"
},
{
"input": "945916\n1403881488",
"output": "NO"
},
{
"input": "68269\n365689065",
"output": "NO"
},
{
"input": "30\n900",
"output": "YES\n1"
},
{
"input": "6\n1296",
"output": "YES\n3"
},
{
"input": "1470193122\n1420950405",
"output": "NO"
},
{
"input": "90750\n1793111557",
"output": "NO"
},
{
"input": "1950054\n1664545956",
"output": "NO"
},
{
"input": "6767692\n123762320",
"output": "NO"
},
{
"input": "1437134\n1622348229",
"output": "NO"
},
{
"input": "444103\n1806462642",
"output": "NO"
},
{
"input": "2592\n6718464",
"output": "YES\n1"
},
{
"input": "50141\n366636234",
"output": "NO"
},
{
"input": "835\n582182875",
"output": "YES\n2"
},
{
"input": "156604\n902492689",
"output": "NO"
},
{
"input": "27385965\n1742270058",
"output": "NO"
},
{
"input": "3\n9",
"output": "YES\n1"
},
{
"input": "35\n1838265625",
"output": "YES\n5"
},
{
"input": "8\n4096",
"output": "YES\n3"
},
{
"input": "85955\n945811082",
"output": "NO"
},
{
"input": "54958832\n956670209",
"output": "NO"
},
{
"input": "1475381\n1348159738",
"output": "NO"
},
{
"input": "7313241\n413670642",
"output": "NO"
},
{
"input": "582470\n2116368165",
"output": "NO"
},
{
"input": "26859739\n595086170",
"output": "NO"
},
{
"input": "249766393\n1582130",
"output": "NO"
},
{
"input": "11734\n137686756",
"output": "YES\n1"
},
{
"input": "925093\n1098566745",
"output": "NO"
},
{
"input": "40\n1600",
"output": "YES\n1"
},
{
"input": "2147483647\n2147483647",
"output": "YES\n0"
},
{
"input": "2147483646\n2147483647",
"output": "NO"
},
{
"input": "2147483647\n2147483646",
"output": "NO"
},
{
"input": "2\n2147483647",
"output": "NO"
},
{
"input": "2\n1073741825",
"output": "NO"
},
{
"input": "2\n1073741824",
"output": "YES\n29"
},
{
"input": "10000\n10",
"output": "NO"
},
{
"input": "10\n10000",
"output": "YES\n3"
},
{
"input": "10\n2000000000",
"output": "NO"
},
{
"input": "10\n1000000000",
"output": "YES\n8"
},
{
"input": "5\n1808548329",
"output": "NO"
},
{
"input": "2\n2147483646",
"output": "NO"
},
{
"input": "25\n125",
"output": "NO"
},
{
"input": "6\n18",
"output": "NO"
},
{
"input": "5\n30",
"output": "NO"
}
] | 1,378,082,689
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
import sys
first_num = int(sys.stdin.readline())
second_num = int(sys.stdin.readline())
count = 0
first = first_num
while first < second_num:
first *= first_num
count += 1
if first == second_num:
print("Yes")
print(count)
else:
print("No")
|
Title: Cifera
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
Input Specification:
The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1).
Output Specification:
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*.
Demo Input:
['5\n25\n', '3\n8\n']
Demo Output:
['YES\n1\n', 'NO\n']
Note:
none
|
```python
import sys
first_num = int(sys.stdin.readline())
second_num = int(sys.stdin.readline())
count = 0
first = first_num
while first < second_num:
first *= first_num
count += 1
if first == second_num:
print("Yes")
print(count)
else:
print("No")
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
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"output": "11111110000000100101000100110111001100011001"
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},
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"output": "0101010111101001011011110110011101010101010100011"
},
{
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"output": "11001011010010111000010110011101100100001110111111"
},
{
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},
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},
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"output": "01011001110111010111001100010011010100010000111011000"
},
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},
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"output": "1100110010000101101010111111101001001001110101110010110"
},
{
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"output": "01000111100111001011110010100011111111110010101100001101"
},
{
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"output": "110001010001000011000101110101000100001011111001011001001"
},
{
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"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
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"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
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"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
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"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
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"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
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"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
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"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
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"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
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"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
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"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
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"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
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"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
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"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
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"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
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"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
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"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
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"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
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"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
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"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
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"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
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"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
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"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
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"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
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"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
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"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
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"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
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"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
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"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
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"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,692,524,173
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 0
|
num1 = input()
num2 = input()
answer = ''.join('1' if num1[i] != num2[i] else '0' for i in range(len(num1)))
print(answer)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
num1 = input()
num2 = input()
answer = ''.join('1' if num1[i] != num2[i] else '0' for i in range(len(num1)))
print(answer)
```
| 3.9885
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,685,887,022
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
def replace_word(word):
length = len(word)
if length <= 10:
return word
else:
return word[0] + str(length - 2) + word[length - 1]
S = int(input())
for i in range(S):
word = input()
result = replace_word(word)
print(result)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
def replace_word(word):
length = len(word)
if length <= 10:
return word
else:
return word[0] + str(length - 2) + word[length - 1]
S = int(input())
for i in range(S):
word = input()
result = replace_word(word)
print(result)
```
| 3.977
|
897
|
A
|
Scarborough Fair
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
|
Output string *s* after performing *m* operations described above.
|
[
"3 1\nioi\n1 1 i n\n",
"5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n"
] |
[
"noi",
"gaaak"
] |
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak.
| 500
|
[
{
"input": "3 1\nioi\n1 1 i n",
"output": "noi"
},
{
"input": "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g",
"output": "gaaak"
},
{
"input": "9 51\nbhfbdcgff\n2 3 b b\n2 8 e f\n3 8 g f\n5 7 d a\n1 5 e b\n3 4 g b\n6 7 c d\n3 6 e g\n3 6 e h\n5 6 a e\n7 9 a c\n4 9 a h\n3 7 c b\n6 9 b g\n1 7 h b\n4 5 a e\n3 9 f a\n1 2 c h\n4 8 a c\n3 5 e d\n3 4 g f\n2 3 d h\n2 3 d e\n1 7 d g\n2 6 e g\n2 3 d g\n5 5 h h\n2 8 g d\n8 9 a f\n5 9 c e\n1 7 f d\n1 6 e e\n5 7 c a\n8 9 b b\n2 6 e b\n6 6 g h\n1 2 b b\n1 5 a f\n5 8 f h\n1 5 e g\n3 9 f h\n6 8 g a\n4 6 h g\n1 5 f a\n5 6 a c\n4 8 e d\n1 4 d g\n7 8 b f\n5 6 h b\n3 9 c e\n1 9 b a",
"output": "aahaddddh"
},
{
"input": "28 45\ndcbbaddjhbeefjadjchgkhgggfha\n10 25 c a\n13 19 a f\n12 28 e d\n12 27 e a\n9 20 b e\n7 17 g d\n22 26 j j\n8 16 c g\n14 16 a d\n3 10 f c\n10 26 d b\n8 17 i e\n10 19 d i\n6 21 c j\n7 22 b k\n17 19 a i\n4 18 j k\n8 25 a g\n10 27 j e\n9 18 g d\n16 23 h a\n17 26 k e\n8 16 h f\n1 15 d f\n22 28 k k\n11 20 c k\n6 11 b h\n17 17 e i\n15 22 g h\n8 18 c f\n4 16 e a\n8 25 b c\n6 24 d g\n5 9 f j\n12 19 i h\n4 25 e f\n15 25 c j\n15 27 e e\n11 20 b f\n19 27 e k\n2 21 d a\n9 27 k e\n14 24 b a\n3 6 i g\n2 26 k f",
"output": "fcbbajjfjaaefefehfahfagggfha"
},
{
"input": "87 5\nnfinedeojadjmgafnaogekfjkjfncnliagfchjfcmellgigjjcaaoeakdolchjcecljdeblmheimkibkgdkcdml\n47 56 a k\n51 81 o d\n5 11 j h\n48 62 j d\n16 30 k m",
"output": "nfinedeohadjmgafnaogemfjmjfncnliagfchjfcmellgigddckkdekkddlchdcecljdeblmheimkibkgdkcdml"
},
{
"input": "5 16\nacfbb\n1 2 e f\n2 5 a f\n2 3 b e\n4 4 f a\n2 3 f a\n1 2 b e\n4 5 c d\n2 4 e c\n1 4 e a\n1 3 d c\n3 5 e b\n3 5 e b\n2 2 e d\n1 3 e c\n3 3 a e\n1 5 a a",
"output": "acebb"
},
{
"input": "94 13\nbcaaaaaaccacddcdaacbdaabbcbaddbccbccbbbddbadddcccbddadddaadbdababadaacdcdbcdadabdcdcbcbcbcbbcd\n52 77 d d\n21 92 d b\n45 48 c b\n20 25 d a\n57 88 d b\n3 91 b d\n64 73 a a\n5 83 b d\n2 69 c c\n28 89 a b\n49 67 c b\n41 62 a c\n49 87 b c",
"output": "bcaaaaaaccacddcdaacddaaddcdbdddccdccddddddbdddddcdddcdddccdddcdcdcdcccdcddcdcdcddcdcdcdcdcdbcd"
},
{
"input": "67 39\nacbcbccccbabaabcabcaaaaaaccbcbbcbaaaacbbcccbcbabbcacccbbabbabbabaac\n4 36 a b\n25 38 a a\n3 44 b c\n35 57 b a\n4 8 a c\n20 67 c a\n30 66 b b\n27 40 a a\n2 56 a b\n10 47 c a\n22 65 c b\n29 42 a b\n1 46 c b\n57 64 b c\n20 29 b a\n14 51 c a\n12 55 b b\n20 20 a c\n2 57 c a\n22 60 c b\n16 51 c c\n31 64 a c\n17 30 c a\n23 36 c c\n28 67 a c\n37 40 a c\n37 50 b c\n29 48 c b\n2 34 b c\n21 53 b a\n26 63 a c\n23 28 c a\n51 56 c b\n32 61 b b\n64 67 b b\n21 67 b c\n8 53 c c\n40 62 b b\n32 38 c c",
"output": "accccccccaaaaaaaaaaaaaaaaaaaccccccccccccccccccccccccccccccccccccccc"
},
{
"input": "53 33\nhhcbhfafeececbhadfbdbehdfacfchbhdbfebdfeghebfcgdhehfh\n27 41 h g\n18 35 c b\n15 46 h f\n48 53 e g\n30 41 b c\n12 30 b f\n10 37 e f\n18 43 a h\n10 52 d a\n22 48 c e\n40 53 f d\n7 12 b h\n12 51 f a\n3 53 g a\n19 41 d h\n22 29 b h\n2 30 a b\n26 28 e h\n25 35 f a\n19 31 h h\n44 44 d e\n19 22 e c\n29 44 d h\n25 33 d h\n3 53 g c\n18 44 h b\n19 28 f e\n3 22 g h\n8 17 c a\n37 51 d d\n3 28 e h\n27 50 h h\n27 46 f b",
"output": "hhcbhfbfhfababbbbbbbbbbbbbbbbbeaaeaaeaaeabebdeaahahdh"
},
{
"input": "83 10\nfhbecdgadecabbbecedcgfdcefcbgechbedagecgdgfgdaahchdgchbeaedgafdefecdchceececfcdhcdh\n9 77 e e\n26 34 b g\n34 70 b a\n40 64 e g\n33 78 h f\n14 26 a a\n17 70 d g\n56 65 a c\n8 41 d c\n11 82 c b",
"output": "fhbecdgacebabbbebegbgfgbefbggebhgegagebgggfggaafbfggbfagbgggbfggfebgbfbeebebfbdhbdh"
},
{
"input": "1 4\ne\n1 1 c e\n1 1 e a\n1 1 e c\n1 1 d a",
"output": "a"
},
{
"input": "71 21\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n61 61 a a\n32 56 a a\n10 67 a a\n7 32 a a\n26 66 a a\n41 55 a a\n49 55 a a\n4 61 a a\n53 59 a a\n37 58 a a\n7 63 a a\n39 40 a a\n51 64 a a\n27 37 a a\n22 71 a a\n4 45 a a\n7 8 a a\n43 46 a a\n19 28 a a\n51 54 a a\n14 67 a a",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "30 4\neaaddabedcbbcccddbabdecadcecce\n2 17 c a\n16 29 e e\n16 21 c b\n7 11 b c",
"output": "eaaddacedacbaaaddbabdecadcecce"
},
{
"input": "48 30\naaaabaabbaababbbaabaabaababbabbbaabbbaabaaaaaaba\n3 45 a b\n1 14 a a\n15 32 a b\n37 47 a b\n9 35 a b\n36 39 b b\n6 26 a b\n36 44 a a\n28 44 b a\n29 31 b a\n20 39 a a\n45 45 a b\n21 32 b b\n7 43 a b\n14 48 a b\n14 33 a b\n39 44 a a\n9 36 b b\n4 23 b b\n9 42 b b\n41 41 b a\n30 47 a b\n8 42 b a\n14 38 b b\n3 15 a a\n35 47 b b\n14 34 a b\n38 43 a b\n1 35 b a\n16 28 b a",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbb"
},
{
"input": "89 29\nbabaabaaabaaaababbbbbbbabbbaaaaababbaababababbababaaabbababaaabbbbaaabaaaaaabaaabaabbabab\n39 70 b b\n3 56 b b\n5 22 b a\n4 39 a b\n41 87 b b\n34 41 a a\n10 86 a b\n29 75 a b\n2 68 a a\n27 28 b b\n42 51 b a\n18 61 a a\n6 67 b a\n47 63 a a\n8 68 a b\n4 74 b a\n19 65 a b\n8 55 a b\n5 30 a a\n3 65 a b\n16 57 a b\n34 56 b a\n1 70 a b\n59 68 b b\n29 57 b a\n47 49 b b\n49 73 a a\n32 61 b b\n29 42 a a",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaabbbbbbbbbbbbbab"
},
{
"input": "59 14\nfbebcfabdefbaaedcefdeecababcabebadfbccaaedaebfdaefdbbcbebbe\n5 32 e f\n8 46 e e\n31 43 e f\n3 10 e a\n53 54 f d\n55 59 d a\n39 58 e b\n54 56 f a\n9 40 b e\n28 37 d a\n7 35 e b\n7 56 c f\n23 26 e a\n15 44 e d",
"output": "fbabcfabdffbaafdfffdfffababfabfbaafdffaafdabbfdabfdbbfbbbbe"
},
{
"input": "7 17\nbbaabab\n3 5 a b\n5 7 a a\n5 5 a a\n4 4 b a\n7 7 a a\n5 6 b b\n1 3 b a\n6 7 a b\n4 6 a b\n6 6 a a\n2 4 b a\n1 7 b a\n4 6 b b\n2 5 b b\n2 5 a b\n1 4 a a\n4 4 b a",
"output": "abbabaa"
},
{
"input": "100 1\ndebaaagbfdgehagadabfgheegggfghghgeeeabgceffeffggcbcegfgebbdhebhfagcgadcbdbabddbcadgbgdebdfehceehcaef\n13 99 f c",
"output": "debaaagbfdgehagadabcgheegggcghghgeeeabgcecceccggcbcegcgebbdhebhcagcgadcbdbabddbcadgbgdebdcehceehcaef"
},
{
"input": "1 1\na\n1 1 a b",
"output": "b"
},
{
"input": "100 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n1 100 a b",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2 2\naa\n1 2 a b\n1 2 b c",
"output": "cc"
},
{
"input": "3 3\naaa\n1 3 a b\n1 3 b c\n1 3 c d",
"output": "ddd"
},
{
"input": "2 2\naa\n2 2 a b\n1 1 a b",
"output": "bb"
}
] | 1,550,042,675
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 109
| 0
|
n, m = [int(x) for x in input().split()]
s = '$'+input()
for i in range(m):
arr = input().split()
for j in range(int(arr[0]), int(arr[1])+1):
if s[j] == arr[2]:
s = s[:j] + arr[3] + s[j+1:]
print(s[1:])
|
Title: Scarborough Fair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
Output Specification:
Output string *s* after performing *m* operations described above.
Demo Input:
['3 1\nioi\n1 1 i n\n', '5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n']
Demo Output:
['noi', 'gaaak']
Note:
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak.
|
```python
n, m = [int(x) for x in input().split()]
s = '$'+input()
for i in range(m):
arr = input().split()
for j in range(int(arr[0]), int(arr[1])+1):
if s[j] == arr[2]:
s = s[:j] + arr[3] + s[j+1:]
print(s[1:])
```
| 3
|
|
992
|
A
|
Nastya and an Array
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
|
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
|
[
"5\n1 1 1 1 1\n",
"3\n2 0 -1\n",
"4\n5 -6 -5 1\n"
] |
[
"1\n",
"2\n",
"4\n"
] |
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
| 500
|
[
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "3\n2 0 -1",
"output": "2"
},
{
"input": "4\n5 -6 -5 1",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n21794 -79194",
"output": "2"
},
{
"input": "3\n-63526 95085 -5239",
"output": "3"
},
{
"input": "3\n0 53372 -20572",
"output": "2"
},
{
"input": "13\n-2075 -32242 27034 -37618 -96962 82203 64846 48249 -71761 28908 -21222 -61370 46899",
"output": "13"
},
{
"input": "5\n806 0 1308 1954 683",
"output": "4"
},
{
"input": "8\n-26 0 -249 -289 -126 -206 288 -11",
"output": "7"
},
{
"input": "10\n2 2 2 1 2 -1 0 2 -1 1",
"output": "3"
},
{
"input": "1\n8",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "5\n2 0 -1 0 0",
"output": "2"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n-1 0",
"output": "1"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "8\n10 9 -1 0 0 3 2 3",
"output": "5"
},
{
"input": "5\n5 0 1 2 3",
"output": "4"
},
{
"input": "3\n1 1 0",
"output": "1"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "5\n1 2 0 0 0",
"output": "2"
},
{
"input": "5\n1 0 0 0 0",
"output": "1"
},
{
"input": "5\n4 5 6 0 0",
"output": "3"
},
{
"input": "4\n-1 0 0 1",
"output": "2"
},
{
"input": "5\n3 0 0 4 5",
"output": "3"
},
{
"input": "3\n0 0 2",
"output": "1"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "4\n0 0 0 4",
"output": "1"
},
{
"input": "5\n-1 0 0 0 0",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 2 3 0 0",
"output": "3"
},
{
"input": "4\n0 1 2 3",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "2"
},
{
"input": "4\n2 0 0 -1",
"output": "2"
}
] | 1,574,553,586
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 140
| 0
|
n=int(input())
L=list(int(x) for x in input().split())
if L==[0 for i in range(n)]:
print(0)
else:
L.sort()
m=n
c=-1
for i in range(m):
if L[0]!=0:
k=L[0]
z=L.count(0)
for p in range(z):
L.remove(0)
m=n-z
L=[x-k for x in L]
c+=1
if L==[]:
break
print(c)
|
Title: Nastya and an Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
Output Specification:
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
Demo Input:
['5\n1 1 1 1 1\n', '3\n2 0 -1\n', '4\n5 -6 -5 1\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
|
```python
n=int(input())
L=list(int(x) for x in input().split())
if L==[0 for i in range(n)]:
print(0)
else:
L.sort()
m=n
c=-1
for i in range(m):
if L[0]!=0:
k=L[0]
z=L.count(0)
for p in range(z):
L.remove(0)
m=n-z
L=[x-k for x in L]
c+=1
if L==[]:
break
print(c)
```
| 0
|
|
929
|
A
|
Прокат велосипедов
|
PROGRAMMING
| 1,400
|
[
"*special",
"greedy",
"implementation"
] | null | null |
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
|
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
|
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
|
[
"4 4\n3 6 8 10\n",
"2 9\n10 20\n",
"12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n"
] |
[
"2\n",
"-1\n",
"6\n"
] |
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
| 500
|
[
{
"input": "4 4\n3 6 8 10",
"output": "2"
},
{
"input": "2 9\n10 20",
"output": "-1"
},
{
"input": "12 3\n4 6 7 9 10 11 13 15 17 18 20 21",
"output": "6"
},
{
"input": "2 1\n11164 11165",
"output": "1"
},
{
"input": "3 7\n45823 45825 45829",
"output": "1"
},
{
"input": "2 100000\n0 100000",
"output": "1"
},
{
"input": "50 15001\n1269 1580 5431 5916 6642 14145 15760 19922 20589 22062 24138 33454 33505 35916 37012 42577 43140 49457 54720 55188 56205 56639 56870 57997 58379 59088 59297 61805 61861 63005 64013 68848 71182 71497 72715 73008 75616 76042 76614 80690 83089 85033 86408 92392 92763 93833 95085 95815 97152 99379",
"output": "8"
},
{
"input": "5 7\n6219 6222 6229 6231 6236",
"output": "3"
},
{
"input": "10 448\n78449 78573 78599 78742 78748 78759 78853 79091 79298 79324",
"output": "3"
},
{
"input": "20 19191\n11272 12386 14108 24663 24932 26547 29656 44677 45436 45654 48562 62367 71049 71238 78182 84042 88736 92026 96836 99343",
"output": "6"
},
{
"input": "2 100000\n1 100000",
"output": "1"
},
{
"input": "2 99999\n0 100000",
"output": "-1"
},
{
"input": "2 2\n1 3",
"output": "1"
},
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "2 2\n1 4",
"output": "-1"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
}
] | 1,520,007,284
| 2,384
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 7
| 77
| 5,632,000
|
n, k = map(int, input().split())
arr = list(map(int, input().split()))
xs = arr[0]
ans = 1
for i in range(len(arr)):
if arr[i] - xs > k:
xs = arr[i-1]
ans += 1
if arr[-1] - xs <= k:
print(ans)
else:
print(-1)
|
Title: Прокат велосипедов
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Как известно, в теплую погоду многие жители крупных городов пользуются сервисами городского велопроката. Вот и Аркадий сегодня будет добираться от школы до дома, используя городские велосипеды.
Школа и дом находятся на одной прямой улице, кроме того, на той же улице есть *n* точек, где можно взять велосипед в прокат или сдать его. Первый велопрокат находится в точке *x*1 километров вдоль улицы, второй — в точке *x*2 и так далее, *n*-й велопрокат находится в точке *x**n*. Школа Аркадия находится в точке *x*1 (то есть там же, где и первый велопрокат), а дом — в точке *x**n* (то есть там же, где и *n*-й велопрокат). Известно, что *x**i*<=<<=*x**i*<=+<=1 для всех 1<=≤<=*i*<=<<=*n*.
Согласно правилам пользования велопроката, Аркадий может брать велосипед в прокат только на ограниченное время, после этого он должен обязательно вернуть его в одной из точек велопроката, однако, он тут же может взять новый велосипед, и отсчет времени пойдет заново. Аркадий может брать не более одного велосипеда в прокат одновременно. Если Аркадий решает взять велосипед в какой-то точке проката, то он сдаёт тот велосипед, на котором он до него доехал, берёт ровно один новый велосипед и продолжает на нём своё движение.
За отведенное время, независимо от выбранного велосипеда, Аркадий успевает проехать не больше *k* километров вдоль улицы.
Определите, сможет ли Аркадий доехать на велосипедах от школы до дома, и если да, то какое минимальное число раз ему необходимо будет взять велосипед в прокат, включая первый велосипед? Учтите, что Аркадий не намерен сегодня ходить пешком.
Input Specification:
В первой строке следуют два целых числа *n* и *k* (2<=≤<=*n*<=≤<=1<=000, 1<=≤<=*k*<=≤<=100<=000) — количество велопрокатов и максимальное расстояние, которое Аркадий может проехать на одном велосипеде.
В следующей строке следует последовательность целых чисел *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=≤<=100<=000) — координаты точек, в которых находятся велопрокаты. Гарантируется, что координаты велопрокатов заданы в порядке возрастания.
Output Specification:
Если Аркадий не сможет добраться от школы до дома только на велосипедах, выведите -1. В противном случае, выведите минимальное количество велосипедов, которые Аркадию нужно взять в точках проката.
Demo Input:
['4 4\n3 6 8 10\n', '2 9\n10 20\n', '12 3\n4 6 7 9 10 11 13 15 17 18 20 21\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
В первом примере Аркадий должен взять первый велосипед в первом велопрокате и доехать на нём до второго велопроката. Во втором велопрокате он должен взять новый велосипед, на котором он сможет добраться до четвертого велопроката, рядом с которым и находится его дом. Поэтому Аркадию нужно всего два велосипеда, чтобы добраться от школы до дома.
Во втором примере всего два велопроката, расстояние между которыми 10. Но максимальное расстояние, которое можно проехать на одном велосипеде, равно 9. Поэтому Аркадий не сможет добраться от школы до дома только на велосипедах.
|
```python
n, k = map(int, input().split())
arr = list(map(int, input().split()))
xs = arr[0]
ans = 1
for i in range(len(arr)):
if arr[i] - xs > k:
xs = arr[i-1]
ans += 1
if arr[-1] - xs <= k:
print(ans)
else:
print(-1)
```
| 0
|
|
729
|
B
|
Spotlights
|
PROGRAMMING
| 1,200
|
[
"dp",
"implementation"
] | null | null |
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
|
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
|
Print one integer — the number of good positions for placing the spotlight.
|
[
"2 4\n0 1 0 0\n1 0 1 0\n",
"4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n"
] |
[
"9\n",
"20\n"
] |
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
| 1,000
|
[
{
"input": "2 4\n0 1 0 0\n1 0 1 0",
"output": "9"
},
{
"input": "4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0",
"output": "20"
},
{
"input": "1 5\n1 1 0 0 0",
"output": "3"
},
{
"input": "2 10\n0 0 0 0 0 0 0 1 0 0\n1 0 0 0 0 0 0 0 0 0",
"output": "20"
},
{
"input": "3 1\n1\n0\n0",
"output": "2"
},
{
"input": "5 7\n0 0 0 0 0 0 1\n0 0 0 0 0 0 1\n0 0 0 1 0 0 0\n0 0 0 0 0 0 0\n0 0 0 0 0 0 0",
"output": "25"
},
{
"input": "10 20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "5 7\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1\n0 1 1 1 1 1 1\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1",
"output": "26"
},
{
"input": "10 20\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "121"
},
{
"input": "1 2\n0 1",
"output": "1"
},
{
"input": "1 2\n1 0",
"output": "1"
},
{
"input": "1 2\n1 1",
"output": "0"
},
{
"input": "2 1\n1\n0",
"output": "1"
},
{
"input": "2 1\n0\n1",
"output": "1"
},
{
"input": "2 1\n1\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "4 4\n1 1 1 1\n1 0 0 1\n1 0 0 1\n1 1 1 1",
"output": "16"
}
] | 1,582,461,140
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 72
| 514
| 28,364,800
|
from sys import stdin
n,m = [int(x) for x in stdin.readline().split()]
plan = [[int(x) for x in stdin.readline().split()] for _ in range(n)]
prowsum = [[0 for _ in range(m)] for _ in range(n)]
for i in range(n):
prowsum[i][0] = plan[i][0]
for j in range(1,m):
prowsum[i][j] = prowsum[i][j-1] or plan[i][j]
srowsum = [[0 for _ in range(m)] for _ in range(n)]
for i in range(n):
srowsum[i][m-1] = plan[i][m-1]
for j in range(m-2,-1,-1):
srowsum[i][j] = srowsum[i][j+1] or plan[i][j]
pcolsum = [[0 for _ in range(m)] for _ in range(n)]
for j in range(m):
pcolsum[0][j] = plan[0][j]
for i in range(1,n):
pcolsum[i][j] = pcolsum[i-1][j] or plan[i][j]
scolsum = [[0 for _ in range(m)] for _ in range(n)]
for j in range(m):
scolsum[n-1][j] = plan[n-1][j]
for i in range(n-2,-1,-1):
scolsum[i][j] = scolsum[i+1][j] or plan[i][j]
ans = 0
for i in range(n):
for j in range(m):
if plan[i][j]==1: continue
if i>0 and pcolsum[i-1][j]: ans += 1
if i<n-1 and scolsum[i+1][j]: ans += 1
if j>0 and prowsum[i][j-1]: ans += 1
if j<m-1 and srowsum[i][j+1]: ans += 1
print(ans)
|
Title: Spotlights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
Output Specification:
Print one integer — the number of good positions for placing the spotlight.
Demo Input:
['2 4\n0 1 0 0\n1 0 1 0\n', '4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n']
Demo Output:
['9\n', '20\n']
Note:
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
|
```python
from sys import stdin
n,m = [int(x) for x in stdin.readline().split()]
plan = [[int(x) for x in stdin.readline().split()] for _ in range(n)]
prowsum = [[0 for _ in range(m)] for _ in range(n)]
for i in range(n):
prowsum[i][0] = plan[i][0]
for j in range(1,m):
prowsum[i][j] = prowsum[i][j-1] or plan[i][j]
srowsum = [[0 for _ in range(m)] for _ in range(n)]
for i in range(n):
srowsum[i][m-1] = plan[i][m-1]
for j in range(m-2,-1,-1):
srowsum[i][j] = srowsum[i][j+1] or plan[i][j]
pcolsum = [[0 for _ in range(m)] for _ in range(n)]
for j in range(m):
pcolsum[0][j] = plan[0][j]
for i in range(1,n):
pcolsum[i][j] = pcolsum[i-1][j] or plan[i][j]
scolsum = [[0 for _ in range(m)] for _ in range(n)]
for j in range(m):
scolsum[n-1][j] = plan[n-1][j]
for i in range(n-2,-1,-1):
scolsum[i][j] = scolsum[i+1][j] or plan[i][j]
ans = 0
for i in range(n):
for j in range(m):
if plan[i][j]==1: continue
if i>0 and pcolsum[i-1][j]: ans += 1
if i<n-1 and scolsum[i+1][j]: ans += 1
if j>0 and prowsum[i][j-1]: ans += 1
if j<m-1 and srowsum[i][j+1]: ans += 1
print(ans)
```
| 3
|
|
414
|
A
|
Mashmokh and Numbers
|
PROGRAMMING
| 1,500
|
[
"constructive algorithms",
"number theory"
] | null | null |
It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
In this game Mashmokh writes sequence of *n* distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers *x* and *y* from the board, he gets *gcd*(*x*,<=*y*) points. At the beginning of the game Bimokh has zero points.
Mashmokh wants to win in the game. For this reason he wants his boss to get exactly *k* points in total. But the guy doesn't know how choose the initial sequence in the right way.
Please, help him. Find *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* such that his boss will score exactly *k* points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109.
|
The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*<=≤<=105; 0<=≤<=*k*<=≤<=108).
|
If such sequence doesn't exist output -1 otherwise output *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
[
"5 2\n",
"5 3",
"7 2\n"
] |
[
"1 2 3 4 5\n",
"2 4 3 7 1",
"-1\n"
] |
*gcd*(*x*, *y*) is greatest common divisor of *x* and *y*.
| 500
|
[
{
"input": "5 2",
"output": "1 2 3 4 5"
},
{
"input": "5 3",
"output": "2 4 5 6 7"
},
{
"input": "7 2",
"output": "-1"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "2 0",
"output": "-1"
},
{
"input": "1 10",
"output": "-1"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "7 3",
"output": "1 2 3 4 5 6 7"
},
{
"input": "7 6",
"output": "4 8 1 2 5 6 7"
},
{
"input": "7 7",
"output": "5 10 1 2 3 4 6"
},
{
"input": "100000 100000000",
"output": "99950001 199900002 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 ..."
},
{
"input": "3455 2792393",
"output": "2790667 5581334 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151..."
},
{
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"output": "16467569 32935138 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 1..."
},
{
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},
{
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},
{
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},
{
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},
{
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},
{
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},
{
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{
"input": "57534 85681593",
"output": "85652827 171305654 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 ..."
},
{
"input": "28652 18840000",
"output": "18825675 37651350 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 1..."
},
{
"input": "18247 23541343",
"output": "23532221 47064442 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 1..."
},
{
"input": "89529 95022203",
"output": "94977440 189954880 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 ..."
},
{
"input": "42775 89315917",
"output": "89294531 178589062 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 ..."
},
{
"input": "946 93333203",
"output": "93332731 186665462 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 ..."
},
{
"input": "93595 48782905",
"output": "48736109 97472218 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 1..."
},
{
"input": "87371 60145723",
"output": "60102039 120204078 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 ..."
},
{
"input": "7695 94816808",
"output": "94812962 189625924 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 ..."
},
{
"input": "21846 16967905",
"output": "16956983 33913966 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 1..."
},
{
"input": "10 3",
"output": "-1"
},
{
"input": "6 1000003",
"output": "1000001 2000002 1 2 3 4"
},
{
"input": "100000 549999",
"output": "500000 1000000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 ..."
},
{
"input": "10 4",
"output": "-1"
},
{
"input": "8 10",
"output": "7 14 1 2 3 4 5 6"
},
{
"input": "6 10000003",
"output": "10000001 20000002 1 2 3 4"
},
{
"input": "50 50000030",
"output": "50000006 100000012 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48"
},
{
"input": "7 11",
"output": "9 18 1 2 3 4 5"
},
{
"input": "2 96996900",
"output": "96996900 193993800"
},
{
"input": "3 99999997",
"output": "99999997 199999994 1"
},
{
"input": "10000 10",
"output": "-1"
},
{
"input": "5 100000000",
"output": "99999999 199999998 1 2 3"
},
{
"input": "20 15",
"output": "6 12 1 2 3 4 7 8 9 10 13 14 15 16 17 18 19 20 21 22"
},
{
"input": "10 50000006",
"output": "50000002 100000004 1 2 3 4 5 6 7 8"
},
{
"input": "4 1257",
"output": "1256 2512 1 2"
},
{
"input": "100 1",
"output": "-1"
},
{
"input": "6 1",
"output": "-1"
},
{
"input": "10 1000004",
"output": "1000000 2000000 1 2 3 4 5 6 7 8"
},
{
"input": "100000 100000",
"output": "50001 100002 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 15..."
},
{
"input": "10 3000004",
"output": "3000000 6000000 1 2 3 4 5 6 7 8"
},
{
"input": "99999 149998",
"output": "100000 200000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 1..."
},
{
"input": "11 1434567",
"output": "1434563 2869126 1 2 3 4 5 6 7 8 9"
},
{
"input": "205 110",
"output": "9 18 1 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "11 14342267",
"output": "14342263 28684526 1 2 3 4 5 6 7 8 9"
}
] | 1,584,010,454
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 140
| 0
|
n, k = map(int, input().split())
mn = n // 2
if mn > k:
print(-1)
exit()
ret = []
for i in range(1, (n - 2) // 2 + 1):
ret.extend((i * 2 - 1, i * 2))
v = k - mn + 1
print(*(ret + [v * 2903, v * 3533] + [10 ** 9] * (n & 1)))
|
Title: Mashmokh and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
In this game Mashmokh writes sequence of *n* distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers *x* and *y* from the board, he gets *gcd*(*x*,<=*y*) points. At the beginning of the game Bimokh has zero points.
Mashmokh wants to win in the game. For this reason he wants his boss to get exactly *k* points in total. But the guy doesn't know how choose the initial sequence in the right way.
Please, help him. Find *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* such that his boss will score exactly *k* points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109.
Input Specification:
The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*<=≤<=105; 0<=≤<=*k*<=≤<=108).
Output Specification:
If such sequence doesn't exist output -1 otherwise output *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Demo Input:
['5 2\n', '5 3', '7 2\n']
Demo Output:
['1 2 3 4 5\n', '2 4 3 7 1', '-1\n']
Note:
*gcd*(*x*, *y*) is greatest common divisor of *x* and *y*.
|
```python
n, k = map(int, input().split())
mn = n // 2
if mn > k:
print(-1)
exit()
ret = []
for i in range(1, (n - 2) // 2 + 1):
ret.extend((i * 2 - 1, i * 2))
v = k - mn + 1
print(*(ret + [v * 2903, v * 3533] + [10 ** 9] * (n & 1)))
```
| 0
|
|
160
|
A
|
Twins
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] | null | null |
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
|
In the single line print the single number — the minimum needed number of coins.
|
[
"2\n3 3\n",
"3\n2 1 2\n"
] |
[
"2\n",
"2\n"
] |
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
| 500
|
[
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "4"
},
{
"input": "7\n10 10 5 5 5 5 1",
"output": "3"
},
{
"input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4",
"output": "8"
},
{
"input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40",
"output": "8"
},
{
"input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90",
"output": "6"
},
{
"input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28",
"output": "14"
},
{
"input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42",
"output": "22"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "51"
},
{
"input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2",
"output": "37"
},
{
"input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "36"
},
{
"input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3",
"output": "33"
},
{
"input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "30"
},
{
"input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2",
"output": "29"
},
{
"input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "30"
},
{
"input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "29"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "30"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "4\n4 1 2 3",
"output": "2"
},
{
"input": "5\n2 2 2 4 4",
"output": "2"
},
{
"input": "4\n100 99 99 100",
"output": "2"
}
] | 1,698,231,210
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 92
| 0
|
n = int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
s = sum(a)
f=0
t=0
while(t<=s):
s-=a[f]
t+=a[f]
f+=1
print(f)
|
Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
|
```python
n = int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
s = sum(a)
f=0
t=0
while(t<=s):
s-=a[f]
t+=a[f]
f+=1
print(f)
```
| 3
|
|
383
|
A
|
Milking cows
|
PROGRAMMING
| 1,600
|
[
"data structures",
"greedy"
] | null | null |
Iahub helps his grandfather at the farm. Today he must milk the cows. There are *n* cows sitting in a row, numbered from 1 to *n* from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, where *a**i* is 0 if the cow number *i* is facing left, and 1 if it is facing right.
|
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"4\n0 0 1 0\n",
"5\n1 0 1 0 1\n"
] |
[
"1",
"3"
] |
In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost.
| 500
|
[
{
"input": "4\n0 0 1 0",
"output": "1"
},
{
"input": "5\n1 0 1 0 1",
"output": "3"
},
{
"input": "50\n1 1 0 1 1 1 1 1 1 0 0 1 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 1 1 0 1 0 0 1 0",
"output": "416"
},
{
"input": "100\n1 1 0 0 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 1 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 1 1 0 0 1 0 0",
"output": "1446"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n0 1",
"output": "0"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "4\n1 1 1 1",
"output": "0"
}
] | 1,599,914,965
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 249
| 7,577,600
|
def solve():
n = int(input())
arr = list(map(int, input().split()))
count_arr = [0] * n
zero_count = 0
for _ in range(n):
if arr[n - _ - 1] == 0:
zero_count += 1
count_arr[n - _ - 1] = zero_count
ans = 0
for _ in range(n):
if arr[_] == 1:
ans += count_arr[_]
print(ans)
solve()
|
Title: Milking cows
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub helps his grandfather at the farm. Today he must milk the cows. There are *n* cows sitting in a row, numbered from 1 to *n* from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, where *a**i* is 0 if the cow number *i* is facing left, and 1 if it is facing right.
Output Specification:
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['4\n0 0 1 0\n', '5\n1 0 1 0 1\n']
Demo Output:
['1', '3']
Note:
In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost.
|
```python
def solve():
n = int(input())
arr = list(map(int, input().split()))
count_arr = [0] * n
zero_count = 0
for _ in range(n):
if arr[n - _ - 1] == 0:
zero_count += 1
count_arr[n - _ - 1] = zero_count
ans = 0
for _ in range(n):
if arr[_] == 1:
ans += count_arr[_]
print(ans)
solve()
```
| 3
|
|
873
|
A
|
Chores
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted.
Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* ().
Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously.
|
The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself.
The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore.
It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1.
|
Print one number — minimum time Luba needs to do all *n* chores.
|
[
"4 2 2\n3 6 7 10\n",
"5 2 1\n100 100 100 100 100\n"
] |
[
"13\n",
"302\n"
] |
In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, respectively. Then the answer is 3 + 6 + 2 + 2 = 13.
In the second example Luba can choose any two chores to spend *x* time on them instead of *a*<sub class="lower-index">*i*</sub>. So the answer is 100·3 + 2·1 = 302.
| 0
|
[
{
"input": "4 2 2\n3 6 7 10",
"output": "13"
},
{
"input": "5 2 1\n100 100 100 100 100",
"output": "302"
},
{
"input": "1 1 1\n100",
"output": "1"
},
{
"input": "100 1 99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "9999"
},
{
"input": "100 100 1\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "100 50 50\n51 51 52 53 55 55 55 55 56 56 56 57 57 58 58 59 59 59 60 60 61 61 62 62 63 64 64 64 64 65 65 65 65 66 66 66 67 68 68 68 69 69 70 70 70 70 71 71 71 71 71 71 72 72 76 76 76 76 77 79 79 81 81 81 81 82 82 82 82 83 84 85 86 87 87 88 88 88 89 89 89 90 90 90 91 91 91 92 92 93 95 95 96 96 96 97 97 98 99 100",
"output": "5618"
},
{
"input": "100 100 1\n2 4 4 4 5 5 5 6 10 10 11 11 12 12 13 13 13 14 17 18 20 20 21 21 22 22 23 24 24 25 26 29 29 32 32 34 34 35 38 39 39 40 40 42 42 43 45 47 48 49 51 52 52 54 57 59 59 60 61 61 62 63 63 64 65 65 68 70 70 72 74 75 75 76 76 77 77 78 78 78 79 80 81 82 82 83 83 83 84 89 90 92 92 93 94 96 96 97 98 99",
"output": "100"
},
{
"input": "100 1 1\n3 3 5 7 8 8 8 9 9 9 11 13 14 15 18 18 19 20 21 22 22 25 27 27 29 31 32 33 33 34 36 37 37 38 40 42 44 44 46 47 47 48 48 48 50 50 51 51 54 54 54 55 55 56 56 56 60 61 62 62 63 64 65 65 68 70 70 71 71 71 71 75 75 76 76 79 79 79 79 81 81 82 82 86 86 86 86 88 90 90 92 96 97 97 98 98 98 98 100 100",
"output": "5202"
},
{
"input": "100 50 49\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51",
"output": "4950"
},
{
"input": "100 50 1\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "151"
},
{
"input": "100 1 1\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "199"
}
] | 1,695,651,598
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 11
| 46
| 0
|
b, k, x = input().split()
b = int(b)
x = int(x)
k = int(k)
a = [int(i) for i in input().split()]
sum = 0
for i in range(b-1, -1, -1):
if k > 0:
sum += x
k -= 1
else:
sum += a[i]
print(sum)
|
Title: Chores
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted.
Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* ().
Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously.
Input Specification:
The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself.
The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore.
It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1.
Output Specification:
Print one number — minimum time Luba needs to do all *n* chores.
Demo Input:
['4 2 2\n3 6 7 10\n', '5 2 1\n100 100 100 100 100\n']
Demo Output:
['13\n', '302\n']
Note:
In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, respectively. Then the answer is 3 + 6 + 2 + 2 = 13.
In the second example Luba can choose any two chores to spend *x* time on them instead of *a*<sub class="lower-index">*i*</sub>. So the answer is 100·3 + 2·1 = 302.
|
```python
b, k, x = input().split()
b = int(b)
x = int(x)
k = int(k)
a = [int(i) for i in input().split()]
sum = 0
for i in range(b-1, -1, -1):
if k > 0:
sum += x
k -= 1
else:
sum += a[i]
print(sum)
```
| 3
|
|
939
|
A
|
Love Triangle
|
PROGRAMMING
| 800
|
[
"graphs"
] | null | null |
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
|
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
|
[
"5\n2 4 5 1 3\n",
"5\n5 5 5 5 1\n"
] |
[
"YES\n",
"NO\n"
] |
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles.
| 500
|
[
{
"input": "5\n2 4 5 1 3",
"output": "YES"
},
{
"input": "5\n5 5 5 5 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "10\n4 10 9 5 3 1 5 10 6 4",
"output": "NO"
},
{
"input": "10\n5 5 4 9 10 9 9 5 3 1",
"output": "YES"
},
{
"input": "100\n50 40 60 87 39 58 44 84 46 68 16 57 77 87 92 95 42 31 74 15 36 84 30 3 47 15 87 90 76 66 6 63 74 19 40 49 6 84 41 9 77 34 7 12 11 73 58 24 81 14 81 29 65 100 1 85 64 32 38 4 54 67 32 81 80 7 100 71 29 80 4 52 47 7 78 56 52 75 81 37 16 41 27 28 58 60 62 47 29 40 37 14 59 91 12 54 25 58 12 43",
"output": "NO"
},
{
"input": "100\n25 6 46 37 87 99 70 31 46 12 94 40 87 56 28 8 94 39 13 12 67 13 71 39 83 48 40 14 62 41 16 71 20 41 83 41 68 98 23 82 62 83 62 35 49 22 31 21 66 98 54 39 34 52 11 28 47 89 25 44 68 36 91 46 82 86 88 48 27 93 7 9 53 36 16 100 84 84 44 25 58 66 16 46 72 21 91 78 4 17 44 17 47 67 93 89 75 44 56 50",
"output": "YES"
},
{
"input": "2\n2 1",
"output": "NO"
},
{
"input": "3\n2 3 1",
"output": "YES"
},
{
"input": "5\n2 1 4 5 3",
"output": "YES"
},
{
"input": "5\n5 4 5 5 2",
"output": "YES"
}
] | 1,651,141,249
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 16
| 62
| 2,764,800
|
a=int(input())
b=list(map(int,input().split()))
d={}
for i in range(len(b)):
d[i+1]=b[i]
f=0
for key,value in d.items():
if value in d:
if d[value] in d:
if d[d[value]]==key:
print("YES")
f=1
break
if f!=1:
print("NO")
|
Title: Love Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
Output Specification:
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Demo Input:
['5\n2 4 5 1 3\n', '5\n5 5 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles.
|
```python
a=int(input())
b=list(map(int,input().split()))
d={}
for i in range(len(b)):
d[i+1]=b[i]
f=0
for key,value in d.items():
if value in d:
if d[value] in d:
if d[d[value]]==key:
print("YES")
f=1
break
if f!=1:
print("NO")
```
| 3
|
|
192
|
A
|
Funky Numbers
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force",
"implementation"
] | null | null |
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
|
The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).
|
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
|
[
"256\n",
"512\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
| 500
|
[
{
"input": "256",
"output": "YES"
},
{
"input": "512",
"output": "NO"
},
{
"input": "80",
"output": "NO"
},
{
"input": "828",
"output": "YES"
},
{
"input": "6035",
"output": "NO"
},
{
"input": "39210",
"output": "YES"
},
{
"input": "79712",
"output": "NO"
},
{
"input": "190492",
"output": "YES"
},
{
"input": "5722367",
"output": "NO"
},
{
"input": "816761542",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "2",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "NO"
},
{
"input": "9",
"output": "YES"
},
{
"input": "10",
"output": "NO"
},
{
"input": "12",
"output": "YES"
},
{
"input": "13",
"output": "YES"
},
{
"input": "14",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "16",
"output": "YES"
},
{
"input": "17",
"output": "NO"
},
{
"input": "18",
"output": "YES"
},
{
"input": "19",
"output": "NO"
},
{
"input": "20",
"output": "YES"
},
{
"input": "41",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "69",
"output": "YES"
},
{
"input": "82",
"output": "NO"
},
{
"input": "85",
"output": "NO"
},
{
"input": "736",
"output": "NO"
},
{
"input": "895",
"output": "YES"
},
{
"input": "934",
"output": "YES"
},
{
"input": "6213",
"output": "YES"
},
{
"input": "7405",
"output": "NO"
},
{
"input": "9919",
"output": "NO"
},
{
"input": "40942",
"output": "YES"
},
{
"input": "41992",
"output": "NO"
},
{
"input": "68535",
"output": "NO"
},
{
"input": "405718",
"output": "NO"
},
{
"input": "1046146",
"output": "YES"
},
{
"input": "3761248",
"output": "YES"
},
{
"input": "6195181",
"output": "YES"
},
{
"input": "35354345",
"output": "NO"
},
{
"input": "81282830",
"output": "NO"
},
{
"input": "187719774",
"output": "NO"
},
{
"input": "296798673",
"output": "NO"
},
{
"input": "938938476",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "999887464",
"output": "YES"
},
{
"input": "999111944",
"output": "NO"
},
{
"input": "999966520",
"output": "YES"
},
{
"input": "999912080",
"output": "NO"
},
{
"input": "999992017",
"output": "YES"
},
{
"input": "999990474",
"output": "NO"
},
{
"input": "999999190",
"output": "YES"
},
{
"input": "999999125",
"output": "NO"
},
{
"input": "999999940",
"output": "YES"
},
{
"input": "999999995",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "83495494",
"output": "NO"
},
{
"input": "968022000",
"output": "YES"
},
{
"input": "399980000",
"output": "YES"
},
{
"input": "4",
"output": "YES"
},
{
"input": "999999998",
"output": "NO"
}
] | 1,586,677,326
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 2,000
| 12,595,200
|
n=int(input())
a=[i*(i+1)//2 for i in range(200000)]
s=set(a)
for i in a:
if n-i in a:
print("YES")
exit()
if i>n:
break
print("NO")
|
Title: Funky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
Input Specification:
The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
Demo Input:
['256\n', '512\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
|
```python
n=int(input())
a=[i*(i+1)//2 for i in range(200000)]
s=set(a)
for i in a:
if n-i in a:
print("YES")
exit()
if i>n:
break
print("NO")
```
| 0
|
|
456
|
A
|
Laptops
|
PROGRAMMING
| 1,100
|
[
"sortings"
] | null | null |
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
|
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
|
[
"2\n1 2\n2 1\n"
] |
[
"Happy Alex\n"
] |
none
| 500
|
[
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n2 3\n1 1\n3 2",
"output": "Happy Alex"
},
{
"input": "4\n4 1\n3 2\n2 3\n1 4",
"output": "Happy Alex"
},
{
"input": "2\n2 1\n1 2",
"output": "Happy Alex"
},
{
"input": "3\n3 2\n1 1\n2 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 1\n3 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
}
] | 1,674,207,999
| 2,199
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 3
| 1,000
| 7,680,000
|
n = int(input())
l = ""
a=[]
k =0
h =0
p =n
for i in range (n):
m= input().split()
for j in m:
a.append(j)
# Cach nhap vao mang cac so
while(p > 0):
for i in range (n*2):
if i % 2 ==0 and i < (n*2-2):
if int(a[i]) < int(a[i+2]) and int(a[i+1])> int(a[i+3]):
h +=1
else:
k += 1
p -=1
if h > k:
print("Happy Alex")
else:
print("Poor Alex")
|
Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none
|
```python
n = int(input())
l = ""
a=[]
k =0
h =0
p =n
for i in range (n):
m= input().split()
for j in m:
a.append(j)
# Cach nhap vao mang cac so
while(p > 0):
for i in range (n*2):
if i % 2 ==0 and i < (n*2-2):
if int(a[i]) < int(a[i+2]) and int(a[i+1])> int(a[i+3]):
h +=1
else:
k += 1
p -=1
if h > k:
print("Happy Alex")
else:
print("Poor Alex")
```
| 0
|
|
56
|
D
|
Changing a String
|
PROGRAMMING
| 2,100
|
[
"dp"
] |
D. Changing a String
|
2
|
256
|
There is a string *s*, consisting of capital Latin letters. Let's denote its current length as |*s*|. During one move it is allowed to apply one of the following operations to it:
- INSERT *pos* *ch* — insert a letter *ch* in the string *s* in the position *pos* (1<=≤<=*pos*<=≤<=|*s*|<=+<=1,<=*A*<=≤<=*ch*<=≤<=*Z*). The letter *ch* becomes the *pos*-th symbol of the string *s*, at that the letters shift aside and the length of the string increases by 1. - DELETE *pos* — delete a character number *pos* (1<=≤<=*pos*<=≤<=|*s*|) from the string *s*. At that the letters shift together and the length of the string decreases by 1. - REPLACE *pos* *ch* — the letter in the position *pos* of the line *s* is replaced by *ch* (1<=≤<=*pos*<=≤<=|*s*|,<=*A*<=≤<=*ch*<=≤<=*Z*). At that the length of the string does not change.
Your task is to find in which minimal number of moves one can get a *t* string from an *s* string. You should also find the sequence of actions leading to the required results.
|
The first line contains *s*, the second line contains *t*. The lines consist only of capital Latin letters, their lengths are positive numbers from 1 to 1000.
|
In the first line print the number of moves *k* in the given sequence of operations. The number should be the minimal possible one. Then print *k* lines containing one operation each. Print the operations in the format, described above. If there are several solutions, print any of them.
|
[
"ABA\nABBBA\n",
"ACCEPTED\nWRONGANSWER\n"
] |
[
"2\nINSERT 3 B\nINSERT 4 B\n",
"10\nREPLACE 1 W\nREPLACE 2 R\nREPLACE 3 O\nREPLACE 4 N\nREPLACE 5 G\nREPLACE 6 A\nINSERT 7 N\nINSERT 8 S\nINSERT 9 W\nREPLACE 11 R\n"
] |
none
| 2,000
|
[
{
"input": "ABA\nABBBA",
"output": "2\nINSERT 3 B\nINSERT 4 B"
},
{
"input": "ACCEPTED\nWRONGANSWER",
"output": "10\nREPLACE 1 W\nREPLACE 2 R\nREPLACE 3 O\nREPLACE 4 N\nREPLACE 5 G\nREPLACE 6 A\nINSERT 7 N\nINSERT 8 S\nINSERT 9 W\nREPLACE 11 R"
},
{
"input": "V\nBNBNE",
"output": "5\nREPLACE 1 B\nINSERT 2 N\nINSERT 3 B\nINSERT 4 N\nINSERT 5 E"
},
{
"input": "UB\nPWL",
"output": "3\nREPLACE 1 P\nREPLACE 2 W\nINSERT 3 L"
},
{
"input": "JOYXNKYPF\nGDV",
"output": "9\nREPLACE 1 G\nREPLACE 2 D\nREPLACE 3 V\nDELETE 4\nDELETE 4\nDELETE 4\nDELETE 4\nDELETE 4\nDELETE 4"
},
{
"input": "SZDAWSVGK\nUM",
"output": "9\nREPLACE 1 U\nREPLACE 2 M\nDELETE 3\nDELETE 3\nDELETE 3\nDELETE 3\nDELETE 3\nDELETE 3\nDELETE 3"
},
{
"input": "TJHGFKKCDOHRNAXZROCWIYFF\nZBWUEHEVEOUATECAGLZIQMUDXEMHRSOZMAUJRWLQMPPZOUMXHAMWUGEDIKVKBLVMXWUOFMPAFDPRBCFTEWOULCZWRQHCTBTBXRHHODWBCXWIMNCNEXOSKSUJLISGCLLLXOKRSBNOZTHAJNNLILYFFMSYKOFPTXRNEFBSOUHFOLTIQAINRPXWRQ",
"output": "164\nINSERT 1 Z\nINSERT 2 B\nINSERT 3 W\nINSERT 4 U\nINSERT 5 E\nINSERT 6 H\nINSERT 7 E\nINSERT 8 V\nINSERT 9 E\nINSERT 10 O\nINSERT 11 U\nINSERT 12 A\nINSERT 14 E\nINSERT 15 C\nINSERT 16 A\nINSERT 17 G\nINSERT 18 L\nINSERT 19 Z\nINSERT 20 I\nINSERT 21 Q\nINSERT 22 M\nINSERT 23 U\nINSERT 24 D\nINSERT 25 X\nINSERT 26 E\nINSERT 27 M\nINSERT 28 H\nINSERT 29 R\nINSERT 30 S\nINSERT 31 O\nINSERT 32 Z\nINSERT 33 M\nINSERT 34 A\nINSERT 35 U\nINSERT 37 R\nINSERT 38 W\nINSERT 39 L\nINSERT 40 Q\nINSERT 41 M\nINSERT 4..."
},
{
"input": "GXPLMUNZIRBHFJOOJHOMQNIKHVQSGFYSJLSWJQBUWYWHLQHLZYLPZWBOMPOLOLUGSBMHHLYGEIOUWDKPFIAAIRKYRLXTIFAZOPOLLPSNZHCIZDRTJPCYSCDTXBTMFSGEPRNOHJHNXZFYJPAMSHNOVZZYWCVEXLLOVHGAJBANAXVNTWCYTCUMQPEUUESQZTSDANIMUVJGDJCDLIAZKIAYAUQKPXRYKFRBVQJDRSUQZQZTIHWZDXQND\nYTJUCSBGESVMVRIDTBJTMPVBCWWDWKBPEBVMGDXGIVLWQXVEETNSDXKTVJPXOPERWSGDPPMKNMWDIGEHFXNUQADIS",
"output": "209\nREPLACE 1 Y\nREPLACE 2 T\nREPLACE 3 J\nDELETE 4\nDELETE 4\nREPLACE 5 C\nREPLACE 6 S\nDELETE 7\nDELETE 7\nREPLACE 8 G\nREPLACE 9 E\nREPLACE 10 S\nREPLACE 11 V\nDELETE 12\nDELETE 12\nDELETE 12\nDELETE 12\nREPLACE 13 V\nREPLACE 14 R\nREPLACE 16 D\nREPLACE 17 T\nREPLACE 18 B\nDELETE 19\nDELETE 19\nDELETE 19\nDELETE 19\nDELETE 19\nDELETE 19\nREPLACE 20 T\nREPLACE 21 M\nREPLACE 22 P\nREPLACE 23 V\nDELETE 24\nREPLACE 25 C\nDELETE 27\nREPLACE 28 D\nREPLACE 29 W\nREPLACE 30 K\nREPLACE 31 B\nDELETE 32\nDELETE 3..."
},
{
"input": "BPYEOOTCVXAZPTHUEAIUZURZPHBWOEHGVHSDBYNXLHGOPBPCLPWQWVRYJGILKOOLASFSFYFGIMPVFGRRINJOXENOIMETWXUWCKXYBPBPPYRTMOZSBMBVJENPUSAEUPMQDGAQXUDIFUPOHYXPUWZLJP\nYRWPQARABPMMBLUZJHDVOUODHRNLMOOVGVJIZDDLXEWCPUFYYPKCDDYGYOLTNHAVURLLEKNOLC",
"output": "123\nDELETE 1\nDELETE 1\nREPLACE 2 R\nREPLACE 3 W\nREPLACE 4 P\nREPLACE 5 Q\nREPLACE 6 A\nREPLACE 7 R\nDELETE 8\nREPLACE 9 B\nREPLACE 11 M\nREPLACE 12 M\nREPLACE 13 B\nREPLACE 14 L\nDELETE 15\nDELETE 15\nREPLACE 17 J\nDELETE 18\nDELETE 18\nDELETE 18\nREPLACE 19 D\nREPLACE 20 V\nREPLACE 22 U\nREPLACE 23 O\nDELETE 24\nDELETE 24\nDELETE 24\nDELETE 24\nDELETE 25\nDELETE 25\nDELETE 25\nDELETE 25\nDELETE 25\nDELETE 26\nDELETE 26\nDELETE 26\nDELETE 26\nDELETE 26\nDELETE 26\nDELETE 26\nDELETE 26\nDELETE 26\nDELETE..."
},
{
"input": "YWJJBIRAPVSKOZUBVXOEMSCFWLSGNDUUBAOGTFAIOWJIZLVJCUKUYCYSAAHVLGCUAXDGYKJQMNOIFUDWFOSAVKRMAKLEKMPPISHARHVGEWRKUADEQZCJUXDVRQIIGO\nYWJJBIRAPVSKOZUBVXOEMSCFWLSGNDUUBAOGTFAIOWJIZLVJCUKUYCYSAAHVLGCUAXDGYKJQMNOIFUDWFOSAVKRMAKLEKMPISHARHVGEWRKUADEQZCJUXDVRQIIGO",
"output": "1\nDELETE 96"
},
{
"input": "A\nC",
"output": "1\nREPLACE 1 C"
},
{
"input": "Z\nZ",
"output": "0"
}
] | 1,671,882,693
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 60
| 102,400
|
from functools import lru_cache
def main():
s = input()
t = input()
@lru_cache(None)
def dp(i, j):
if i == len(s) and t == len(t):
return (0, tuple())
if i == len(s):
changes = len(t) - j
ans = []
for k in range(j, len(t)):
ans.append("INSERT " + str(k + 1) + " " + t[k])
return (changes, tuple(ans))
if j == len(t):
changes = len(s) - i
ans = []
for k in range(i, len(s)):
ans.append("REMOVE "+ str(k + 1) + " " + s[k])
return (changes, tuple(ans))
changes = 1
if s[i] == t[j]:
return dp(i + 1, j + 1)
# remove
one = dp(i + 1, j)
# INSERT
two = dp(i, j + 1)
# REPLACE
three = dp(i + 1, j + 1)
res = min(one, two, three)
if res == one:
return (res[0] + 1, tuple(list(res[1]) + ["REMOVE " + str(i + 1) + " " + s[i] ]))
elif res == two:
return (res[0] + 1, tuple(list(res[1]) + ["INSERT " + str(i + 1) + " " + t[j] ]))
else:
return (res[0] + 1, tuple(list(res[1]) + ["REPLACE " + str(i + 1) + " " + t[j] ]))
ANS = dp(0, 0)
print(ANS[0])
second = ANS[1][::-1]
# print(second)
diff = 0
for line in second:
action, num , word = line.split(" ")
n = int(num) + diff
print("{action} {n} {word}".format(action = action, n = n, word = word ))
if action == "INSERT":
diff += 1
main()
|
Title: Changing a String
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a string *s*, consisting of capital Latin letters. Let's denote its current length as |*s*|. During one move it is allowed to apply one of the following operations to it:
- INSERT *pos* *ch* — insert a letter *ch* in the string *s* in the position *pos* (1<=≤<=*pos*<=≤<=|*s*|<=+<=1,<=*A*<=≤<=*ch*<=≤<=*Z*). The letter *ch* becomes the *pos*-th symbol of the string *s*, at that the letters shift aside and the length of the string increases by 1. - DELETE *pos* — delete a character number *pos* (1<=≤<=*pos*<=≤<=|*s*|) from the string *s*. At that the letters shift together and the length of the string decreases by 1. - REPLACE *pos* *ch* — the letter in the position *pos* of the line *s* is replaced by *ch* (1<=≤<=*pos*<=≤<=|*s*|,<=*A*<=≤<=*ch*<=≤<=*Z*). At that the length of the string does not change.
Your task is to find in which minimal number of moves one can get a *t* string from an *s* string. You should also find the sequence of actions leading to the required results.
Input Specification:
The first line contains *s*, the second line contains *t*. The lines consist only of capital Latin letters, their lengths are positive numbers from 1 to 1000.
Output Specification:
In the first line print the number of moves *k* in the given sequence of operations. The number should be the minimal possible one. Then print *k* lines containing one operation each. Print the operations in the format, described above. If there are several solutions, print any of them.
Demo Input:
['ABA\nABBBA\n', 'ACCEPTED\nWRONGANSWER\n']
Demo Output:
['2\nINSERT 3 B\nINSERT 4 B\n', '10\nREPLACE 1 W\nREPLACE 2 R\nREPLACE 3 O\nREPLACE 4 N\nREPLACE 5 G\nREPLACE 6 A\nINSERT 7 N\nINSERT 8 S\nINSERT 9 W\nREPLACE 11 R\n']
Note:
none
|
```python
from functools import lru_cache
def main():
s = input()
t = input()
@lru_cache(None)
def dp(i, j):
if i == len(s) and t == len(t):
return (0, tuple())
if i == len(s):
changes = len(t) - j
ans = []
for k in range(j, len(t)):
ans.append("INSERT " + str(k + 1) + " " + t[k])
return (changes, tuple(ans))
if j == len(t):
changes = len(s) - i
ans = []
for k in range(i, len(s)):
ans.append("REMOVE "+ str(k + 1) + " " + s[k])
return (changes, tuple(ans))
changes = 1
if s[i] == t[j]:
return dp(i + 1, j + 1)
# remove
one = dp(i + 1, j)
# INSERT
two = dp(i, j + 1)
# REPLACE
three = dp(i + 1, j + 1)
res = min(one, two, three)
if res == one:
return (res[0] + 1, tuple(list(res[1]) + ["REMOVE " + str(i + 1) + " " + s[i] ]))
elif res == two:
return (res[0] + 1, tuple(list(res[1]) + ["INSERT " + str(i + 1) + " " + t[j] ]))
else:
return (res[0] + 1, tuple(list(res[1]) + ["REPLACE " + str(i + 1) + " " + t[j] ]))
ANS = dp(0, 0)
print(ANS[0])
second = ANS[1][::-1]
# print(second)
diff = 0
for line in second:
action, num , word = line.split(" ")
n = int(num) + diff
print("{action} {n} {word}".format(action = action, n = n, word = word ))
if action == "INSERT":
diff += 1
main()
```
| 0
|
820
|
A
|
Mister B and Book Reading
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
|
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
|
Print one integer — the number of days Mister B needed to finish the book.
|
[
"5 5 10 5 4\n",
"12 4 12 4 1\n",
"15 1 100 0 0\n"
] |
[
"1\n",
"3\n",
"15\n"
] |
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.
| 500
|
[
{
"input": "5 5 10 5 4",
"output": "1"
},
{
"input": "12 4 12 4 1",
"output": "3"
},
{
"input": "15 1 100 0 0",
"output": "15"
},
{
"input": "1 1 1 0 0",
"output": "1"
},
{
"input": "1000 999 1000 1000 998",
"output": "2"
},
{
"input": "1000 2 2 5 1",
"output": "999"
},
{
"input": "1000 1 1 1000 0",
"output": "1000"
},
{
"input": "737 41 74 12 11",
"output": "13"
},
{
"input": "1000 1000 1000 0 999",
"output": "1"
},
{
"input": "765 12 105 5 7",
"output": "17"
},
{
"input": "15 2 2 1000 0",
"output": "8"
},
{
"input": "1000 1 1000 1000 0",
"output": "2"
},
{
"input": "20 3 7 1 2",
"output": "6"
},
{
"input": "1000 500 500 1000 499",
"output": "501"
},
{
"input": "1 1000 1000 1000 0",
"output": "1"
},
{
"input": "1000 2 1000 56 0",
"output": "7"
},
{
"input": "1000 2 1000 802 0",
"output": "3"
},
{
"input": "16 1 8 2 0",
"output": "4"
},
{
"input": "20 6 10 2 2",
"output": "3"
},
{
"input": "8 2 12 4 1",
"output": "3"
},
{
"input": "8 6 13 2 5",
"output": "2"
},
{
"input": "70 4 20 87 0",
"output": "5"
},
{
"input": "97 8 13 234 5",
"output": "13"
},
{
"input": "16 4 23 8 3",
"output": "3"
},
{
"input": "65 7 22 7 4",
"output": "5"
},
{
"input": "93 10 18 11 7",
"output": "9"
},
{
"input": "86 13 19 15 9",
"output": "9"
},
{
"input": "333 17 50 10 16",
"output": "12"
},
{
"input": "881 16 55 10 12",
"output": "23"
},
{
"input": "528 11 84 3 9",
"output": "19"
},
{
"input": "896 2 184 8 1",
"output": "16"
},
{
"input": "236 10 930 9 8",
"output": "8"
},
{
"input": "784 1 550 14 0",
"output": "12"
},
{
"input": "506 1 10 4 0",
"output": "53"
},
{
"input": "460 1 3 2 0",
"output": "154"
},
{
"input": "701 1 3 1 0",
"output": "235"
},
{
"input": "100 49 50 1000 2",
"output": "3"
},
{
"input": "100 1 100 100 0",
"output": "2"
},
{
"input": "12 1 4 2 0",
"output": "4"
},
{
"input": "22 10 12 0 0",
"output": "3"
},
{
"input": "20 10 15 1 4",
"output": "3"
},
{
"input": "1000 5 10 1 4",
"output": "169"
},
{
"input": "1000 1 1000 1 0",
"output": "45"
},
{
"input": "4 1 2 2 0",
"output": "3"
},
{
"input": "1 5 5 1 1",
"output": "1"
},
{
"input": "19 10 11 0 2",
"output": "3"
},
{
"input": "1 2 3 0 0",
"output": "1"
},
{
"input": "10 1 4 10 0",
"output": "4"
},
{
"input": "20 3 100 1 1",
"output": "5"
},
{
"input": "1000 5 9 5 0",
"output": "112"
},
{
"input": "1 11 12 0 10",
"output": "1"
},
{
"input": "1 1 1 1 0",
"output": "1"
},
{
"input": "1000 1 20 1 0",
"output": "60"
},
{
"input": "9 1 4 2 0",
"output": "4"
},
{
"input": "129 2 3 4 0",
"output": "44"
},
{
"input": "4 2 2 0 1",
"output": "3"
},
{
"input": "1000 1 10 100 0",
"output": "101"
},
{
"input": "100 1 100 1 0",
"output": "14"
},
{
"input": "8 3 4 2 0",
"output": "3"
},
{
"input": "20 1 6 4 0",
"output": "5"
},
{
"input": "8 2 4 2 0",
"output": "3"
},
{
"input": "11 5 6 7 2",
"output": "3"
},
{
"input": "100 120 130 120 0",
"output": "1"
},
{
"input": "7 1 4 1 0",
"output": "4"
},
{
"input": "5 3 10 0 2",
"output": "3"
},
{
"input": "5 2 2 0 0",
"output": "3"
},
{
"input": "1000 10 1000 10 0",
"output": "14"
},
{
"input": "25 3 50 4 2",
"output": "4"
},
{
"input": "9 10 10 10 9",
"output": "1"
},
{
"input": "17 10 12 6 5",
"output": "2"
},
{
"input": "15 5 10 3 0",
"output": "3"
},
{
"input": "8 3 5 1 0",
"output": "3"
},
{
"input": "19 1 12 5 0",
"output": "4"
},
{
"input": "1000 10 1000 1 0",
"output": "37"
},
{
"input": "100 1 2 1000 0",
"output": "51"
},
{
"input": "20 10 11 1000 9",
"output": "6"
},
{
"input": "16 2 100 1 1",
"output": "5"
},
{
"input": "18 10 13 2 5",
"output": "3"
},
{
"input": "12 3 5 3 1",
"output": "4"
},
{
"input": "17 3 11 2 0",
"output": "4"
},
{
"input": "4 2 100 1 1",
"output": "2"
},
{
"input": "7 4 5 2 3",
"output": "3"
},
{
"input": "100 1 2 2 0",
"output": "51"
},
{
"input": "50 4 5 5 0",
"output": "11"
},
{
"input": "1 2 2 0 1",
"output": "1"
},
{
"input": "1000 2 3 10 1",
"output": "500"
},
{
"input": "500 10 500 1000 0",
"output": "2"
},
{
"input": "1000 4 12 1 0",
"output": "87"
},
{
"input": "18 10 13 1 5",
"output": "3"
},
{
"input": "7 3 6 2 2",
"output": "3"
},
{
"input": "15 5 100 1 2",
"output": "4"
},
{
"input": "100 1 10 1 0",
"output": "15"
},
{
"input": "8 2 7 5 1",
"output": "2"
},
{
"input": "11 2 4 1 1",
"output": "5"
},
{
"input": "1000 500 900 100 300",
"output": "3"
},
{
"input": "7 1 2 5 0",
"output": "4"
},
{
"input": "7 3 5 3 2",
"output": "3"
},
{
"input": "7 3 10 2 1",
"output": "2"
},
{
"input": "1000 501 510 1 499",
"output": "50"
},
{
"input": "1000 1 1000 2 0",
"output": "32"
},
{
"input": "1 5 5 0 0",
"output": "1"
},
{
"input": "18 10 15 1 5",
"output": "3"
},
{
"input": "100 4 1000 1 2",
"output": "13"
},
{
"input": "20 2 40 1 1",
"output": "6"
},
{
"input": "1 11 1000 100 1",
"output": "1"
},
{
"input": "6 4 4 1 2",
"output": "2"
},
{
"input": "8 3 5 3 1",
"output": "3"
},
{
"input": "10 5 7 1 2",
"output": "3"
},
{
"input": "400 100 198 1 99",
"output": "25"
},
{
"input": "3 1 2 5 0",
"output": "2"
}
] | 1,498,864,891
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 61
| 5,529,600
|
c, v0, v1, a, l = map(int, input().split())
speed = v0
pages = v0
days = 1
while(pages < c):
if speed + a <= v1:
speed += a
pages = pages + speed - l
days += 1
print(days)
|
Title: Mister B and Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
Input Specification:
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
Output Specification:
Print one integer — the number of days Mister B needed to finish the book.
Demo Input:
['5 5 10 5 4\n', '12 4 12 4 1\n', '15 1 100 0 0\n']
Demo Output:
['1\n', '3\n', '15\n']
Note:
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.
|
```python
c, v0, v1, a, l = map(int, input().split())
speed = v0
pages = v0
days = 1
while(pages < c):
if speed + a <= v1:
speed += a
pages = pages + speed - l
days += 1
print(days)
```
| 0
|
|
959
|
A
|
Mahmoud and Ehab and the even-odd game
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
|
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
|
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
|
[
"1\n",
"2\n"
] |
[
"Ehab",
"Mahmoud"
] |
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
| 500
|
[
{
"input": "1",
"output": "Ehab"
},
{
"input": "2",
"output": "Mahmoud"
},
{
"input": "10000",
"output": "Mahmoud"
},
{
"input": "33333",
"output": "Ehab"
},
{
"input": "5",
"output": "Ehab"
},
{
"input": "1000000000",
"output": "Mahmoud"
},
{
"input": "999999999",
"output": "Ehab"
},
{
"input": "123123123",
"output": "Ehab"
},
{
"input": "22222221",
"output": "Ehab"
},
{
"input": "22222220",
"output": "Mahmoud"
},
{
"input": "3",
"output": "Ehab"
},
{
"input": "4",
"output": "Mahmoud"
},
{
"input": "6",
"output": "Mahmoud"
},
{
"input": "7",
"output": "Ehab"
},
{
"input": "8",
"output": "Mahmoud"
},
{
"input": "9",
"output": "Ehab"
},
{
"input": "10",
"output": "Mahmoud"
},
{
"input": "536870912",
"output": "Mahmoud"
},
{
"input": "536870913",
"output": "Ehab"
},
{
"input": "536870911",
"output": "Ehab"
}
] | 1,618,676,827
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 61
| 0
|
n=int(input())
print("Mahmoud") if n%2 == 0 else print("Ehab")
|
Title: Mahmoud and Ehab and the even-odd game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
Output Specification:
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Demo Input:
['1\n', '2\n']
Demo Output:
['Ehab', 'Mahmoud']
Note:
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
|
```python
n=int(input())
print("Mahmoud") if n%2 == 0 else print("Ehab")
```
| 3
|
|
302
|
B
|
Eugeny and Play List
|
PROGRAMMING
| 1,200
|
[
"binary search",
"implementation",
"two pointers"
] | null | null |
Eugeny loves listening to music. He has *n* songs in his play list. We know that song number *i* has the duration of *t**i* minutes. Eugeny listens to each song, perhaps more than once. He listens to song number *i* *c**i* times. Eugeny's play list is organized as follows: first song number 1 plays *c*1 times, then song number 2 plays *c*2 times, ..., in the end the song number *n* plays *c**n* times.
Eugeny took a piece of paper and wrote out *m* moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment *x* means that Eugeny wants to know which song was playing during the *x*-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
|
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The next *n* lines contain pairs of integers. The *i*-th line contains integers *c**i*,<=*t**i* (1<=≤<=*c**i*,<=*t**i*<=≤<=109) — the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 .
The next line contains *m* positive integers *v*1,<=*v*2,<=...,<=*v**m*, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time *v**i*, when the music doesn't play any longer. It is guaranteed that *v**i*<=<<=*v**i*<=+<=1 (*i*<=<<=*m*).
The moment of time *v**i* means that Eugeny wants to know which song was playing during the *v**i*-th munite from the start of listening to the playlist.
|
Print *m* integers — the *i*-th number must equal the number of the song that was playing during the *v**i*-th minute after Eugeny started listening to the play list.
|
[
"1 2\n2 8\n1 16\n",
"4 9\n1 2\n2 1\n1 1\n2 2\n1 2 3 4 5 6 7 8 9\n"
] |
[
"1\n1\n",
"1\n1\n2\n2\n3\n4\n4\n4\n4\n"
] |
none
| 1,000
|
[
{
"input": "1 2\n2 8\n1 16",
"output": "1\n1"
},
{
"input": "4 9\n1 2\n2 1\n1 1\n2 2\n1 2 3 4 5 6 7 8 9",
"output": "1\n1\n2\n2\n3\n4\n4\n4\n4"
},
{
"input": "3 3\n2 8\n5 1\n10 5\n13 16 62",
"output": "1\n1\n3"
},
{
"input": "4 4\n2 8\n2 2\n6 3\n8 7\n13 23 29 85",
"output": "1\n3\n3\n4"
},
{
"input": "5 5\n9 6\n8 7\n2 9\n10 3\n8 10\n69 95 146 162 177",
"output": "2\n2\n4\n5\n5"
},
{
"input": "6 6\n4 9\n8 5\n3 8\n8 10\n4 2\n10 9\n15 45 97 197 231 265",
"output": "1\n2\n3\n6\n6\n6"
},
{
"input": "7 7\n1 10\n1 1\n7 2\n4 9\n10 4\n5 5\n7 1\n48 71 86 87 110 113 127",
"output": "4\n5\n5\n5\n6\n6\n7"
},
{
"input": "8 8\n4 6\n10 9\n5 1\n8 7\n4 7\n2 6\n5 3\n1 10\n21 91 93 142 145 157 181 206",
"output": "1\n2\n2\n4\n4\n4\n5\n6"
},
{
"input": "9 9\n2 5\n7 1\n8 2\n8 8\n8 8\n4 4\n6 10\n10 9\n2 9\n1 10 36 48 76 151 229 276 310",
"output": "1\n1\n4\n4\n4\n5\n7\n8\n8"
},
{
"input": "10 10\n3 1\n2 7\n5 1\n7 2\n9 10\n9 5\n2 5\n4 10\n9 9\n1 9\n26 34 37 141 146 201 239 245 296 299",
"output": "4\n4\n5\n6\n6\n8\n9\n9\n9\n9"
},
{
"input": "2 3\n1 500000000\n1 500000000\n499999999 500000000 500000001",
"output": "1\n1\n2"
},
{
"input": "2 3\n500000000 1\n1 500000000\n499999999 500000000 500000001",
"output": "1\n1\n2"
},
{
"input": "2 3\n500000000 1\n500000000 1\n499999999 500000000 500000001",
"output": "1\n1\n2"
},
{
"input": "3 1\n2 50\n1 50\n1 50\n160",
"output": "3"
}
] | 1,635,927,339
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 11
| 2,000
| 15,052,800
|
def songs(moments, songs):
answer = [0 for _ in range(len(moments))]
ptr = 0
so_far = 0
for i in range(len(songs)):
j = songs[i].split()
ci = j[0]
ti = j[1]
total = int(ci)*int(ti)
for k in range(len(moments)):
if(int(moments[k]) in range(ptr, total+1+so_far)):
answer[k] = i+1
elif(int(moments[k]) >= total+1+so_far):
break
ptr = total+1+so_far
so_far+= total
for ans in answer:
print(ans)
if __name__ == '__main__':
n = (input().rstrip()).split()
m = int(n[1])
n = int(n[0])
output = []
for i in range(n):
first_input = input().rstrip()
output.append(first_input)
second_input = input().rstrip().split()
songs(second_input, output)
|
Title: Eugeny and Play List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eugeny loves listening to music. He has *n* songs in his play list. We know that song number *i* has the duration of *t**i* minutes. Eugeny listens to each song, perhaps more than once. He listens to song number *i* *c**i* times. Eugeny's play list is organized as follows: first song number 1 plays *c*1 times, then song number 2 plays *c*2 times, ..., in the end the song number *n* plays *c**n* times.
Eugeny took a piece of paper and wrote out *m* moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment *x* means that Eugeny wants to know which song was playing during the *x*-th minute of his listening to the play list.
Help Eugeny and calculate the required numbers of songs.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The next *n* lines contain pairs of integers. The *i*-th line contains integers *c**i*,<=*t**i* (1<=≤<=*c**i*,<=*t**i*<=≤<=109) — the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 109 .
The next line contains *m* positive integers *v*1,<=*v*2,<=...,<=*v**m*, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time *v**i*, when the music doesn't play any longer. It is guaranteed that *v**i*<=<<=*v**i*<=+<=1 (*i*<=<<=*m*).
The moment of time *v**i* means that Eugeny wants to know which song was playing during the *v**i*-th munite from the start of listening to the playlist.
Output Specification:
Print *m* integers — the *i*-th number must equal the number of the song that was playing during the *v**i*-th minute after Eugeny started listening to the play list.
Demo Input:
['1 2\n2 8\n1 16\n', '4 9\n1 2\n2 1\n1 1\n2 2\n1 2 3 4 5 6 7 8 9\n']
Demo Output:
['1\n1\n', '1\n1\n2\n2\n3\n4\n4\n4\n4\n']
Note:
none
|
```python
def songs(moments, songs):
answer = [0 for _ in range(len(moments))]
ptr = 0
so_far = 0
for i in range(len(songs)):
j = songs[i].split()
ci = j[0]
ti = j[1]
total = int(ci)*int(ti)
for k in range(len(moments)):
if(int(moments[k]) in range(ptr, total+1+so_far)):
answer[k] = i+1
elif(int(moments[k]) >= total+1+so_far):
break
ptr = total+1+so_far
so_far+= total
for ans in answer:
print(ans)
if __name__ == '__main__':
n = (input().rstrip()).split()
m = int(n[1])
n = int(n[0])
output = []
for i in range(n):
first_input = input().rstrip()
output.append(first_input)
second_input = input().rstrip().split()
songs(second_input, output)
```
| 0
|
|
981
|
B
|
Businessmen Problems
|
PROGRAMMING
| 1,000
|
[
"sortings"
] | null | null |
Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies.
In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies.
All elements are enumerated with integers. The ChemForces company has discovered $n$ distinct chemical elements with indices $a_1, a_2, \ldots, a_n$, and will get an income of $x_i$ Berland rubles if the $i$-th element from this list is in the set of this company.
The TopChemist company discovered $m$ distinct chemical elements with indices $b_1, b_2, \ldots, b_m$, and it will get an income of $y_j$ Berland rubles for including the $j$-th element from this list to its set.
In other words, the first company can present any subset of elements from $\{a_1, a_2, \ldots, a_n\}$ (possibly empty subset), the second company can present any subset of elements from $\{b_1, b_2, \ldots, b_m\}$ (possibly empty subset). There shouldn't be equal elements in the subsets.
Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible.
|
The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the number of elements discovered by ChemForces.
The $i$-th of the next $n$ lines contains two integers $a_i$ and $x_i$ ($1 \leq a_i \leq 10^9$, $1 \leq x_i \leq 10^9$) — the index of the $i$-th element and the income of its usage on the exhibition. It is guaranteed that all $a_i$ are distinct.
The next line contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of chemicals invented by TopChemist.
The $j$-th of the next $m$ lines contains two integers $b_j$ and $y_j$, ($1 \leq b_j \leq 10^9$, $1 \leq y_j \leq 10^9$) — the index of the $j$-th element and the income of its usage on the exhibition. It is guaranteed that all $b_j$ are distinct.
|
Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets.
|
[
"3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4\n",
"1\n1000000000 239\n3\n14 15\n92 65\n35 89\n"
] |
[
"24\n",
"408\n"
] |
In the first example ChemForces can choose the set ($3, 7$), while TopChemist can choose ($1, 2, 4$). This way the total income is $(10 + 2) + (4 + 4 + 4) = 24$.
In the second example ChemForces can choose the only element $10^9$, while TopChemist can choose ($14, 92, 35$). This way the total income is $(239) + (15 + 65 + 89) = 408$.
| 750
|
[
{
"input": "3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4",
"output": "24"
},
{
"input": "1\n1000000000 239\n3\n14 15\n92 65\n35 89",
"output": "408"
},
{
"input": "10\n598654597 488228616\n544064902 21923894\n329635457 980089248\n988262691 654502493\n967529230 543358150\n835120075 128123793\n809901567 613170206\n152157661 479980560\n859252956 318029856\n614959273 735298093\n10\n849172317 361325206\n341655282 740807372\n453949939 809030434\n813199219 765838311\n725032511 427644854\n631233366 115983627\n825067417 721190169\n463026366 953335259\n881501578 987288937\n500099903 427659916",
"output": "11272808994"
},
{
"input": "10\n598654597 488228616\n544064902 21923894\n329635457 980089248\n988262691 654502493\n967529230 543358150\n835120075 128123793\n809901567 613170206\n152157661 479980560\n859252956 318029856\n614959273 735298093\n10\n544064902 740807372\n859252956 765838311\n835120075 115983627\n825067417 463026366\n809901567 500099903\n614959273 322932832\n33955432 657532637\n816892960 927933923\n916770606 754371712\n967529230 566149252",
"output": "8955052582"
},
{
"input": "1\n259021863 682413763\n1\n554146726 693647287",
"output": "1376061050"
},
{
"input": "1\n259021863 682413763\n1\n554146726 425883728",
"output": "1108297491"
},
{
"input": "1\n1000000000 1\n1\n1000000000 1000000000",
"output": "1000000000"
},
{
"input": "2\n1 4\n2 4\n2\n1 4\n2 4",
"output": "8"
},
{
"input": "1\n14 15\n3\n35 89\n92 65\n1000000000 239",
"output": "408"
}
] | 1,680,710,794
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 483
| 13,926,400
|
n=int(input())
d1={}
d2={}
for i in range(n):
a1,x1=map(int,input().split())
d1[a1]=x1
m=int(input())
for i in range(m):
b1,y1=map(int,input().split())
d2[b1]=y1
p=sum(d1.values())+sum(d2.values())
for ai in d1.keys():
if ai in d2.keys():
p-=min(d1[ai],d2[ai])
print(p)
|
Title: Businessmen Problems
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies.
In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies.
All elements are enumerated with integers. The ChemForces company has discovered $n$ distinct chemical elements with indices $a_1, a_2, \ldots, a_n$, and will get an income of $x_i$ Berland rubles if the $i$-th element from this list is in the set of this company.
The TopChemist company discovered $m$ distinct chemical elements with indices $b_1, b_2, \ldots, b_m$, and it will get an income of $y_j$ Berland rubles for including the $j$-th element from this list to its set.
In other words, the first company can present any subset of elements from $\{a_1, a_2, \ldots, a_n\}$ (possibly empty subset), the second company can present any subset of elements from $\{b_1, b_2, \ldots, b_m\}$ (possibly empty subset). There shouldn't be equal elements in the subsets.
Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible.
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the number of elements discovered by ChemForces.
The $i$-th of the next $n$ lines contains two integers $a_i$ and $x_i$ ($1 \leq a_i \leq 10^9$, $1 \leq x_i \leq 10^9$) — the index of the $i$-th element and the income of its usage on the exhibition. It is guaranteed that all $a_i$ are distinct.
The next line contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of chemicals invented by TopChemist.
The $j$-th of the next $m$ lines contains two integers $b_j$ and $y_j$, ($1 \leq b_j \leq 10^9$, $1 \leq y_j \leq 10^9$) — the index of the $j$-th element and the income of its usage on the exhibition. It is guaranteed that all $b_j$ are distinct.
Output Specification:
Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets.
Demo Input:
['3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4\n', '1\n1000000000 239\n3\n14 15\n92 65\n35 89\n']
Demo Output:
['24\n', '408\n']
Note:
In the first example ChemForces can choose the set ($3, 7$), while TopChemist can choose ($1, 2, 4$). This way the total income is $(10 + 2) + (4 + 4 + 4) = 24$.
In the second example ChemForces can choose the only element $10^9$, while TopChemist can choose ($14, 92, 35$). This way the total income is $(239) + (15 + 65 + 89) = 408$.
|
```python
n=int(input())
d1={}
d2={}
for i in range(n):
a1,x1=map(int,input().split())
d1[a1]=x1
m=int(input())
for i in range(m):
b1,y1=map(int,input().split())
d2[b1]=y1
p=sum(d1.values())+sum(d2.values())
for ai in d1.keys():
if ai in d2.keys():
p-=min(d1[ai],d2[ai])
print(p)
```
| 3
|
|
450
|
A
|
Jzzhu and Children
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
|
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
Output a single integer, representing the number of the last child.
|
[
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] |
[
"4\n",
"6\n"
] |
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home.
| 500
|
[
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "100"
},
{
"input": "9 3\n9 5 2 3 7 1 8 4 6",
"output": "7"
},
{
"input": "20 10\n58 4 32 10 73 7 30 39 47 6 59 21 24 66 79 79 46 13 29 58",
"output": "16"
},
{
"input": "50 5\n89 56 3 2 40 37 56 52 83 59 43 83 43 59 29 74 22 58 53 41 53 67 78 30 57 32 58 29 95 46 45 85 60 49 41 82 8 71 52 40 45 26 6 71 84 91 4 93 40 54",
"output": "48"
},
{
"input": "50 1\n4 3 9 7 6 8 3 7 10 9 8 8 10 2 9 3 2 4 4 10 4 6 8 10 9 9 4 2 8 9 4 4 9 5 1 5 2 4 4 9 10 2 5 10 7 2 8 6 8 1",
"output": "44"
},
{
"input": "50 5\n3 9 10 8 3 3 4 6 8 2 9 9 3 1 2 10 6 8 7 2 7 4 2 7 5 10 2 2 2 5 10 5 6 6 8 7 10 4 3 2 10 8 6 6 8 6 4 4 1 3",
"output": "46"
},
{
"input": "50 2\n56 69 72 15 95 92 51 1 74 87 100 29 46 54 18 81 84 72 84 83 20 63 71 27 45 74 50 89 48 8 21 15 47 3 39 73 80 84 6 99 17 25 56 3 74 64 71 39 89 78",
"output": "40"
},
{
"input": "50 3\n31 39 64 16 86 3 1 9 25 54 98 42 20 3 49 41 73 37 55 62 33 77 64 22 33 82 26 13 10 13 7 40 48 18 46 79 94 72 19 12 11 61 16 37 10 49 14 94 48 69",
"output": "11"
},
{
"input": "50 100\n67 67 61 68 42 29 70 77 12 61 71 27 4 73 87 52 59 38 93 90 31 27 87 47 26 57 76 6 28 72 81 68 50 84 69 79 39 93 52 6 88 12 46 13 90 68 71 38 90 95",
"output": "50"
},
{
"input": "100 3\n4 14 20 11 19 11 14 20 5 7 6 12 11 17 5 11 7 6 2 10 13 5 12 8 5 17 20 18 7 19 11 7 7 20 20 8 10 17 17 19 20 5 15 16 19 7 11 16 4 17 2 10 1 20 20 16 19 9 9 11 5 7 12 9 9 6 20 18 13 19 8 4 8 1 2 4 10 11 15 14 1 7 17 12 13 19 12 2 3 14 15 15 5 17 14 12 17 14 16 9",
"output": "86"
},
{
"input": "100 5\n16 8 14 16 12 11 17 19 19 2 8 9 5 6 19 9 11 18 6 9 14 16 14 18 17 17 17 5 15 20 19 7 7 10 10 5 14 20 5 19 11 16 16 19 17 9 7 12 14 10 2 11 14 5 20 8 10 11 19 2 14 14 19 17 5 10 8 8 4 2 1 10 20 12 14 11 7 6 6 15 1 5 9 15 3 17 16 17 5 14 11 9 16 15 1 11 10 6 15 7",
"output": "93"
},
{
"input": "100 1\n58 94 18 50 17 14 96 62 83 80 75 5 9 22 25 41 3 96 74 45 66 37 2 37 13 85 68 54 77 11 85 19 25 21 52 59 90 61 72 89 82 22 10 16 3 68 61 29 55 76 28 85 65 76 27 3 14 10 56 37 86 18 35 38 56 68 23 88 33 38 52 87 55 83 94 34 100 41 83 56 91 77 32 74 97 13 67 31 57 81 53 39 5 88 46 1 79 4 49 42",
"output": "77"
},
{
"input": "100 2\n1 51 76 62 34 93 90 43 57 59 52 78 3 48 11 60 57 48 5 54 28 81 87 23 44 77 67 61 14 73 29 53 21 89 67 41 47 9 63 37 1 71 40 85 4 14 77 40 78 75 89 74 4 70 32 65 81 95 49 90 72 41 76 55 69 83 73 84 85 93 46 6 74 90 62 37 97 7 7 37 83 30 37 88 34 16 11 59 85 19 57 63 85 20 63 97 97 65 61 48",
"output": "97"
},
{
"input": "100 3\n30 83 14 55 61 66 34 98 90 62 89 74 45 93 33 31 75 35 82 100 63 69 48 18 99 2 36 71 14 30 70 76 96 85 97 90 49 36 6 76 37 94 70 3 63 73 75 48 39 29 13 2 46 26 9 56 1 18 54 53 85 34 2 12 1 93 75 67 77 77 14 26 33 25 55 9 57 70 75 6 87 66 18 3 41 69 73 24 49 2 20 72 39 58 91 54 74 56 66 78",
"output": "20"
},
{
"input": "100 4\n69 92 76 3 32 50 15 38 21 22 14 3 67 41 95 12 10 62 83 52 78 1 18 58 94 35 62 71 58 75 13 73 60 34 50 97 50 70 19 96 53 10 100 26 20 39 62 59 88 26 24 83 70 68 66 8 6 38 16 93 2 91 81 89 78 74 21 8 31 56 28 53 77 5 81 5 94 42 77 75 92 15 59 36 61 18 55 45 69 68 81 51 12 42 85 74 98 31 17 41",
"output": "97"
},
{
"input": "100 5\n2 72 10 60 6 50 72 34 97 77 35 43 80 64 40 53 46 6 90 22 29 70 26 68 52 19 72 88 83 18 55 32 99 81 11 21 39 42 41 63 60 97 30 23 55 78 89 35 24 50 99 52 27 76 24 8 20 27 51 37 17 82 69 18 46 19 26 77 52 83 76 65 43 66 84 84 13 30 66 88 84 23 37 1 17 26 11 50 73 56 54 37 40 29 35 8 1 39 50 82",
"output": "51"
},
{
"input": "100 7\n6 73 7 54 92 33 66 65 80 47 2 53 28 59 61 16 54 89 37 48 77 40 49 59 27 52 17 22 78 80 81 80 8 93 50 7 87 57 29 16 89 55 20 7 51 54 30 98 44 96 27 70 1 1 32 61 22 92 84 98 31 89 91 90 28 56 49 25 86 49 55 16 19 1 18 8 88 47 16 18 73 86 2 96 16 91 74 49 38 98 94 25 34 85 29 27 99 31 31 58",
"output": "97"
},
{
"input": "100 9\n36 4 45 16 19 6 10 87 44 82 71 49 70 35 83 19 40 76 45 94 44 96 10 54 82 77 86 63 11 37 21 3 15 89 80 88 89 16 72 23 25 9 51 25 10 45 96 5 6 18 51 31 42 57 41 51 42 15 89 61 45 82 16 48 61 67 19 40 9 33 90 36 78 36 79 79 16 10 83 87 9 22 84 12 23 76 36 14 2 81 56 33 56 23 57 84 76 55 35 88",
"output": "47"
},
{
"input": "100 10\n75 81 39 64 90 58 92 28 75 9 96 78 92 83 77 68 76 71 14 46 58 60 80 25 78 11 13 63 22 82 65 68 47 6 33 63 90 50 85 43 73 94 80 48 67 11 83 17 22 15 94 80 66 99 66 4 46 35 52 1 62 39 96 57 37 47 97 49 64 12 36 63 90 16 4 75 85 82 85 56 13 4 92 45 44 93 17 35 22 46 18 44 29 7 52 4 100 98 87 51",
"output": "98"
},
{
"input": "100 20\n21 19 61 70 54 97 98 14 61 72 25 94 24 56 55 25 12 80 76 11 35 17 80 26 11 94 52 47 84 61 10 2 74 25 10 21 2 79 55 50 30 75 10 64 44 5 60 96 52 16 74 41 20 77 20 44 8 86 74 36 49 61 99 13 54 64 19 99 50 43 12 73 48 48 83 55 72 73 63 81 30 27 95 9 97 82 24 3 89 90 33 14 47 88 22 78 12 75 58 67",
"output": "94"
},
{
"input": "100 30\n56 79 59 23 11 23 67 82 81 80 99 79 8 58 93 36 98 81 46 39 34 67 3 50 4 68 70 71 2 21 52 30 75 23 33 21 16 100 56 43 8 27 40 8 56 24 17 40 94 10 67 49 61 36 95 87 17 41 7 94 33 19 17 50 26 11 94 54 38 46 77 9 53 35 98 42 50 20 43 6 78 6 38 24 100 45 43 16 1 50 16 46 14 91 95 88 10 1 50 19",
"output": "95"
},
{
"input": "100 40\n86 11 97 17 38 95 11 5 13 83 67 75 50 2 46 39 84 68 22 85 70 23 64 46 59 93 39 80 35 78 93 21 83 19 64 1 49 59 99 83 44 81 70 58 15 82 83 47 55 65 91 10 2 92 4 77 37 32 12 57 78 11 42 8 59 21 96 69 61 30 44 29 12 70 91 14 10 83 11 75 14 10 19 39 8 98 5 81 66 66 79 55 36 29 22 45 19 24 55 49",
"output": "88"
},
{
"input": "100 50\n22 39 95 69 94 53 80 73 33 90 40 60 2 4 84 50 70 38 92 12 36 74 87 70 51 36 57 5 54 6 35 81 52 17 55 100 95 81 32 76 21 1 100 1 95 1 40 91 98 59 84 19 11 51 79 19 47 86 45 15 62 2 59 77 31 68 71 92 17 33 10 33 85 57 5 2 88 97 91 99 63 20 63 54 79 93 24 62 46 27 30 87 3 64 95 88 16 50 79 1",
"output": "99"
},
{
"input": "100 70\n61 48 89 17 97 6 93 13 64 50 66 88 24 52 46 99 6 65 93 64 82 37 57 41 47 1 84 5 97 83 79 46 16 35 40 7 64 15 44 96 37 17 30 92 51 67 26 3 14 56 27 68 66 93 36 39 51 6 40 55 79 26 71 54 8 48 18 2 71 12 55 60 29 37 31 97 26 37 25 68 67 70 3 87 100 41 5 82 65 92 24 66 76 48 89 8 40 93 31 95",
"output": "100"
},
{
"input": "100 90\n87 32 30 15 10 52 93 63 84 1 82 41 27 51 75 32 42 94 39 53 70 13 4 22 99 35 44 38 5 23 18 100 61 80 9 12 42 93 9 77 3 7 60 95 66 78 95 42 69 8 1 88 93 66 96 20 76 63 15 36 92 52 2 72 36 57 48 63 29 20 74 88 49 47 81 61 94 74 70 93 47 3 19 52 59 41 5 40 22 3 76 97 91 37 95 88 91 99 76 15",
"output": "98"
},
{
"input": "100 100\n79 75 7 28 6 96 38 35 57 95 41 74 24 96 32 78 81 13 63 84 24 95 3 23 66 1 60 6 96 49 41 5 14 18 31 97 66 19 49 89 49 70 51 28 20 99 18 1 28 77 24 46 69 21 40 32 31 66 28 6 66 97 9 16 70 90 91 30 34 82 93 41 65 11 39 52 1 88 63 43 80 50 60 49 28 56 18 76 24 57 74 1 28 99 36 35 79 54 18 16",
"output": "100"
},
{
"input": "1 3\n5",
"output": "1"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 3\n4 2",
"output": "1"
},
{
"input": "2 5\n99 97",
"output": "2"
},
{
"input": "3 4\n7 5 2",
"output": "2"
},
{
"input": "3 50\n47 86 51",
"output": "3"
},
{
"input": "5 100\n82 100 85 1 37",
"output": "5"
},
{
"input": "5 20\n40 39 21 5 20",
"output": "3"
},
{
"input": "1 27\n81",
"output": "1"
},
{
"input": "20 13\n7 8 29 83 74 28 93 85 7 8 3 9 8 70 49 50 39 41 57 1",
"output": "7"
},
{
"input": "2 1\n100 2",
"output": "1"
},
{
"input": "2 2\n6 4",
"output": "1"
},
{
"input": "5 2\n6 4 4 1 1",
"output": "1"
},
{
"input": "1 4\n3",
"output": "1"
},
{
"input": "3 2\n1 5 3",
"output": "2"
},
{
"input": "3 1\n3 2 2",
"output": "1"
},
{
"input": "3 1\n2 3 2",
"output": "2"
},
{
"input": "5 1\n5 1 1 1 1",
"output": "1"
},
{
"input": "2 3\n7 4",
"output": "1"
}
] | 1,678,535,862
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
# t = int(input())
t = 1
def find_num_2(x: int):
count = 0
while (x%2 == 0 and x!=1):
count += 1
x = x//2
while (x%3 == 0 and x!=1):
count += 1
x = x//3
if x==1:
return count
elif x%2 or x%3:
return -1
def cp():
# code here
x,y = tuple(map(int,input().split()))
index = 0
mylst = list(map(int,input().split()))
max_value = max(mylst)
for i in range(-1,-len(mylst)-1,-1):
if mylst[i] == max_value:
print(x + i + 1)
break
for i in range(t):
cp()
|
Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home.
|
```python
# t = int(input())
t = 1
def find_num_2(x: int):
count = 0
while (x%2 == 0 and x!=1):
count += 1
x = x//2
while (x%3 == 0 and x!=1):
count += 1
x = x//3
if x==1:
return count
elif x%2 or x%3:
return -1
def cp():
# code here
x,y = tuple(map(int,input().split()))
index = 0
mylst = list(map(int,input().split()))
max_value = max(mylst)
for i in range(-1,-len(mylst)-1,-1):
if mylst[i] == max_value:
print(x + i + 1)
break
for i in range(t):
cp()
```
| 0
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,676,211,028
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 31
| 0
|
x = input()
r = {"h", "l", "o"}
s = set(x)
#s
if len(r.union(s)) <= 4:
print("NO")
else:
print("YES")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
x = input()
r = {"h", "l", "o"}
s = set(x)
#s
if len(r.union(s)) <= 4:
print("NO")
else:
print("YES")
```
| 0
|
139
|
A
|
Petr and Book
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
|
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
|
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
|
[
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] |
[
"6\n",
"1\n"
] |
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
| 500
|
[
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,667,658,749
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 92
| 0
|
n = int(input())
day = list(map(int, input().split()))
cnt = 0
p = day[0]
while (p < n):
cnt += 1
cnt %= 7
p += day[cnt]
print(cnt + 1)
|
Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
|
```python
n = int(input())
day = list(map(int, input().split()))
cnt = 0
p = day[0]
while (p < n):
cnt += 1
cnt %= 7
p += day[cnt]
print(cnt + 1)
```
| 3
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,685,598,840
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
N = int(input())
i =1
Capacity = 0
for i in range(1,N+1):
a,b = [int(a) for a in input().split()]
if(a<b):
Capacity = Capacity + (b-a)
print(Capacity)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
N = int(input())
i =1
Capacity = 0
for i in range(1,N+1):
a,b = [int(a) for a in input().split()]
if(a<b):
Capacity = Capacity + (b-a)
print(Capacity)
```
| 0
|
|
749
|
A
|
Bachgold Problem
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"math",
"number theory"
] | null | null |
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
|
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
|
[
"5\n",
"6\n"
] |
[
"2\n2 3\n",
"3\n2 2 2\n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "100000",
"output": "50000\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7",
"output": "3\n2 2 3"
},
{
"input": "4",
"output": "2\n2 2"
},
{
"input": "8",
"output": "4\n2 2 2 2"
},
{
"input": "9",
"output": "4\n2 2 2 3"
},
{
"input": "99995",
"output": "49997\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99996",
"output": "49998\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "10",
"output": "5\n2 2 2 2 2"
},
{
"input": "11",
"output": "5\n2 2 2 2 3"
},
{
"input": "99997",
"output": "49998\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99998",
"output": "49999\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "12",
"output": "6\n2 2 2 2 2 2"
},
{
"input": "13",
"output": "6\n2 2 2 2 2 3"
},
{
"input": "99993",
"output": "49996\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99994",
"output": "49997\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "14",
"output": "7\n2 2 2 2 2 2 2"
},
{
"input": "15",
"output": "7\n2 2 2 2 2 2 3"
},
{
"input": "53",
"output": "26\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "57",
"output": "28\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "61",
"output": "30\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "774",
"output": "387\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "202",
"output": "101\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2"
},
{
"input": "530",
"output": "265\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7166",
"output": "3583\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "9294",
"output": "4647\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "2422",
"output": "1211\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "15326",
"output": "7663\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "11454",
"output": "5727\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "14878",
"output": "7439\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "90672",
"output": "45336\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99544",
"output": "49772\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "90472",
"output": "45236\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "23",
"output": "11\n2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "93",
"output": "46\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "19",
"output": "9\n2 2 2 2 2 2 2 2 3"
},
{
"input": "11110",
"output": "5555\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
}
] | 1,642,173,013
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 4
| 77
| 102,400
|
x = int(input())
a = x//2
b = x - 3
if x % 2 == 0:
print(a)
print("2 " * a)
else:
y = b//2
print(y + 1)
print("2" * y, "3")
|
Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none
|
```python
x = int(input())
a = x//2
b = x - 3
if x % 2 == 0:
print(a)
print("2 " * a)
else:
y = b//2
print(y + 1)
print("2" * y, "3")
```
| 0
|
|
854
|
B
|
Maxim Buys an Apartment
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"math"
] | null | null |
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has *n* apartments that are numbered from 1 to *n* and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly *k* already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
|
The only line of the input contains two integers: *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=*n*).
|
Print the minimum possible and the maximum possible number of apartments good for Maxim.
|
[
"6 3\n"
] |
[
"1 3\n"
] |
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
| 1,000
|
[
{
"input": "6 3",
"output": "1 3"
},
{
"input": "10 1",
"output": "1 2"
},
{
"input": "10 9",
"output": "1 1"
},
{
"input": "8 0",
"output": "0 0"
},
{
"input": "8 8",
"output": "0 0"
},
{
"input": "966871928 890926970",
"output": "1 75944958"
},
{
"input": "20 2",
"output": "1 4"
},
{
"input": "1 0",
"output": "0 0"
},
{
"input": "1 1",
"output": "0 0"
},
{
"input": "2 0",
"output": "0 0"
},
{
"input": "2 1",
"output": "1 1"
},
{
"input": "2 2",
"output": "0 0"
},
{
"input": "7 2",
"output": "1 4"
},
{
"input": "8 3",
"output": "1 5"
},
{
"input": "9 4",
"output": "1 5"
},
{
"input": "10 3",
"output": "1 6"
},
{
"input": "10 4",
"output": "1 6"
},
{
"input": "10 5",
"output": "1 5"
},
{
"input": "1000 1000",
"output": "0 0"
},
{
"input": "1000 333",
"output": "1 666"
},
{
"input": "1000 334",
"output": "1 666"
},
{
"input": "999 333",
"output": "1 666"
},
{
"input": "999 334",
"output": "1 665"
},
{
"input": "998 332",
"output": "1 664"
},
{
"input": "998 333",
"output": "1 665"
},
{
"input": "89 4",
"output": "1 8"
},
{
"input": "66 50",
"output": "1 16"
},
{
"input": "88 15",
"output": "1 30"
},
{
"input": "95 43",
"output": "1 52"
},
{
"input": "900 344",
"output": "1 556"
},
{
"input": "777 113",
"output": "1 226"
},
{
"input": "964 42",
"output": "1 84"
},
{
"input": "982 867",
"output": "1 115"
},
{
"input": "1000000000 0",
"output": "0 0"
},
{
"input": "1000000000 1000000000",
"output": "0 0"
},
{
"input": "1000000000 333333333",
"output": "1 666666666"
},
{
"input": "1000000000 333333334",
"output": "1 666666666"
},
{
"input": "999999999 333333333",
"output": "1 666666666"
},
{
"input": "999999999 333333334",
"output": "1 666666665"
},
{
"input": "999999998 333333332",
"output": "1 666666664"
},
{
"input": "999999998 333333333",
"output": "1 666666665"
},
{
"input": "78602604 42160832",
"output": "1 36441772"
},
{
"input": "35679021 9137902",
"output": "1 18275804"
},
{
"input": "41949373 13173511",
"output": "1 26347022"
},
{
"input": "77855558 49163875",
"output": "1 28691683"
},
{
"input": "87187123 2851901",
"output": "1 5703802"
},
{
"input": "66849627 25004217",
"output": "1 41845410"
},
{
"input": "873046672 517064947",
"output": "1 355981725"
},
{
"input": "639857373 1393427",
"output": "1 2786854"
},
{
"input": "637563683 69636269",
"output": "1 139272538"
},
{
"input": "911669737 141068293",
"output": "1 282136586"
},
{
"input": "547575919 313272818",
"output": "1 234303101"
},
{
"input": "955020006 297895809",
"output": "1 595791618"
},
{
"input": "10 4",
"output": "1 6"
},
{
"input": "11 3",
"output": "1 6"
},
{
"input": "10 3",
"output": "1 6"
},
{
"input": "4 1",
"output": "1 2"
},
{
"input": "9 3",
"output": "1 6"
},
{
"input": "7 2",
"output": "1 4"
},
{
"input": "7 3",
"output": "1 4"
},
{
"input": "12 5",
"output": "1 7"
},
{
"input": "8 3",
"output": "1 5"
},
{
"input": "1000 8",
"output": "1 16"
}
] | 1,519,670,763
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 19,968,000
|
n, k = [int(f) for f in input().split(' ') if f]
blocks = n // 3
if blocks > k:
print(2*k)
else:
print([0, 1][k+1 <= n], max(2 * blocks - (k - blocks), 0))
|
Title: Maxim Buys an Apartment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has *n* apartments that are numbered from 1 to *n* and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.
Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly *k* already inhabited apartments, but he doesn't know their indices yet.
Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.
Input Specification:
The only line of the input contains two integers: *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=*n*).
Output Specification:
Print the minimum possible and the maximum possible number of apartments good for Maxim.
Demo Input:
['6 3\n']
Demo Output:
['1 3\n']
Note:
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
|
```python
n, k = [int(f) for f in input().split(' ') if f]
blocks = n // 3
if blocks > k:
print(2*k)
else:
print([0, 1][k+1 <= n], max(2 * blocks - (k - blocks), 0))
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,584,946,743
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 124
| 0
|
p=input()
q=input()
for i in range(len(p)):
m=int(p[i])^int(q[i])
print(m,end="")
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
p=input()
q=input()
for i in range(len(p)):
m=int(p[i])^int(q[i])
print(m,end="")
```
| 3.969
|
368
|
B
|
Sereja and Suffixes
|
PROGRAMMING
| 1,100
|
[
"data structures",
"dp"
] | null | null |
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
|
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
|
[
"10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n"
] |
[
"6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n"
] |
none
| 1,000
|
[
{
"input": "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "6\n6\n6\n6\n6\n5\n4\n3\n2\n1"
},
{
"input": "8 3\n8 6 4 3 4 2 4 8\n6\n4\n2",
"output": "3\n4\n5"
},
{
"input": "7 10\n1 3 8 6 2 2 7\n4\n2\n6\n3\n4\n4\n6\n2\n7\n4",
"output": "3\n5\n2\n4\n3\n3\n2\n5\n1\n3"
},
{
"input": "10 2\n2 6 5 7 2 2 3 2 4 8\n1\n2",
"output": "7\n7"
},
{
"input": "7 1\n68346 10956 76708 23018 84063 34833 80407\n1",
"output": "7"
},
{
"input": "2 2\n8 4\n1\n1",
"output": "2\n2"
},
{
"input": "1 5\n5\n1\n1\n1\n1\n1",
"output": "1\n1\n1\n1\n1"
},
{
"input": "4 7\n3 1 4 2\n4\n1\n2\n3\n2\n4\n4",
"output": "1\n4\n3\n2\n3\n1\n1"
},
{
"input": "4 3\n9 1 7 1\n1\n4\n2",
"output": "3\n1\n2"
},
{
"input": "8 3\n9280 6676 2720 6172 8329 10413 3975 1394\n5\n6\n7",
"output": "4\n3\n2"
},
{
"input": "1 1\n1\n1",
"output": "1"
}
] | 1,668,499,232
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,000
| 5,120,000
|
n,m = map(int,input().split())
a = list(map(int,input().split()))
for i in range(m):
l = int(input())
aset = set(a[j] for j in range(l-1,n))
print(len(aset))
|
Title: Sereja and Suffixes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
Output Specification:
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
Demo Input:
['10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n']
Demo Output:
['6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n']
Note:
none
|
```python
n,m = map(int,input().split())
a = list(map(int,input().split()))
for i in range(m):
l = int(input())
aset = set(a[j] for j in range(l-1,n))
print(len(aset))
```
| 0
|
|
742
|
A
|
Arpa’s hard exam and Mehrdad’s naive cheat
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math",
"number theory"
] | null | null |
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
|
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
|
Print single integer — the last digit of 1378*n*.
|
[
"1\n",
"2\n"
] |
[
"8",
"4"
] |
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4.
| 500
|
[
{
"input": "1",
"output": "8"
},
{
"input": "2",
"output": "4"
},
{
"input": "1000",
"output": "6"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "6"
},
{
"input": "1000000000",
"output": "6"
},
{
"input": "5",
"output": "8"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "2"
},
{
"input": "1378",
"output": "4"
},
{
"input": "13781378",
"output": "4"
},
{
"input": "51202278",
"output": "4"
},
{
"input": "999999998",
"output": "4"
},
{
"input": "999999997",
"output": "8"
},
{
"input": "12193721",
"output": "8"
},
{
"input": "0",
"output": "1"
},
{
"input": "989898989",
"output": "8"
},
{
"input": "7",
"output": "2"
},
{
"input": "8",
"output": "6"
},
{
"input": "9",
"output": "8"
},
{
"input": "10",
"output": "4"
},
{
"input": "11",
"output": "2"
},
{
"input": "12",
"output": "6"
},
{
"input": "13",
"output": "8"
},
{
"input": "14",
"output": "4"
},
{
"input": "15",
"output": "2"
},
{
"input": "16",
"output": "6"
},
{
"input": "999999996",
"output": "6"
},
{
"input": "999999995",
"output": "2"
},
{
"input": "999999994",
"output": "4"
}
] | 1,675,873,185
| 2,147,483,647
|
PyPy 3-64
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
print(pow(8, int(input()), 10))
|
Title: Arpa’s hard exam and Mehrdad’s naive cheat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
Input Specification:
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
Output Specification:
Print single integer — the last digit of 1378*n*.
Demo Input:
['1\n', '2\n']
Demo Output:
['8', '4']
Note:
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4.
|
```python
print(pow(8, int(input()), 10))
```
| -1
|
|
15
|
A
|
Cottage Village
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] |
A. Cottage Village
|
2
|
64
|
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
|
The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000).
|
Output the amount of possible positions of the new house.
|
[
"2 2\n0 4\n6 2\n",
"2 2\n0 4\n5 2\n",
"2 3\n0 4\n5 2\n"
] |
[
"4\n",
"3\n",
"2\n"
] |
It is possible for the *x*-coordinate of the new house to have non-integer value.
| 0
|
[
{
"input": "2 2\n0 4\n6 2",
"output": "4"
},
{
"input": "2 2\n0 4\n5 2",
"output": "3"
},
{
"input": "2 3\n0 4\n5 2",
"output": "2"
},
{
"input": "1 1\n1 1",
"output": "2"
},
{
"input": "1 2\n2 1",
"output": "2"
},
{
"input": "2 1\n2 1\n1 1",
"output": "2"
},
{
"input": "2 2\n0 4\n7 4",
"output": "4"
},
{
"input": "4 1\n-12 1\n-14 1\n4 1\n-11 1",
"output": "5"
},
{
"input": "6 15\n19 1\n2 3\n6 2\n-21 2\n-15 2\n23 1",
"output": "2"
},
{
"input": "10 21\n-61 6\n55 2\n-97 1\n37 1\n-39 1\n26 2\n21 1\n64 3\n-68 1\n-28 6",
"output": "6"
},
{
"input": "26 51\n783 54\n-850 6\n-997 59\n573 31\n-125 20\n472 52\n101 5\n-561 4\n625 35\n911 14\n-47 33\n677 55\n-410 54\n13 53\n173 31\n968 30\n-497 7\n832 42\n271 59\n-638 52\n-301 51\n378 36\n-813 7\n-206 22\n-737 37\n-911 9",
"output": "35"
},
{
"input": "14 101\n121 88\n-452 91\n635 28\n-162 59\n-872 26\n-996 8\n468 86\n742 63\n892 89\n-249 107\n300 51\n-753 17\n-620 31\n-13 34",
"output": "16"
},
{
"input": "3 501\n827 327\n-85 480\n-999 343",
"output": "6"
},
{
"input": "2 999\n-999 471\n530 588",
"output": "4"
},
{
"input": "22 54\n600 43\n806 19\n-269 43\n-384 78\n222 34\n392 10\n318 30\n488 73\n-756 49\n-662 22\n-568 50\n-486 16\n-470 2\n96 66\n864 16\n934 15\n697 43\n-154 30\n775 5\n-876 71\n-33 78\n-991 31",
"output": "30"
},
{
"input": "17 109\n52 7\n216 24\n-553 101\n543 39\n391 92\n-904 67\n95 34\n132 14\n730 103\n952 118\n-389 41\n-324 36\n-74 2\n-147 99\n-740 33\n233 1\n-995 3",
"output": "16"
},
{
"input": "4 512\n-997 354\n-568 216\n-234 221\n603 403",
"output": "4"
},
{
"input": "3 966\n988 5\n15 2\n-992 79",
"output": "6"
},
{
"input": "2 1000\n-995 201\n206 194",
"output": "4"
},
{
"input": "50 21\n-178 1\n49 1\n-98 1\n-220 1\n152 1\n-160 3\n17 2\n77 1\n-24 1\n214 2\n-154 2\n-141 1\n79 1\n206 1\n8 1\n-208 1\n36 1\n231 3\n-2 2\n-130 2\n-14 2\n34 1\n-187 2\n14 1\n-83 2\n-241 1\n149 2\n73 1\n-233 3\n-45 1\n197 1\n145 2\n-127 2\n-229 4\n-85 1\n-66 1\n-76 2\n104 1\n175 1\n70 1\n131 3\n-108 1\n-5 4\n140 1\n33 1\n248 3\n-36 3\n134 1\n-183 1\n56 2",
"output": "9"
},
{
"input": "50 1\n37 1\n-38 1\n7 1\n47 1\n-4 1\n24 1\n-32 1\n-23 1\n-3 1\n-19 1\n5 1\n-50 1\n11 1\n-11 1\n49 1\n-39 1\n0 1\n43 1\n-10 1\n6 1\n19 1\n1 1\n27 1\n29 1\n-47 1\n-40 1\n-46 1\n-26 1\n-42 1\n-37 1\n13 1\n-29 1\n-30 1\n3 1\n44 1\n10 1\n4 1\n-14 1\n-2 1\n34 1\n18 1\n-33 1\n-44 1\n9 1\n-36 1\n-7 1\n25 1\n22 1\n-20 1\n-41 1",
"output": "43"
},
{
"input": "50 1\n-967 7\n696 7\n-366 4\n557 1\n978 2\n800 4\n-161 2\n-773 2\n-248 2\n134 3\n869 6\n-932 2\n-262 14\n191 3\n669 2\n72 5\n0 1\n757 8\n859 2\n-131 8\n-169 3\n543 10\n-120 2\n-87 8\n-936 6\n-620 3\n-281 11\n684 3\n886 10\n497 4\n380 4\n833 1\n-727 6\n470 11\n584 9\n66 6\n-609 12\n-661 4\n-57 8\n628 7\n635 4\n-924 3\n-982 4\n-201 7\n-9 8\n-560 9\n712 7\n-330 8\n-191 1\n-892 7",
"output": "96"
},
{
"input": "1 1000\n0 1000",
"output": "2"
}
] | 1,643,914,467
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
n, t = [int(item) for item in input().split(' ')]
cont,ans = [],2
for i in range(n):
hcenter, hlen = [float(item) for item in input().split(' ')]
cont.append([hcenter - hlen / 2, hcenter + hlen / 2])
cont.sort(key=lambda arr: arr[0])
for i in range(n-1):
gap = cont[i+1][0] - cont[i][1]
if gap > t:
ans += 2
elif gap == t:
ans += 1
print(ans)
|
Title: Cottage Village
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input Specification:
The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000).
Output Specification:
Output the amount of possible positions of the new house.
Demo Input:
['2 2\n0 4\n6 2\n', '2 2\n0 4\n5 2\n', '2 3\n0 4\n5 2\n']
Demo Output:
['4\n', '3\n', '2\n']
Note:
It is possible for the *x*-coordinate of the new house to have non-integer value.
|
```python
n, t = [int(item) for item in input().split(' ')]
cont,ans = [],2
for i in range(n):
hcenter, hlen = [float(item) for item in input().split(' ')]
cont.append([hcenter - hlen / 2, hcenter + hlen / 2])
cont.sort(key=lambda arr: arr[0])
for i in range(n-1):
gap = cont[i+1][0] - cont[i][1]
if gap > t:
ans += 2
elif gap == t:
ans += 1
print(ans)
```
| 3.977
|
450
|
B
|
Jzzhu and Sequences
|
PROGRAMMING
| 1,300
|
[
"implementation",
"math"
] | null | null |
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
|
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
|
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
|
[
"2 3\n3\n",
"0 -1\n2\n"
] |
[
"1\n",
"1000000006\n"
] |
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
| 1,000
|
[
{
"input": "2 3\n3",
"output": "1"
},
{
"input": "0 -1\n2",
"output": "1000000006"
},
{
"input": "-9 -11\n12345",
"output": "1000000005"
},
{
"input": "0 0\n1000000000",
"output": "0"
},
{
"input": "-1000000000 1000000000\n2000000000",
"output": "1000000000"
},
{
"input": "-12345678 12345678\n1912345678",
"output": "12345678"
},
{
"input": "728374857 678374857\n1928374839",
"output": "950000007"
},
{
"input": "278374837 992837483\n1000000000",
"output": "721625170"
},
{
"input": "-693849384 502938493\n982838498",
"output": "502938493"
},
{
"input": "-783928374 983738273\n992837483",
"output": "16261734"
},
{
"input": "-872837483 -682738473\n999999999",
"output": "190099010"
},
{
"input": "-892837483 -998273847\n999283948",
"output": "892837483"
},
{
"input": "-283938494 738473848\n1999999999",
"output": "716061513"
},
{
"input": "-278374857 819283838\n1",
"output": "721625150"
},
{
"input": "-1000000000 123456789\n1",
"output": "7"
},
{
"input": "-529529529 -524524524\n2",
"output": "475475483"
},
{
"input": "1 2\n2000000000",
"output": "2"
},
{
"input": "-1 -2\n2000000000",
"output": "1000000005"
},
{
"input": "1 2\n1999999999",
"output": "1"
},
{
"input": "1 2\n1999999998",
"output": "1000000006"
},
{
"input": "1 2\n1999999997",
"output": "1000000005"
},
{
"input": "1 2\n1999999996",
"output": "1000000006"
},
{
"input": "69975122 366233206\n1189460676",
"output": "703741923"
},
{
"input": "812229413 904420051\n806905621",
"output": "812229413"
},
{
"input": "872099024 962697902\n1505821695",
"output": "90598878"
},
{
"input": "887387283 909670917\n754835014",
"output": "112612724"
},
{
"input": "37759824 131342932\n854621399",
"output": "868657075"
},
{
"input": "-246822123 800496170\n626323615",
"output": "753177884"
},
{
"input": "-861439463 974126967\n349411083",
"output": "835566423"
},
{
"input": "-69811049 258093841\n1412447",
"output": "741906166"
},
{
"input": "844509330 -887335829\n123329059",
"output": "844509330"
},
{
"input": "83712471 -876177148\n1213284777",
"output": "40110388"
},
{
"input": "598730524 -718984219\n1282749880",
"output": "401269483"
},
{
"input": "-474244697 -745885656\n1517883612",
"output": "271640959"
},
{
"input": "-502583588 -894906953\n1154189557",
"output": "497416419"
},
{
"input": "-636523651 -873305815\n154879215",
"output": "763217843"
},
{
"input": "721765550 594845720\n78862386",
"output": "126919830"
},
{
"input": "364141461 158854993\n1337196589",
"output": "364141461"
},
{
"input": "878985260 677031952\n394707801",
"output": "798046699"
},
{
"input": "439527072 -24854079\n1129147002",
"output": "464381151"
},
{
"input": "840435009 -612103127\n565968986",
"output": "387896880"
},
{
"input": "875035447 -826471373\n561914518",
"output": "124964560"
},
{
"input": "-342526698 305357084\n70776744",
"output": "352116225"
},
{
"input": "-903244186 899202229\n1527859274",
"output": "899202229"
},
{
"input": "-839482546 815166320\n1127472130",
"output": "839482546"
},
{
"input": "-976992569 -958313041\n1686580818",
"output": "981320479"
},
{
"input": "-497338894 -51069176\n737081851",
"output": "502661113"
},
{
"input": "-697962643 -143148799\n1287886520",
"output": "856851208"
},
{
"input": "-982572938 -482658433\n1259858332",
"output": "982572938"
},
{
"input": "123123 78817\n2000000000",
"output": "78817"
},
{
"input": "1000000000 -1000000000\n3",
"output": "14"
},
{
"input": "-1000000000 1000000000\n6",
"output": "14"
},
{
"input": "2 3\n6",
"output": "1000000006"
},
{
"input": "0 -1\n6",
"output": "1"
},
{
"input": "500000000 -1000000000\n600000003",
"output": "500000014"
},
{
"input": "-1000000000 1000000000\n3",
"output": "999999993"
},
{
"input": "1 3\n6",
"output": "1000000005"
},
{
"input": "1 2\n12",
"output": "1000000006"
},
{
"input": "7 -1000000000\n3",
"output": "0"
},
{
"input": "-999999997 999999997\n6",
"output": "20"
},
{
"input": "3 4\n6",
"output": "1000000006"
},
{
"input": "-1 2\n6",
"output": "1000000004"
},
{
"input": "2 3\n12",
"output": "1000000006"
},
{
"input": "4 18\n6",
"output": "999999993"
},
{
"input": "1 2\n6",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n6",
"output": "999999993"
},
{
"input": "999999999 -999999999\n3",
"output": "16"
},
{
"input": "-1 0\n1",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n9",
"output": "14"
},
{
"input": "999999999 -1000000000\n12",
"output": "999999992"
},
{
"input": "1000000000 -7\n3",
"output": "0"
},
{
"input": "-5 5\n6",
"output": "999999997"
},
{
"input": "5 9\n6",
"output": "1000000003"
},
{
"input": "-15 -10\n1",
"output": "999999992"
}
] | 1,653,403,233
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 74
| 77
| 0
|
from sys import stdin,stdout
input = lambda : stdin.readline().rstrip()
print = stdout.write
x, y = map(int, input().split())
n = int(input())
print(str((x,y,y-x,-x,-y,x-y,x,y,y-x)[(n-1)%6]%1000000007))
|
Title: Jzzhu and Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
Output Specification:
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
Demo Input:
['2 3\n3\n', '0 -1\n2\n']
Demo Output:
['1\n', '1000000006\n']
Note:
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
|
```python
from sys import stdin,stdout
input = lambda : stdin.readline().rstrip()
print = stdout.write
x, y = map(int, input().split())
n = int(input())
print(str((x,y,y-x,-x,-y,x-y,x,y,y-x)[(n-1)%6]%1000000007))
```
| 3
|
|
609
|
B
|
The Best Gift
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"implementation"
] | null | null |
Emily's birthday is next week and Jack has decided to buy a present for her. He knows she loves books so he goes to the local bookshop, where there are *n* books on sale from one of *m* genres.
In the bookshop, Jack decides to buy two books of different genres.
Based on the genre of books on sale in the shop, find the number of options available to Jack for choosing two books of different genres for Emily. Options are considered different if they differ in at least one book.
The books are given by indices of their genres. The genres are numbered from 1 to *m*.
|
The first line contains two positive integers *n* and *m* (2<=≤<=*n*<=≤<=2·105,<=2<=≤<=*m*<=≤<=10) — the number of books in the bookstore and the number of genres.
The second line contains a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (1<=≤<=*a**i*<=≤<=*m*) equals the genre of the *i*-th book.
It is guaranteed that for each genre there is at least one book of that genre.
|
Print the only integer — the number of ways in which Jack can choose books.
It is guaranteed that the answer doesn't exceed the value 2·109.
|
[
"4 3\n2 1 3 1\n",
"7 4\n4 2 3 1 2 4 3\n"
] |
[
"5\n",
"18\n"
] |
The answer to the first test sample equals 5 as Sasha can choose:
1. the first and second books, 1. the first and third books, 1. the first and fourth books, 1. the second and third books, 1. the third and fourth books.
| 0
|
[
{
"input": "4 3\n2 1 3 1",
"output": "5"
},
{
"input": "7 4\n4 2 3 1 2 4 3",
"output": "18"
},
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "3 2\n1 2 2",
"output": "2"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 10",
"output": "45"
},
{
"input": "9 2\n1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "12 3\n1 2 3 1 2 3 1 2 3 1 2 3",
"output": "48"
},
{
"input": "100 3\n2 1 1 1 3 2 3 3 2 3 3 1 3 3 1 3 3 1 1 1 2 3 1 2 3 1 2 3 3 1 3 1 1 2 3 2 3 3 2 3 3 1 2 2 1 2 3 2 3 2 2 1 1 3 1 3 2 1 3 1 3 1 3 1 1 3 3 3 2 3 2 2 2 2 1 3 3 3 1 2 1 2 3 2 1 3 1 3 2 1 3 1 2 1 2 3 1 3 2 3",
"output": "3296"
},
{
"input": "100 5\n5 5 2 4 5 4 4 4 4 2 5 3 4 2 4 4 1 1 5 3 2 2 1 3 3 2 5 3 4 5 1 3 5 4 4 4 3 1 4 4 3 4 5 2 5 4 2 1 2 2 3 5 5 5 1 4 5 3 1 4 2 2 5 1 5 3 4 1 5 1 2 2 3 5 1 3 2 4 2 4 2 2 4 1 3 5 2 2 2 3 3 4 3 2 2 5 5 4 2 5",
"output": "3953"
},
{
"input": "100 10\n7 4 5 5 10 10 5 8 5 7 4 5 4 6 8 8 2 6 3 3 10 7 10 8 6 2 7 3 9 7 7 2 4 5 2 4 9 5 10 1 10 5 10 4 1 3 4 2 6 9 9 9 10 6 2 5 6 1 8 10 4 10 3 4 10 5 5 4 10 4 5 3 7 10 2 7 3 6 9 6 1 6 5 5 4 6 6 4 4 1 5 1 6 6 6 8 8 6 2 6",
"output": "4428"
}
] | 1,575,655,062
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 140
| 3,584,000
|
import sys
from collections import Counter
def solve(nb_books, nb_genres, books):
counter = Counter(books)
nb_possibilities = 0
nb_seen = 0
for genre, count in counter.items():
nb_remaining_books = max(0, nb_books - nb_seen - count)
nb_possibilities += (count * nb_remaining_books)
nb_seen += count
return nb_possibilities
assert solve(4, 3, [2, 1, 3, 1]) == 5
assert solve(7, 4, [4, 2, 3, 1, 2, 4, 3]) == 18
nb_books, nb_genres = map(int, input().strip().split(' '))
books = input().strip().split(' ')
print(solve(nb_books, nb_genres, books))
|
Title: The Best Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emily's birthday is next week and Jack has decided to buy a present for her. He knows she loves books so he goes to the local bookshop, where there are *n* books on sale from one of *m* genres.
In the bookshop, Jack decides to buy two books of different genres.
Based on the genre of books on sale in the shop, find the number of options available to Jack for choosing two books of different genres for Emily. Options are considered different if they differ in at least one book.
The books are given by indices of their genres. The genres are numbered from 1 to *m*.
Input Specification:
The first line contains two positive integers *n* and *m* (2<=≤<=*n*<=≤<=2·105,<=2<=≤<=*m*<=≤<=10) — the number of books in the bookstore and the number of genres.
The second line contains a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (1<=≤<=*a**i*<=≤<=*m*) equals the genre of the *i*-th book.
It is guaranteed that for each genre there is at least one book of that genre.
Output Specification:
Print the only integer — the number of ways in which Jack can choose books.
It is guaranteed that the answer doesn't exceed the value 2·109.
Demo Input:
['4 3\n2 1 3 1\n', '7 4\n4 2 3 1 2 4 3\n']
Demo Output:
['5\n', '18\n']
Note:
The answer to the first test sample equals 5 as Sasha can choose:
1. the first and second books, 1. the first and third books, 1. the first and fourth books, 1. the second and third books, 1. the third and fourth books.
|
```python
import sys
from collections import Counter
def solve(nb_books, nb_genres, books):
counter = Counter(books)
nb_possibilities = 0
nb_seen = 0
for genre, count in counter.items():
nb_remaining_books = max(0, nb_books - nb_seen - count)
nb_possibilities += (count * nb_remaining_books)
nb_seen += count
return nb_possibilities
assert solve(4, 3, [2, 1, 3, 1]) == 5
assert solve(7, 4, [4, 2, 3, 1, 2, 4, 3]) == 18
nb_books, nb_genres = map(int, input().strip().split(' '))
books = input().strip().split(' ')
print(solve(nb_books, nb_genres, books))
```
| 3
|
|
298
|
B
|
Sail
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)?
|
The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
|
If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).
|
[
"5 0 0 1 1\nSESNW\n",
"10 5 3 3 6\nNENSWESNEE\n"
] |
[
"4\n",
"-1\n"
] |
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination.
| 500
|
[
{
"input": "5 0 0 1 1\nSESNW",
"output": "4"
},
{
"input": "10 5 3 3 6\nNENSWESNEE",
"output": "-1"
},
{
"input": "19 -172106364 -468680119 -172106365 -468680119\nSSEEESSSESESWSEESSS",
"output": "13"
},
{
"input": "39 -1000000000 -1000000000 -999999997 -1000000000\nENEENWSWSSWESNSSEESNSESWSWNSWESNENWNWEE",
"output": "4"
},
{
"input": "41 -264908123 -86993764 -264908123 -86993723\nNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN",
"output": "41"
},
{
"input": "34 -1000000000 -1000000000 -999999983 -1000000000\nEEEEESSEWNSSSESWEWSWESEWSEESNEWEEE",
"output": "-1"
},
{
"input": "1 0 0 0 -1\nS",
"output": "1"
},
{
"input": "1 5 5 5 6\nE",
"output": "-1"
},
{
"input": "15 1 1 1 2\nNNNNNNNNNNNNNNN",
"output": "1"
},
{
"input": "11 1 1 1 2\nNNNNNNNNNNN",
"output": "1"
}
] | 1,638,967,672
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 92
| 204,800
|
n,sx,sy,ex,ey = map(int,input().split())
st = input()
stpsx = ex-sx;stpsy= ey-sy
dirlst = []
va = 0
if stpsx < 0:
if st.count("W") < stpsx*-1:
print(-1)
exit()
else:
dirlst.extend(["W"]*(stpsx*-1))
elif stpsx >0:
if st.count("E") < stpsx:
print(-1)
exit()
else:
dirlst.extend(["E"]*stpsx)
if stpsy < 0:
if st.count("E") < stpsy*-1:
print(-1)
exit()
else:
dirlst.extend(["S"]*(stpsy*-1))
elif stpsy > 0:
if st.count("N") < stpsy:
print(-1)
exit()
else:
dirlst.extend(["N"]*stpsy)
midx = 0
for i in dirlst:
if midx < st.index(i)+1:
midx = st.index(i)+1
st = st.replace(i,"X",1)
print(midx)
|
Title: Sail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)?
Input Specification:
The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
Output Specification:
If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).
Demo Input:
['5 0 0 1 1\nSESNW\n', '10 5 3 3 6\nNENSWESNEE\n']
Demo Output:
['4\n', '-1\n']
Note:
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination.
|
```python
n,sx,sy,ex,ey = map(int,input().split())
st = input()
stpsx = ex-sx;stpsy= ey-sy
dirlst = []
va = 0
if stpsx < 0:
if st.count("W") < stpsx*-1:
print(-1)
exit()
else:
dirlst.extend(["W"]*(stpsx*-1))
elif stpsx >0:
if st.count("E") < stpsx:
print(-1)
exit()
else:
dirlst.extend(["E"]*stpsx)
if stpsy < 0:
if st.count("E") < stpsy*-1:
print(-1)
exit()
else:
dirlst.extend(["S"]*(stpsy*-1))
elif stpsy > 0:
if st.count("N") < stpsy:
print(-1)
exit()
else:
dirlst.extend(["N"]*stpsy)
midx = 0
for i in dirlst:
if midx < st.index(i)+1:
midx = st.index(i)+1
st = st.replace(i,"X",1)
print(midx)
```
| 0
|
|
365
|
A
|
Good Number
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
|
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the number of *k*-good numbers in *a*.
|
[
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] |
[
"10\n",
"1\n"
] |
none
| 500
|
[
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "6 0\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "1 3\n1000000000",
"output": "0"
},
{
"input": "1 9\n1000000000",
"output": "0"
},
{
"input": "1 0\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 3\n1",
"output": "0"
},
{
"input": "1 9\n1",
"output": "0"
},
{
"input": "2 8\n123456780\n123",
"output": "1"
},
{
"input": "2 8\n12345678\n1230",
"output": "0"
},
{
"input": "6 1\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "6 2\n10\n102\n120\n1032\n1212103\n1999999",
"output": "4"
},
{
"input": "6 3\n10\n102\n120\n1032\n1212103\n1999999",
"output": "2"
},
{
"input": "6 9\n10\n102\n120\n1032\n1212103\n1999999",
"output": "0"
},
{
"input": "1 2\n2220",
"output": "0"
}
] | 1,643,032,617
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 37
| 109
| 0
|
def listGen(k):
l = []
for i in range(k+1):
l.append(str(i))
return l
arr = list(map(int,input().split()))
n = arr[0]
k = arr[1]
count = 0
x = listGen(k)
for i in range(n):
arr1 = list(input())
isGood = True
for y in x:
if y not in arr1:
isGood = False
break
if isGood:
count+=1
print(count)
|
Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none
|
```python
def listGen(k):
l = []
for i in range(k+1):
l.append(str(i))
return l
arr = list(map(int,input().split()))
n = arr[0]
k = arr[1]
count = 0
x = listGen(k)
for i in range(n):
arr1 = list(input())
isGood = True
for y in x:
if y not in arr1:
isGood = False
break
if isGood:
count+=1
print(count)
```
| 3
|
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
|
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,695,217,107
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
n, w = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
if a[0] >= 0:
print(0)
else:
print(-sum(a[:w]))
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
n, w = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
if a[0] >= 0:
print(0)
else:
print(-sum(a[:w]))
```
| 0
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,675,185,309
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 30
| 0
|
s1=input()
s2=input()
n1=int(s1,2)
n2=int(s2,2)
x=n1^n2
y=bin(x).replace("0b","")
if(len(y)==len(s1)):
print(y)
else:
print("0"*(len(s1)-len(x)),end="")
print(y)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
s1=input()
s2=input()
n1=int(s1,2)
n2=int(s2,2)
x=n1^n2
y=bin(x).replace("0b","")
if(len(y)==len(s1)):
print(y)
else:
print("0"*(len(s1)-len(x)),end="")
print(y)
```
| -1
|
454
|
B
|
Little Pony and Sort by Shift
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
|
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
|
[
"2\n2 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] |
[
"1\n",
"-1\n",
"0\n"
] |
none
| 1,000
|
[
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 3 2",
"output": "-1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "6\n3 4 5 6 3 2",
"output": "-1"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n1 1 2 1 1",
"output": "2"
},
{
"input": "4\n5 4 5 4",
"output": "-1"
},
{
"input": "7\n3 4 5 5 5 1 2",
"output": "2"
},
{
"input": "5\n2 2 1 2 2",
"output": "3"
},
{
"input": "5\n5 4 1 2 3",
"output": "-1"
},
{
"input": "4\n6 1 2 7",
"output": "-1"
},
{
"input": "5\n4 5 6 2 3",
"output": "2"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "4\n1 2 2 1",
"output": "1"
},
{
"input": "9\n4 5 6 7 1 2 3 4 10",
"output": "-1"
},
{
"input": "7\n2 3 4 1 2 3 4",
"output": "-1"
},
{
"input": "6\n1 2 1 2 1 2",
"output": "-1"
},
{
"input": "3\n3 2 1",
"output": "-1"
},
{
"input": "4\n1 4 4 1",
"output": "1"
},
{
"input": "5\n1 2 1 1 1",
"output": "3"
},
{
"input": "5\n4 6 7 3 5",
"output": "-1"
},
{
"input": "4\n2 3 1 4",
"output": "-1"
},
{
"input": "5\n5 4 3 2 1",
"output": "-1"
},
{
"input": "4\n2 4 1 4",
"output": "-1"
},
{
"input": "6\n4 5 6 1 2 7",
"output": "-1"
},
{
"input": "6\n1 2 3 1 1 1",
"output": "3"
},
{
"input": "5\n1 3 3 3 1",
"output": "1"
},
{
"input": "6\n5 6 7 5 5 5",
"output": "3"
},
{
"input": "5\n3 4 2 1 2",
"output": "-1"
},
{
"input": "3\n3 4 2",
"output": "1"
},
{
"input": "6\n1 1 2 2 1 1",
"output": "2"
},
{
"input": "4\n2 3 4 2",
"output": "1"
},
{
"input": "5\n3 5 7 7 3",
"output": "1"
},
{
"input": "4\n1 1 4 1",
"output": "1"
},
{
"input": "7\n1 5 6 1 1 1 1",
"output": "4"
},
{
"input": "5\n7 8 6 7 8",
"output": "-1"
},
{
"input": "4\n2 4 1 3",
"output": "-1"
}
] | 1,687,707,486
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 77
| 12,390,400
|
n = int(input())
lst = list(map(int, input().split()))
decreasing = False
increasing = False
for i in range(n - 1):
first = lst[i]
second = lst[i + 1]
if(second < first):
decreasing = True
if(second > first):
increasing = True
if((increasing and decreasing) or (n > 2 and decreasing)):
print(-1)
elif(increasing and not decreasing):
print(0)
else:
print(1)
|
Title: Little Pony and Sort by Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, Twilight Sparkle is interested in how to sort a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105). The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
Demo Input:
['2\n2 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['1\n', '-1\n', '0\n']
Note:
none
|
```python
n = int(input())
lst = list(map(int, input().split()))
decreasing = False
increasing = False
for i in range(n - 1):
first = lst[i]
second = lst[i + 1]
if(second < first):
decreasing = True
if(second > first):
increasing = True
if((increasing and decreasing) or (n > 2 and decreasing)):
print(-1)
elif(increasing and not decreasing):
print(0)
else:
print(1)
```
| 0
|
|
1,003
|
C
|
Intense Heat
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation",
"math"
] | null | null |
The heat during the last few days has been really intense. Scientists from all over the Berland study how the temperatures and weather change, and they claim that this summer is abnormally hot. But any scientific claim sounds a lot more reasonable if there are some numbers involved, so they have decided to actually calculate some value which would represent how high the temperatures are.
Mathematicians of Berland State University came up with a special heat intensity value. This value is calculated as follows:
Suppose we want to analyze the segment of $n$ consecutive days. We have measured the temperatures during these $n$ days; the temperature during $i$-th day equals $a_i$.
We denote the average temperature of a segment of some consecutive days as the arithmetic mean of the temperature measures during this segment of days. So, if we want to analyze the average temperature from day $x$ to day $y$, we calculate it as $\frac{\sum \limits_{i = x}^{y} a_i}{y - x + 1}$ (note that division is performed without any rounding). The heat intensity value is the maximum of average temperatures over all segments of not less than $k$ consecutive days. For example, if analyzing the measures $[3, 4, 1, 2]$ and $k = 3$, we are interested in segments $[3, 4, 1]$, $[4, 1, 2]$ and $[3, 4, 1, 2]$ (we want to find the maximum value of average temperature over these segments).
You have been hired by Berland State University to write a program that would compute the heat intensity value of a given period of days. Are you up to this task?
|
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 5000$) — the number of days in the given period, and the minimum number of days in a segment we consider when calculating heat intensity value, respectively.
The second line contains $n$ integers $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 5000$) — the temperature measures during given $n$ days.
|
Print one real number — the heat intensity value, i. e., the maximum of average temperatures over all segments of not less than $k$ consecutive days.
Your answer will be considered correct if the following condition holds: $|res - res_0| < 10^{-6}$, where $res$ is your answer, and $res_0$ is the answer given by the jury's solution.
|
[
"4 3\n3 4 1 2\n"
] |
[
"2.666666666666667\n"
] |
none
| 0
|
[
{
"input": "4 3\n3 4 1 2",
"output": "2.666666666666667"
},
{
"input": "5 1\n3 10 9 10 6",
"output": "10.000000000000000"
},
{
"input": "5 2\n7 3 3 1 8",
"output": "5.000000000000000"
},
{
"input": "5 3\n1 7 6 9 1",
"output": "7.333333333333333"
},
{
"input": "5 4\n5 1 10 6 1",
"output": "5.500000000000000"
},
{
"input": "5 5\n4 6 6 6 2",
"output": "4.800000000000000"
},
{
"input": "3 2\n2 1 2",
"output": "1.666666666666667"
},
{
"input": "1 1\n5000",
"output": "5000.000000000000000"
}
] | 1,624,083,854
| 2,474
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 93
| 0
|
from sys import stdin,stdout
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
for _ in range(1):#nmbr()):
n,k=lst()
a=lst()
ans=sm=0
for i in range(n):
sm=a[i]
for j in range(i+1,n):
sm+=a[j]
if (j-i+1)>=k:
ans=max(ans, sm/(j-i+1))
print("{0:.8f}".format(ans))
|
Title: Intense Heat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The heat during the last few days has been really intense. Scientists from all over the Berland study how the temperatures and weather change, and they claim that this summer is abnormally hot. But any scientific claim sounds a lot more reasonable if there are some numbers involved, so they have decided to actually calculate some value which would represent how high the temperatures are.
Mathematicians of Berland State University came up with a special heat intensity value. This value is calculated as follows:
Suppose we want to analyze the segment of $n$ consecutive days. We have measured the temperatures during these $n$ days; the temperature during $i$-th day equals $a_i$.
We denote the average temperature of a segment of some consecutive days as the arithmetic mean of the temperature measures during this segment of days. So, if we want to analyze the average temperature from day $x$ to day $y$, we calculate it as $\frac{\sum \limits_{i = x}^{y} a_i}{y - x + 1}$ (note that division is performed without any rounding). The heat intensity value is the maximum of average temperatures over all segments of not less than $k$ consecutive days. For example, if analyzing the measures $[3, 4, 1, 2]$ and $k = 3$, we are interested in segments $[3, 4, 1]$, $[4, 1, 2]$ and $[3, 4, 1, 2]$ (we want to find the maximum value of average temperature over these segments).
You have been hired by Berland State University to write a program that would compute the heat intensity value of a given period of days. Are you up to this task?
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 5000$) — the number of days in the given period, and the minimum number of days in a segment we consider when calculating heat intensity value, respectively.
The second line contains $n$ integers $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 5000$) — the temperature measures during given $n$ days.
Output Specification:
Print one real number — the heat intensity value, i. e., the maximum of average temperatures over all segments of not less than $k$ consecutive days.
Your answer will be considered correct if the following condition holds: $|res - res_0| < 10^{-6}$, where $res$ is your answer, and $res_0$ is the answer given by the jury's solution.
Demo Input:
['4 3\n3 4 1 2\n']
Demo Output:
['2.666666666666667\n']
Note:
none
|
```python
from sys import stdin,stdout
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
for _ in range(1):#nmbr()):
n,k=lst()
a=lst()
ans=sm=0
for i in range(n):
sm=a[i]
for j in range(i+1,n):
sm+=a[j]
if (j-i+1)>=k:
ans=max(ans, sm/(j-i+1))
print("{0:.8f}".format(ans))
```
| 0
|
|
599
|
A
|
Patrick and Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
|
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
|
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
|
[
"10 20 30\n",
"1 1 5\n"
] |
[
"60\n",
"4\n"
] |
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
| 500
|
[
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,694,792,142
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
x, y, z = map(int, input().split())
n = min(
x + y + z,
x + x + y + y,
x + y + y + x,
y + x + x + y
)
print(n)
|
Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
|
```python
x, y, z = map(int, input().split())
n = min(
x + y + z,
x + x + y + y,
x + y + y + x,
y + x + x + y
)
print(n)
```
| 0
|
|
91
|
B
|
Queue
|
PROGRAMMING
| 1,500
|
[
"binary search",
"data structures"
] |
B. Queue
|
2
|
256
|
There are *n* walruses standing in a queue in an airport. They are numbered starting from the queue's tail: the 1-st walrus stands at the end of the queue and the *n*-th walrus stands at the beginning of the queue. The *i*-th walrus has the age equal to *a**i*.
The *i*-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such *j* (*i*<=<<=*j*), that *a**i*<=><=*a**j*. The displeasure of the *i*-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the *i*-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.
The airport manager asked you to count for each of *n* walruses in the queue his displeasure.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105) — the number of walruses in the queue. The second line contains integers *a**i* (1<=≤<=*a**i*<=≤<=109).
Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.
|
Print *n* numbers: if the *i*-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the *i*-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.
|
[
"6\n10 8 5 3 50 45\n",
"7\n10 4 6 3 2 8 15\n",
"5\n10 3 1 10 11\n"
] |
[
"2 1 0 -1 0 -1 ",
"4 2 1 0 -1 -1 -1 ",
"1 0 -1 -1 -1 "
] |
none
| 1,000
|
[
{
"input": "6\n10 8 5 3 50 45",
"output": "2 1 0 -1 0 -1 "
},
{
"input": "7\n10 4 6 3 2 8 15",
"output": "4 2 1 0 -1 -1 -1 "
},
{
"input": "5\n10 3 1 10 11",
"output": "1 0 -1 -1 -1 "
},
{
"input": "13\n18 9 8 9 23 20 18 18 33 25 31 37 36",
"output": "2 0 -1 -1 2 1 -1 -1 1 -1 -1 0 -1 "
},
{
"input": "10\n15 21 17 22 27 21 31 26 32 30",
"output": "-1 0 -1 1 2 -1 2 -1 0 -1 "
},
{
"input": "10\n18 20 18 17 17 13 22 20 34 29",
"output": "4 3 2 1 0 -1 0 -1 0 -1 "
},
{
"input": "13\n16 14 12 9 11 28 30 21 35 30 32 31 43",
"output": "3 2 1 -1 -1 1 0 -1 2 -1 0 -1 -1 "
},
{
"input": "15\n18 6 18 21 14 20 13 9 18 20 28 13 19 25 21",
"output": "10 -1 8 8 6 6 0 -1 2 2 3 -1 -1 0 -1 "
},
{
"input": "11\n15 17 18 18 26 22 23 33 33 21 29",
"output": "-1 -1 -1 -1 4 3 2 2 1 -1 -1 "
},
{
"input": "15\n14 4 5 12 6 19 14 19 12 22 23 17 14 21 27",
"output": "7 -1 -1 0 -1 6 1 4 -1 3 2 0 -1 -1 -1 "
},
{
"input": "2\n1 1000000000",
"output": "-1 -1 "
},
{
"input": "2\n1000000000 1",
"output": "0 -1 "
},
{
"input": "5\n15 1 8 15 3",
"output": "3 -1 1 0 -1 "
},
{
"input": "12\n5 1 2 5 100 1 1000 100 10000 20000 10000 20000",
"output": "4 -1 2 1 0 -1 0 -1 -1 0 -1 -1 "
}
] | 1,620,050,785
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 124
| 0
|
# https://codeforces.com/problemset/problem/91/B
n=int(input())
arr=list(map(int,input().split()))
res=[-1]*n
for i in range(n-2,-1,-1):
for j in range(1,n-i):
if(arr[i]>=arr[i+j]):
res[i]=res[i+j]+j
for x in res:
print(x,end=" ")
|
Title: Queue
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses standing in a queue in an airport. They are numbered starting from the queue's tail: the 1-st walrus stands at the end of the queue and the *n*-th walrus stands at the beginning of the queue. The *i*-th walrus has the age equal to *a**i*.
The *i*-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such *j* (*i*<=<<=*j*), that *a**i*<=><=*a**j*. The displeasure of the *i*-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the *i*-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.
The airport manager asked you to count for each of *n* walruses in the queue his displeasure.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=105) — the number of walruses in the queue. The second line contains integers *a**i* (1<=≤<=*a**i*<=≤<=109).
Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.
Output Specification:
Print *n* numbers: if the *i*-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the *i*-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.
Demo Input:
['6\n10 8 5 3 50 45\n', '7\n10 4 6 3 2 8 15\n', '5\n10 3 1 10 11\n']
Demo Output:
['2 1 0 -1 0 -1 ', '4 2 1 0 -1 -1 -1 ', '1 0 -1 -1 -1 ']
Note:
none
|
```python
# https://codeforces.com/problemset/problem/91/B
n=int(input())
arr=list(map(int,input().split()))
res=[-1]*n
for i in range(n-2,-1,-1):
for j in range(1,n-i):
if(arr[i]>=arr[i+j]):
res[i]=res[i+j]+j
for x in res:
print(x,end=" ")
```
| 0
|
272
|
A
|
Dima and Friends
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
|
In a single line print the answer to the problem.
|
[
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] |
[
"3\n",
"2\n",
"3\n"
] |
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
| 500
|
[
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,664,045,518
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
n=int(input())
A=list(map(int,input().split()))
total_persons=n+1
Sum=sum(A)
for i in range(1,6):
if (Sum+i)%(total_persons)!=1:
print(i)
break
else:
continue
|
Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
|
```python
n=int(input())
A=list(map(int,input().split()))
total_persons=n+1
Sum=sum(A)
for i in range(1,6):
if (Sum+i)%(total_persons)!=1:
print(i)
break
else:
continue
```
| 0
|
|
645
|
A
|
Amity Assessment
|
PROGRAMMING
| 1,200
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Bessie the cow and her best friend Elsie each received a sliding puzzle on Pi Day. Their puzzles consist of a 2<=×<=2 grid and three tiles labeled 'A', 'B', and 'C'. The three tiles sit on top of the grid, leaving one grid cell empty. To make a move, Bessie or Elsie can slide a tile adjacent to the empty cell into the empty cell as shown below:
In order to determine if they are truly Best Friends For Life (BFFLs), Bessie and Elsie would like to know if there exists a sequence of moves that takes their puzzles to the same configuration (moves can be performed in both puzzles). Two puzzles are considered to be in the same configuration if each tile is on top of the same grid cell in both puzzles. Since the tiles are labeled with letters, rotations and reflections are not allowed.
|
The first two lines of the input consist of a 2<=×<=2 grid describing the initial configuration of Bessie's puzzle. The next two lines contain a 2<=×<=2 grid describing the initial configuration of Elsie's puzzle. The positions of the tiles are labeled 'A', 'B', and 'C', while the empty cell is labeled 'X'. It's guaranteed that both puzzles contain exactly one tile with each letter and exactly one empty position.
|
Output "YES"(without quotes) if the puzzles can reach the same configuration (and Bessie and Elsie are truly BFFLs). Otherwise, print "NO" (without quotes).
|
[
"AB\nXC\nXB\nAC\n",
"AB\nXC\nAC\nBX\n"
] |
[
"YES\n",
"NO\n"
] |
The solution to the first sample is described by the image. All Bessie needs to do is slide her 'A' tile down.
In the second sample, the two puzzles can never be in the same configuration. Perhaps Bessie and Elsie are not meant to be friends after all...
| 500
|
[
{
"input": "AB\nXC\nXB\nAC",
"output": "YES"
},
{
"input": "AB\nXC\nAC\nBX",
"output": "NO"
},
{
"input": "XC\nBA\nCB\nAX",
"output": "NO"
},
{
"input": "AB\nXC\nAX\nCB",
"output": "YES"
},
{
"input": "CB\nAX\nXA\nBC",
"output": "YES"
},
{
"input": "BC\nXA\nBA\nXC",
"output": "NO"
},
{
"input": "CA\nXB\nBA\nCX",
"output": "NO"
},
{
"input": "CA\nXB\nAC\nBX",
"output": "NO"
},
{
"input": "CB\nAX\nCX\nAB",
"output": "YES"
},
{
"input": "AX\nCB\nBC\nXA",
"output": "YES"
},
{
"input": "CA\nXB\nBA\nXC",
"output": "NO"
},
{
"input": "CX\nAB\nAX\nCB",
"output": "NO"
},
{
"input": "AB\nXC\nAB\nCX",
"output": "YES"
},
{
"input": "XC\nBA\nXC\nAB",
"output": "NO"
},
{
"input": "BA\nXC\nAC\nXB",
"output": "YES"
},
{
"input": "AX\nBC\nAC\nBX",
"output": "YES"
},
{
"input": "XC\nBA\nCB\nXA",
"output": "NO"
},
{
"input": "CB\nAX\nXC\nBA",
"output": "NO"
},
{
"input": "AX\nCB\nBC\nAX",
"output": "YES"
},
{
"input": "AB\nXC\nBX\nAC",
"output": "YES"
},
{
"input": "XA\nCB\nBA\nCX",
"output": "NO"
},
{
"input": "CX\nBA\nBX\nAC",
"output": "YES"
},
{
"input": "AB\nXC\nXC\nAB",
"output": "NO"
},
{
"input": "BA\nCX\nAC\nBX",
"output": "YES"
},
{
"input": "XA\nCB\nAB\nXC",
"output": "YES"
},
{
"input": "XC\nBA\nAC\nBX",
"output": "NO"
},
{
"input": "CA\nBX\nBA\nXC",
"output": "NO"
},
{
"input": "AX\nBC\nCA\nXB",
"output": "NO"
},
{
"input": "BC\nAX\nXC\nBA",
"output": "YES"
},
{
"input": "XB\nAC\nBX\nAC",
"output": "YES"
},
{
"input": "CX\nBA\nAX\nBC",
"output": "NO"
},
{
"input": "XB\nCA\nXC\nBA",
"output": "NO"
},
{
"input": "BX\nCA\nXB\nCA",
"output": "YES"
},
{
"input": "XB\nAC\nXC\nAB",
"output": "NO"
},
{
"input": "CX\nBA\nCX\nBA",
"output": "YES"
},
{
"input": "XB\nAC\nCA\nBX",
"output": "YES"
},
{
"input": "BA\nXC\nBC\nAX",
"output": "NO"
},
{
"input": "AC\nXB\nCX\nBA",
"output": "NO"
},
{
"input": "XB\nCA\nCX\nBA",
"output": "NO"
},
{
"input": "AB\nCX\nXA\nBC",
"output": "NO"
},
{
"input": "CX\nAB\nXB\nAC",
"output": "NO"
},
{
"input": "BC\nAX\nAC\nBX",
"output": "NO"
},
{
"input": "XA\nBC\nCB\nAX",
"output": "YES"
},
{
"input": "XC\nAB\nCB\nAX",
"output": "YES"
},
{
"input": "CX\nBA\nCX\nAB",
"output": "NO"
},
{
"input": "CA\nBX\nXC\nBA",
"output": "YES"
},
{
"input": "CX\nBA\nBA\nXC",
"output": "NO"
},
{
"input": "CA\nBX\nCB\nXA",
"output": "NO"
},
{
"input": "CB\nAX\nBC\nAX",
"output": "NO"
},
{
"input": "CB\nAX\nBC\nXA",
"output": "NO"
},
{
"input": "AC\nXB\nCB\nXA",
"output": "YES"
},
{
"input": "AB\nCX\nXB\nAC",
"output": "YES"
},
{
"input": "CX\nBA\nXB\nAC",
"output": "YES"
},
{
"input": "BX\nAC\nAB\nXC",
"output": "YES"
},
{
"input": "CX\nAB\nXC\nBA",
"output": "NO"
},
{
"input": "XB\nAC\nCX\nAB",
"output": "NO"
},
{
"input": "CB\nAX\nXB\nAC",
"output": "NO"
},
{
"input": "CB\nAX\nCA\nXB",
"output": "NO"
},
{
"input": "XC\nBA\nBA\nXC",
"output": "NO"
},
{
"input": "AC\nBX\nCB\nAX",
"output": "YES"
},
{
"input": "CA\nBX\nAC\nXB",
"output": "NO"
},
{
"input": "BX\nAC\nCX\nBA",
"output": "YES"
},
{
"input": "XB\nCA\nAX\nCB",
"output": "NO"
},
{
"input": "CB\nXA\nBC\nXA",
"output": "NO"
},
{
"input": "AX\nCB\nCX\nAB",
"output": "NO"
},
{
"input": "BC\nAX\nXC\nAB",
"output": "NO"
},
{
"input": "XB\nCA\nBC\nXA",
"output": "NO"
},
{
"input": "XB\nAC\nCX\nBA",
"output": "YES"
},
{
"input": "BC\nXA\nCB\nXA",
"output": "NO"
},
{
"input": "AX\nCB\nAX\nBC",
"output": "NO"
},
{
"input": "CA\nBX\nBX\nCA",
"output": "NO"
},
{
"input": "BA\nXC\nXB\nAC",
"output": "NO"
},
{
"input": "XA\nBC\nBX\nAC",
"output": "NO"
},
{
"input": "BX\nCA\nAC\nBX",
"output": "YES"
},
{
"input": "XB\nAC\nXC\nBA",
"output": "YES"
},
{
"input": "XB\nAC\nAB\nXC",
"output": "YES"
},
{
"input": "BA\nCX\nCX\nBA",
"output": "NO"
},
{
"input": "CA\nXB\nXB\nCA",
"output": "NO"
},
{
"input": "BA\nCX\nBA\nXC",
"output": "YES"
},
{
"input": "BA\nCX\nAB\nCX",
"output": "NO"
},
{
"input": "BX\nCA\nXA\nBC",
"output": "YES"
},
{
"input": "XC\nBA\nBX\nCA",
"output": "NO"
},
{
"input": "XC\nAB\nBC\nXA",
"output": "NO"
},
{
"input": "BC\nXA\nXC\nAB",
"output": "NO"
},
{
"input": "BX\nCA\nXB\nAC",
"output": "NO"
},
{
"input": "BA\nXC\nCA\nXB",
"output": "NO"
},
{
"input": "CX\nBA\nAC\nXB",
"output": "NO"
},
{
"input": "AB\nCX\nAC\nBX",
"output": "NO"
},
{
"input": "BC\nXA\nBX\nCA",
"output": "NO"
},
{
"input": "XA\nBC\nCX\nAB",
"output": "YES"
},
{
"input": "AX\nBC\nAX\nCB",
"output": "NO"
},
{
"input": "CB\nAX\nCA\nBX",
"output": "NO"
},
{
"input": "CB\nAX\nBA\nXC",
"output": "YES"
},
{
"input": "AB\nCX\nXC\nBA",
"output": "YES"
},
{
"input": "AC\nXB\nBA\nCX",
"output": "YES"
},
{
"input": "AX\nCB\nCB\nAX",
"output": "NO"
},
{
"input": "CX\nBA\nCA\nXB",
"output": "YES"
},
{
"input": "AC\nBX\nAB\nXC",
"output": "NO"
},
{
"input": "XA\nCB\nXA\nBC",
"output": "NO"
},
{
"input": "XC\nBA\nCA\nBX",
"output": "YES"
},
{
"input": "XA\nBC\nXB\nCA",
"output": "YES"
},
{
"input": "CA\nBX\nCB\nAX",
"output": "NO"
}
] | 1,458,320,827
| 1,927
|
Python 3
|
OK
|
TESTS
| 102
| 62
| 0
|
puzzle1 = ''
puzzle2 = ''
row = input()
if row[0] != 'X':
puzzle1 += row[0]
if row[1] != 'X':
puzzle1 += row[1]
row = input()
if row[1] != 'X':
puzzle1 += row[1]
if row[0] != 'X':
puzzle1 += row[0]
row = input()
if row[0] != 'X':
puzzle2 += row[0]
if row[1] != 'X':
puzzle2 += row[1]
row = input()
if row[1] != 'X':
puzzle2 += row[1]
if row[0] != 'X':
puzzle2 += row[0]
var = [puzzle2,
puzzle2[2]+puzzle2[0]+puzzle2[1],
puzzle2[1]+puzzle2[2]+puzzle2[0]]
if puzzle1 in var:
print('YES')
else:
print('NO')
|
Title: Amity Assessment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bessie the cow and her best friend Elsie each received a sliding puzzle on Pi Day. Their puzzles consist of a 2<=×<=2 grid and three tiles labeled 'A', 'B', and 'C'. The three tiles sit on top of the grid, leaving one grid cell empty. To make a move, Bessie or Elsie can slide a tile adjacent to the empty cell into the empty cell as shown below:
In order to determine if they are truly Best Friends For Life (BFFLs), Bessie and Elsie would like to know if there exists a sequence of moves that takes their puzzles to the same configuration (moves can be performed in both puzzles). Two puzzles are considered to be in the same configuration if each tile is on top of the same grid cell in both puzzles. Since the tiles are labeled with letters, rotations and reflections are not allowed.
Input Specification:
The first two lines of the input consist of a 2<=×<=2 grid describing the initial configuration of Bessie's puzzle. The next two lines contain a 2<=×<=2 grid describing the initial configuration of Elsie's puzzle. The positions of the tiles are labeled 'A', 'B', and 'C', while the empty cell is labeled 'X'. It's guaranteed that both puzzles contain exactly one tile with each letter and exactly one empty position.
Output Specification:
Output "YES"(without quotes) if the puzzles can reach the same configuration (and Bessie and Elsie are truly BFFLs). Otherwise, print "NO" (without quotes).
Demo Input:
['AB\nXC\nXB\nAC\n', 'AB\nXC\nAC\nBX\n']
Demo Output:
['YES\n', 'NO\n']
Note:
The solution to the first sample is described by the image. All Bessie needs to do is slide her 'A' tile down.
In the second sample, the two puzzles can never be in the same configuration. Perhaps Bessie and Elsie are not meant to be friends after all...
|
```python
puzzle1 = ''
puzzle2 = ''
row = input()
if row[0] != 'X':
puzzle1 += row[0]
if row[1] != 'X':
puzzle1 += row[1]
row = input()
if row[1] != 'X':
puzzle1 += row[1]
if row[0] != 'X':
puzzle1 += row[0]
row = input()
if row[0] != 'X':
puzzle2 += row[0]
if row[1] != 'X':
puzzle2 += row[1]
row = input()
if row[1] != 'X':
puzzle2 += row[1]
if row[0] != 'X':
puzzle2 += row[0]
var = [puzzle2,
puzzle2[2]+puzzle2[0]+puzzle2[1],
puzzle2[1]+puzzle2[2]+puzzle2[0]]
if puzzle1 in var:
print('YES')
else:
print('NO')
```
| 3
|
|
251
|
A
|
Points on Line
|
PROGRAMMING
| 1,300
|
[
"binary search",
"combinatorics",
"two pointers"
] | null | null |
Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter.
|
The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
|
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"4 3\n1 2 3 4\n",
"4 2\n-3 -2 -1 0\n",
"5 19\n1 10 20 30 50\n"
] |
[
"4\n",
"2\n",
"1\n"
] |
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}.
| 500
|
[
{
"input": "4 3\n1 2 3 4",
"output": "4"
},
{
"input": "4 2\n-3 -2 -1 0",
"output": "2"
},
{
"input": "5 19\n1 10 20 30 50",
"output": "1"
},
{
"input": "10 5\n31 36 43 47 48 50 56 69 71 86",
"output": "2"
},
{
"input": "10 50\n1 4 20 27 65 79 82 83 99 100",
"output": "25"
},
{
"input": "10 90\n24 27 40 41 61 69 73 87 95 97",
"output": "120"
},
{
"input": "100 100\n-98 -97 -96 -93 -92 -91 -90 -87 -86 -84 -81 -80 -79 -78 -76 -75 -73 -71 -69 -67 -65 -64 -63 -62 -61 -54 -51 -50 -49 -48 -46 -45 -44 -37 -36 -33 -30 -28 -27 -16 -15 -13 -12 -10 -9 -7 -6 -5 -4 2 3 5 8 9 10 11 13 14 15 16 17 19 22 24 25 26 27 28 30 31 32 36 40 43 45 46 47 50 51 52 53 58 60 63 69 70 73 78 80 81 82 85 88 89 90 91 95 96 97 99",
"output": "79351"
},
{
"input": "1 14751211\n847188590",
"output": "0"
},
{
"input": "2 1000000000\n-907894512 -289906312",
"output": "0"
},
{
"input": "2 1000000000\n-14348867 1760823",
"output": "0"
},
{
"input": "3 1000000000\n-5 -1 1",
"output": "1"
}
] | 1,614,596,048
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 404
| 8,908,800
|
n,d = map(int,input().split())
ar = list(map(int,input().split()))
i,j,s = 0,0,0
while i<n:
while ar[i]-ar[j]>d:
j += 1
N = i-j
s += N*(N-1)//2
i += 1
print(s)
|
Title: Points on Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input Specification:
The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output Specification:
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['4 3\n1 2 3 4\n', '4 2\n-3 -2 -1 0\n', '5 19\n1 10 20 30 50\n']
Demo Output:
['4\n', '2\n', '1\n']
Note:
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}.
|
```python
n,d = map(int,input().split())
ar = list(map(int,input().split()))
i,j,s = 0,0,0
while i<n:
while ar[i]-ar[j]>d:
j += 1
N = i-j
s += N*(N-1)//2
i += 1
print(s)
```
| 3
|
|
851
|
B
|
Arpa and an exam about geometry
|
PROGRAMMING
| 1,400
|
[
"geometry",
"math"
] | null | null |
Arpa is taking a geometry exam. Here is the last problem of the exam.
You are given three points *a*,<=*b*,<=*c*.
Find a point and an angle such that if we rotate the page around the point by the angle, the new position of *a* is the same as the old position of *b*, and the new position of *b* is the same as the old position of *c*.
Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.
|
The only line contains six integers *a**x*,<=*a**y*,<=*b**x*,<=*b**y*,<=*c**x*,<=*c**y* (|*a**x*|,<=|*a**y*|,<=|*b**x*|,<=|*b**y*|,<=|*c**x*|,<=|*c**y*|<=≤<=109). It's guaranteed that the points are distinct.
|
Print "Yes" if the problem has a solution, "No" otherwise.
You can print each letter in any case (upper or lower).
|
[
"0 1 1 1 1 0\n",
"1 1 0 0 1000 1000\n"
] |
[
"Yes\n",
"No\n"
] |
In the first sample test, rotate the page around (0.5, 0.5) by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9d845923f4d356a48d8ede337db0303821311f0c.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, you can't find any solution.
| 1,000
|
[
{
"input": "0 1 1 1 1 0",
"output": "Yes"
},
{
"input": "1 1 0 0 1000 1000",
"output": "No"
},
{
"input": "1 0 2 0 3 0",
"output": "No"
},
{
"input": "3 4 0 0 4 3",
"output": "Yes"
},
{
"input": "-1000000000 1 0 0 1000000000 1",
"output": "Yes"
},
{
"input": "49152 0 0 0 0 81920",
"output": "No"
},
{
"input": "1 -1 4 4 2 -3",
"output": "No"
},
{
"input": "-2 -2 1 4 -2 0",
"output": "No"
},
{
"input": "5 0 4 -2 0 1",
"output": "No"
},
{
"input": "-4 -3 2 -1 -3 4",
"output": "No"
},
{
"input": "-3 -3 5 2 3 -1",
"output": "No"
},
{
"input": "-1000000000 -1000000000 0 0 1000000000 999999999",
"output": "No"
},
{
"input": "-1000000000 -1000000000 0 0 1000000000 1000000000",
"output": "No"
},
{
"input": "-357531221 381512519 -761132895 -224448284 328888775 -237692564",
"output": "No"
},
{
"input": "264193194 -448876521 736684426 -633906160 -328597212 -47935734",
"output": "No"
},
{
"input": "419578772 -125025887 169314071 89851312 961404059 21419450",
"output": "No"
},
{
"input": "-607353321 -620687860 248029390 477864359 728255275 -264646027",
"output": "No"
},
{
"input": "299948862 -648908808 338174789 841279400 -850322448 350263551",
"output": "No"
},
{
"input": "48517753 416240699 7672672 272460100 -917845051 199790781",
"output": "No"
},
{
"input": "-947393823 -495674431 211535284 -877153626 -522763219 -778236665",
"output": "No"
},
{
"input": "-685673792 -488079395 909733355 385950193 -705890324 256550506",
"output": "No"
},
{
"input": "-326038504 547872194 49630307 713863100 303770000 -556852524",
"output": "No"
},
{
"input": "-706921242 -758563024 -588592101 -443440080 858751713 238854303",
"output": "No"
},
{
"input": "-1000000000 -1000000000 0 1000000000 1000000000 -1000000000",
"output": "Yes"
},
{
"input": "1000000000 1000000000 0 -1000000000 -1000000000 1000000000",
"output": "Yes"
},
{
"input": "-999999999 -1000000000 0 0 1000000000 999999999",
"output": "Yes"
},
{
"input": "-1000000000 -999999999 0 0 1000000000 999999999",
"output": "No"
},
{
"input": "-1 -1000000000 0 1000000000 1 -1000000000",
"output": "Yes"
},
{
"input": "0 1000000000 1 0 0 -1000000000",
"output": "Yes"
},
{
"input": "0 1000000000 0 0 0 -1000000000",
"output": "No"
},
{
"input": "0 1 1 2 2 3",
"output": "No"
},
{
"input": "999999999 1000000000 0 0 -1000000000 -999999999",
"output": "Yes"
},
{
"input": "0 0 1 1 2 0",
"output": "Yes"
},
{
"input": "0 0 1 1 2 2",
"output": "No"
},
{
"input": "1 1 2 2 3 3",
"output": "No"
},
{
"input": "0 2 0 3 0 4",
"output": "No"
},
{
"input": "1 1 1 2 1 3",
"output": "No"
},
{
"input": "0 0 3 4 3 9",
"output": "Yes"
},
{
"input": "589824 196608 262144 196608 0 0",
"output": "Yes"
},
{
"input": "0 0 1000000000 1 1000000000 -999999999",
"output": "No"
},
{
"input": "0 0 2 45 0 90",
"output": "Yes"
},
{
"input": "0 0 0 2 0 1",
"output": "No"
},
{
"input": "0 2 4 5 4 0",
"output": "Yes"
},
{
"input": "0 0 2 0 4 0",
"output": "No"
},
{
"input": "1 1 3 3 5 5",
"output": "No"
},
{
"input": "1 1 2 2 3 1",
"output": "Yes"
}
] | 1,521,737,573
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 39
| 124
| 7,475,200
|
import sys
import os
import math
import re
ax,ay,bx,by,cx,cy = map(int,input().split())
a = (ax,ay)
b = (bx,by)
c = (cx,cy)
Dab = math.sqrt((bx - ax)**2 + (by - ay)**2)
Dbc = math.sqrt((cx-bx)**2 + (cy - by)**2)
if (Dab == Dbc and (ay - by)*(cx - bx) != (ax -bx)*(cy-by)):
print('Yes')
else:
print('No')
|
Title: Arpa and an exam about geometry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arpa is taking a geometry exam. Here is the last problem of the exam.
You are given three points *a*,<=*b*,<=*c*.
Find a point and an angle such that if we rotate the page around the point by the angle, the new position of *a* is the same as the old position of *b*, and the new position of *b* is the same as the old position of *c*.
Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.
Input Specification:
The only line contains six integers *a**x*,<=*a**y*,<=*b**x*,<=*b**y*,<=*c**x*,<=*c**y* (|*a**x*|,<=|*a**y*|,<=|*b**x*|,<=|*b**y*|,<=|*c**x*|,<=|*c**y*|<=≤<=109). It's guaranteed that the points are distinct.
Output Specification:
Print "Yes" if the problem has a solution, "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['0 1 1 1 1 0\n', '1 1 0 0 1000 1000\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test, rotate the page around (0.5, 0.5) by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9d845923f4d356a48d8ede337db0303821311f0c.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, you can't find any solution.
|
```python
import sys
import os
import math
import re
ax,ay,bx,by,cx,cy = map(int,input().split())
a = (ax,ay)
b = (bx,by)
c = (cx,cy)
Dab = math.sqrt((bx - ax)**2 + (by - ay)**2)
Dbc = math.sqrt((cx-bx)**2 + (cy - by)**2)
if (Dab == Dbc and (ay - by)*(cx - bx) != (ax -bx)*(cy-by)):
print('Yes')
else:
print('No')
```
| 0
|
|
485
|
B
|
Valuable Resources
|
PROGRAMMING
| 1,300
|
[
"brute force",
"greedy"
] | null | null |
Many computer strategy games require building cities, recruiting army, conquering tribes, collecting resources. Sometimes it leads to interesting problems.
Let's suppose that your task is to build a square city. The world map uses the Cartesian coordinates. The sides of the city should be parallel to coordinate axes. The map contains mines with valuable resources, located at some points with integer coordinates. The sizes of mines are relatively small, i.e. they can be treated as points. The city should be built in such a way that all the mines are inside or on the border of the city square.
Building a city takes large amount of money depending on the size of the city, so you have to build the city with the minimum area. Given the positions of the mines find the minimum possible area of the city.
|
The first line of the input contains number *n* — the number of mines on the map (2<=≤<=*n*<=≤<=1000). Each of the next *n* lines contains a pair of integers *x**i* and *y**i* — the coordinates of the corresponding mine (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109). All points are pairwise distinct.
|
Print the minimum area of the city that can cover all the mines with valuable resources.
|
[
"2\n0 0\n2 2\n",
"2\n0 0\n0 3\n"
] |
[
"4\n",
"9\n"
] |
none
| 500
|
[
{
"input": "2\n0 0\n2 2",
"output": "4"
},
{
"input": "2\n0 0\n0 3",
"output": "9"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "3\n2 2\n1 1\n3 3",
"output": "4"
},
{
"input": "3\n3 1\n1 3\n2 2",
"output": "4"
},
{
"input": "3\n0 1\n1 0\n2 2",
"output": "4"
},
{
"input": "2\n-1000000000 -1000000000\n1000000000 1000000000",
"output": "4000000000000000000"
},
{
"input": "2\n1000000000 -1000000000\n-1000000000 1000000000",
"output": "4000000000000000000"
},
{
"input": "5\n-851545463 -208880322\n-154983867 -781305244\n293363100 785256340\n833468900 -593065920\n-920692803 -637662144",
"output": "3077083280271860209"
},
{
"input": "10\n-260530833 169589238\n-681955770 -35391010\n223450511 24504262\n479795061 -26191863\n-291344265 21153856\n714700263 -328447419\n-858655942 161086142\n-270884153 462537328\n-501424901 977460517\n115284904 -151626824",
"output": "2475449747812002025"
},
{
"input": "10\n917139470 819990899\n-69828590 691215072\n-846815289 112372447\n560780737 -890423729\n243241705 284240970\n-47397355 -263709479\n759162072 709456353\n-330469400 -597545533\n436509256 728506920\n133368867 668789238",
"output": "3111536391798748081"
},
{
"input": "10\n-200157522 -824574736\n299208799 -287211553\n-160170880 148363130\n103709327 245344406\n482860382 547328085\n895537733 -545816336\n671947380 910981768\n-43209851 585461399\n-573679087 427675821\n151452830 27262384",
"output": "3012156378576702016"
},
{
"input": "2\n-2 -2\n-3 -3",
"output": "1"
},
{
"input": "2\n-1000 -1000\n-1100 -1100",
"output": "10000"
},
{
"input": "2\n-5 -5\n-4 -4",
"output": "1"
},
{
"input": "2\n-10 0\n-9 0",
"output": "1"
},
{
"input": "2\n-10 -10\n-20 -20",
"output": "100"
},
{
"input": "2\n-1000000 -1000000\n-100 -100",
"output": "999800010000"
},
{
"input": "2\n100000000 100000000\n200000000 200000000",
"output": "10000000000000000"
},
{
"input": "2\n-10 10\n-2 3",
"output": "64"
},
{
"input": "2\n-999999999 -999999999\n-999999991 -999999991",
"output": "64"
},
{
"input": "2\n-1000 -1000\n-999 -999",
"output": "1"
},
{
"input": "2\n-3 0\n-5 0",
"output": "4"
},
{
"input": "2\n999999999 999999999\n999999991 999999991",
"output": "64"
},
{
"input": "2\n100000012 100000012\n100000012 100000013",
"output": "1"
}
] | 1,415,844,013
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 62
| 0
|
a = int(input(""))
XlistOfCoordinates = []
YlistOfCoordinates = []
for value in range(a):
y = input("").split(" ")
XlistOfCoordinates.append(int(y[0]))
YlistOfCoordinates.append(int(y[1]))
print(max(max(XlistOfCoordinates)-min(XlistOfCoordinates),max(YlistOfCoordinates)-min(YlistOfCoordinates))**2)
|
Title: Valuable Resources
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Many computer strategy games require building cities, recruiting army, conquering tribes, collecting resources. Sometimes it leads to interesting problems.
Let's suppose that your task is to build a square city. The world map uses the Cartesian coordinates. The sides of the city should be parallel to coordinate axes. The map contains mines with valuable resources, located at some points with integer coordinates. The sizes of mines are relatively small, i.e. they can be treated as points. The city should be built in such a way that all the mines are inside or on the border of the city square.
Building a city takes large amount of money depending on the size of the city, so you have to build the city with the minimum area. Given the positions of the mines find the minimum possible area of the city.
Input Specification:
The first line of the input contains number *n* — the number of mines on the map (2<=≤<=*n*<=≤<=1000). Each of the next *n* lines contains a pair of integers *x**i* and *y**i* — the coordinates of the corresponding mine (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109). All points are pairwise distinct.
Output Specification:
Print the minimum area of the city that can cover all the mines with valuable resources.
Demo Input:
['2\n0 0\n2 2\n', '2\n0 0\n0 3\n']
Demo Output:
['4\n', '9\n']
Note:
none
|
```python
a = int(input(""))
XlistOfCoordinates = []
YlistOfCoordinates = []
for value in range(a):
y = input("").split(" ")
XlistOfCoordinates.append(int(y[0]))
YlistOfCoordinates.append(int(y[1]))
print(max(max(XlistOfCoordinates)-min(XlistOfCoordinates),max(YlistOfCoordinates)-min(YlistOfCoordinates))**2)
```
| 3
|
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,685,132,827
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 92
| 0
|
stones = input()
instructions = input()
liss_position = 0
for instruction in instructions:
if instruction == stones[liss_position]:
liss_position += 1
print(liss_position + 1)
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
stones = input()
instructions = input()
liss_position = 0
for instruction in instructions:
if instruction == stones[liss_position]:
liss_position += 1
print(liss_position + 1)
```
| 3
|
|
630
|
A
|
Again Twenty Five!
|
PROGRAMMING
| 800
|
[
"number theory"
] | null | null |
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
|
Output the last two digits of 5*n* without spaces between them.
|
[
"2\n"
] |
[
"25"
] |
none
| 0
|
[
{
"input": "2",
"output": "25"
},
{
"input": "7",
"output": "25"
},
{
"input": "1000000000000000000",
"output": "25"
},
{
"input": "2000000000000000000",
"output": "25"
},
{
"input": "987654321012345678",
"output": "25"
}
] | 1,691,402,384
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 500
| 5,222,400
|
n = int(input())
k = 5**n
s = str(k)
m = s[len(s)-2:]
print(m)
|
Title: Again Twenty Five!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
Output Specification:
Output the last two digits of 5*n* without spaces between them.
Demo Input:
['2\n']
Demo Output:
['25']
Note:
none
|
```python
n = int(input())
k = 5**n
s = str(k)
m = s[len(s)-2:]
print(m)
```
| 0
|
|
32
|
B
|
Borze
|
PROGRAMMING
| 800
|
[
"expression parsing",
"implementation"
] |
B. Borze
|
2
|
256
|
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
|
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
|
Output the decoded ternary number. It can have leading zeroes.
|
[
".-.--\n",
"--.\n",
"-..-.--\n"
] |
[
"012",
"20",
"1012"
] |
none
| 1,000
|
[
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,673,340,203
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
inp = input()
# for i in range(len(inp)):
i = 0
ans = ''
while i<len(inp):
if inp[i]==".":
ans += "0"
i += 1
elif inp[i+1]==".":
ans += "1"
i += 2
else:
ans += "2"
i += 2
print(ans)
|
Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none
|
```python
inp = input()
# for i in range(len(inp)):
i = 0
ans = ''
while i<len(inp):
if inp[i]==".":
ans += "0"
i += 1
elif inp[i+1]==".":
ans += "1"
i += 2
else:
ans += "2"
i += 2
print(ans)
```
| 3.977
|
686
|
A
|
Free Ice Cream
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
|
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
|
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
|
[
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] |
[
"22 1\n",
"3 2\n"
] |
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
| 500
|
[
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,632,995,604
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 77
| 6,758,400
|
n,x=map(int,input().split())
c=0
while n:
n-=1
a,b=input().split()
b=int(b)
if a=='+':
x+=b
else:
if x>=b:
x-=b
else:
c+=1
print(x,c)
|
Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
|
```python
n,x=map(int,input().split())
c=0
while n:
n-=1
a,b=input().split()
b=int(b)
if a=='+':
x+=b
else:
if x>=b:
x-=b
else:
c+=1
print(x,c)
```
| 3
|
|
320
|
A
|
Magic Numbers
|
PROGRAMMING
| 900
|
[
"brute force",
"greedy"
] | null | null |
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
|
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
|
Print "YES" if *n* is a magic number or print "NO" if it's not.
|
[
"114114\n",
"1111\n",
"441231\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "114114",
"output": "YES"
},
{
"input": "1111",
"output": "YES"
},
{
"input": "441231",
"output": "NO"
},
{
"input": "1",
"output": "YES"
},
{
"input": "14",
"output": "YES"
},
{
"input": "114",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "414",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "144144144",
"output": "YES"
},
{
"input": "1444",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "141414141",
"output": "YES"
},
{
"input": "11110111",
"output": "NO"
},
{
"input": "114114144",
"output": "YES"
},
{
"input": "444",
"output": "NO"
},
{
"input": "9999",
"output": "NO"
},
{
"input": "111444",
"output": "NO"
},
{
"input": "11114",
"output": "YES"
},
{
"input": "41111",
"output": "NO"
},
{
"input": "114414441",
"output": "NO"
},
{
"input": "144414441",
"output": "NO"
},
{
"input": "144244144",
"output": "NO"
},
{
"input": "111111111",
"output": "YES"
},
{
"input": "144444444",
"output": "NO"
},
{
"input": "444444444",
"output": "NO"
},
{
"input": "141441441",
"output": "YES"
},
{
"input": "441",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "14444",
"output": "NO"
},
{
"input": "11444",
"output": "NO"
},
{
"input": "144",
"output": "YES"
},
{
"input": "1414414",
"output": "YES"
},
{
"input": "141444",
"output": "NO"
},
{
"input": "14144",
"output": "YES"
},
{
"input": "4",
"output": "NO"
},
{
"input": "1144",
"output": "YES"
},
{
"input": "141111444",
"output": "NO"
},
{
"input": "14414414",
"output": "YES"
},
{
"input": "141414144",
"output": "YES"
},
{
"input": "1414",
"output": "YES"
},
{
"input": "1441",
"output": "YES"
},
{
"input": "12",
"output": "NO"
},
{
"input": "144144141",
"output": "YES"
},
{
"input": "144144",
"output": "YES"
},
{
"input": "14414411",
"output": "YES"
},
{
"input": "14414",
"output": "YES"
}
] | 1,669,876,899
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 17
| 92
| 0
|
c=input()
c=c.replace('144','')
c=c.replace('14','')
c=c.replace('1','')
print('YNEOS'[c!=''::2])
|
Title: Magic Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
Input Specification:
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
Output Specification:
Print "YES" if *n* is a magic number or print "NO" if it's not.
Demo Input:
['114114\n', '1111\n', '441231\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
c=input()
c=c.replace('144','')
c=c.replace('14','')
c=c.replace('1','')
print('YNEOS'[c!=''::2])
```
| 0
|
|
702
|
A
|
Maximum Increase
|
PROGRAMMING
| 800
|
[
"dp",
"greedy",
"implementation"
] | null | null |
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
|
The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print the maximum length of an increasing subarray of the given array.
|
[
"5\n1 7 2 11 15\n",
"6\n100 100 100 100 100 100\n",
"3\n1 2 3\n"
] |
[
"3\n",
"1\n",
"3\n"
] |
none
| 0
|
[
{
"input": "5\n1 7 2 11 15",
"output": "3"
},
{
"input": "6\n100 100 100 100 100 100",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "10\n802030518 598196518 640274071 983359971 71550121 96204862 799843967 446173607 796619138 402690754",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n1 2 3 3 4",
"output": "3"
},
{
"input": "4\n1 2 2 3",
"output": "2"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "3\n1 2 1",
"output": "2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "2"
},
{
"input": "11\n1 2 3 1 2 3 2 1 2 3 4",
"output": "4"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "9"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n3 2 1",
"output": "1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "1\n1234394",
"output": "1"
}
] | 1,669,127,607
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 93
| 8,089,600
|
def main():
solve()
def solve():
N = int(input())
numbers = list(map(int,input().split()))
maxx = tmp = 0
for i in range(N - 1):
if numbers[i] < numbers[i + 1]:
tmp += 1
if tmp > maxx: maxx = tmp
elif numbers[i] >= numbers[i + 1]:
tmp = 0
print(maxx + 1)
main()
|
Title: Maximum Increase
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
Input Specification:
The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print the maximum length of an increasing subarray of the given array.
Demo Input:
['5\n1 7 2 11 15\n', '6\n100 100 100 100 100 100\n', '3\n1 2 3\n']
Demo Output:
['3\n', '1\n', '3\n']
Note:
none
|
```python
def main():
solve()
def solve():
N = int(input())
numbers = list(map(int,input().split()))
maxx = tmp = 0
for i in range(N - 1):
if numbers[i] < numbers[i + 1]:
tmp += 1
if tmp > maxx: maxx = tmp
elif numbers[i] >= numbers[i + 1]:
tmp = 0
print(maxx + 1)
main()
```
| 3
|
|
708
|
A
|
Letters Cyclic Shift
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"greedy",
"implementation",
"strings"
] | null | null |
You are given a non-empty string *s* consisting of lowercase English letters. You have to pick exactly one non-empty substring of *s* and shift all its letters 'z' 'y' 'x' 'b' 'a' 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.
What is the lexicographically minimum string that can be obtained from *s* by performing this shift exactly once?
|
The only line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) consisting of lowercase English letters.
|
Print the lexicographically minimum string that can be obtained from *s* by shifting letters of exactly one non-empty substring.
|
[
"codeforces\n",
"abacaba\n"
] |
[
"bncdenqbdr\n",
"aaacaba\n"
] |
String *s* is lexicographically smaller than some other string *t* of the same length if there exists some 1 ≤ *i* ≤ |*s*|, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, and *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
| 500
|
[
{
"input": "codeforces",
"output": "bncdenqbdr"
},
{
"input": "abacaba",
"output": "aaacaba"
},
{
"input": "babbbabaababbaa",
"output": "aabbbabaababbaa"
},
{
"input": "bcbacaabcababaccccaaaabacbbcbbaa",
"output": "abaacaabcababaccccaaaabacbbcbbaa"
},
{
"input": "cabaccaacccabaacdbdcbcdbccbccbabbdadbdcdcdbdbcdcdbdadcbcda",
"output": "babaccaacccabaacdbdcbcdbccbccbabbdadbdcdcdbdbcdcdbdadcbcda"
},
{
"input": "a",
"output": "z"
},
{
"input": "eeeedddccbceaabdaecaebaeaecccbdeeeaadcecdbeacecdcdcceabaadbcbbadcdaeddbcccaaeebccecaeeeaebcaaccbdaccbdcadadaaeacbbdcbaeeaecedeeeedadec",
"output": "ddddcccbbabdaabdaecaebaeaecccbdeeeaadcecdbeacecdcdcceabaadbcbbadcdaeddbcccaaeebccecaeeeaebcaaccbdaccbdcadadaaeacbbdcbaeeaecedeeeedadec"
},
{
"input": "fddfbabadaadaddfbfecadfaefaefefabcccdbbeeabcbbddefbafdcafdfcbdffeeaffcaebbbedabddeaecdddffcbeaafffcddccccfffdbcddcfccefafdbeaacbdeeebdeaaacdfdecadfeafaeaefbfdfffeeaefebdceebcebbfeaccfafdccdcecedeedadcadbfefccfdedfaaefabbaeebdebeecaadbebcfeafbfeeefcfaecadfe",
"output": "ecceaabadaadaddfbfecadfaefaefefabcccdbbeeabcbbddefbafdcafdfcbdffeeaffcaebbbedabddeaecdddffcbeaafffcddccccfffdbcddcfccefafdbeaacbdeeebdeaaacdfdecadfeafaeaefbfdfffeeaefebdceebcebbfeaccfafdccdcecedeedadcadbfefccfdedfaaefabbaeebdebeecaadbebcfeafbfeeefcfaecadfe"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaz"
},
{
"input": "abbabaaaaa",
"output": "aaaabaaaaa"
},
{
"input": "bbbbbbbbbbbb",
"output": "aaaaaaaaaaaa"
},
{
"input": "aabaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaz"
},
{
"input": "abaabaaaaaabbaaaaaaabaaaaaaaaabaaaabaaaaaaabaaaaaaaaaabaaaaaaaaaaaaaaabaaaabbaaaaabaaaaaaaabaaaaaaaa",
"output": "aaaabaaaaaabbaaaaaaabaaaaaaaaabaaaabaaaaaaabaaaaaaaaaabaaaaaaaaaaaaaaabaaaabbaaaaabaaaaaaaabaaaaaaaa"
},
{
"input": "abbbbbbbabbbbbbbbbbbbbbbbbbbbbbbabbabbbbbabbbbbbbbbbbabbbbbbbbabbabbbbbbbbbbbbbbabbabbbaababbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbbabbbbbbbbbbbbbbbabbbbbbbbbaababbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbabbbbbaabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbaabbbbbbbbbbbbababbabbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbabbbbbbbabbbbbbb",
"output": "aaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbabbabbbbbabbbbbbbbbbbabbbbbbbbabbabbbbbbbbbbbbbbabbabbbaababbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbbabbbbbbbbbbbbbbbabbbbbbbbbaababbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbabbbbbaabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbaabbbbbbbbbbbbababbabbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbabbbbbbbabbbbbbb"
},
{
"input": "aaaaa",
"output": "aaaaz"
},
{
"input": "aaa",
"output": "aaz"
},
{
"input": "aa",
"output": "az"
}
] | 1,553,120,795
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 54
| 155
| 6,860,800
|
s = input().split('a')
n = len(s)
for i in range(n):
if s[i]:
s[i] = ''.join(chr(ord(ch) - 1) for ch in s[i])
print('a'.join(s))
break
else:
print('a' * (n - 2) + 'z')
|
Title: Letters Cyclic Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a non-empty string *s* consisting of lowercase English letters. You have to pick exactly one non-empty substring of *s* and shift all its letters 'z' 'y' 'x' 'b' 'a' 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.
What is the lexicographically minimum string that can be obtained from *s* by performing this shift exactly once?
Input Specification:
The only line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) consisting of lowercase English letters.
Output Specification:
Print the lexicographically minimum string that can be obtained from *s* by shifting letters of exactly one non-empty substring.
Demo Input:
['codeforces\n', 'abacaba\n']
Demo Output:
['bncdenqbdr\n', 'aaacaba\n']
Note:
String *s* is lexicographically smaller than some other string *t* of the same length if there exists some 1 ≤ *i* ≤ |*s*|, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, and *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
|
```python
s = input().split('a')
n = len(s)
for i in range(n):
if s[i]:
s[i] = ''.join(chr(ord(ch) - 1) for ch in s[i])
print('a'.join(s))
break
else:
print('a' * (n - 2) + 'z')
```
| 3
|
|
394
|
A
|
Counting Sticks
|
PROGRAMMING
| 0
|
[
"brute force",
"implementation"
] | null | null |
When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us?
|
The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100.
|
If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples.
|
[
"||+|=|||||\n",
"|||||+||=||\n",
"|+|=||||||\n",
"||||+||=||||||\n"
] |
[
"|||+|=||||\n",
"Impossible\n",
"Impossible\n",
"||||+||=||||||\n"
] |
In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks.
| 500
|
[
{
"input": "||+|=|||||",
"output": "|||+|=||||"
},
{
"input": "|||||+||=||",
"output": "Impossible"
},
{
"input": "|+|=||||||",
"output": "Impossible"
},
{
"input": "||||+||=||||||",
"output": "||||+||=||||||"
},
{
"input": "||||||||||||+|||||||||||=||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||+||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||=|||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "|+|=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+|=|",
"output": "|+|=||"
},
{
"input": "|+||=|",
"output": "|+|=||"
},
{
"input": "|+|=||",
"output": "|+|=||"
},
{
"input": "|||+|=|",
"output": "Impossible"
},
{
"input": "|||+|=|",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|+|=|||",
"output": "Impossible"
},
{
"input": "|+|=||||",
"output": "||+|=|||"
},
{
"input": "|+||=|",
"output": "|+|=||"
},
{
"input": "|+||||||=|||||",
"output": "|+|||||=||||||"
},
{
"input": "|+|||=||",
"output": "|+||=|||"
},
{
"input": "|+||||=|||",
"output": "|+|||=||||"
},
{
"input": "|+|||||=||||",
"output": "|+||||=|||||"
},
{
"input": "||+||=||",
"output": "|+||=|||"
},
{
"input": "||+|||=|||",
"output": "|+|||=||||"
},
{
"input": "|||||+||=||||||||||",
"output": "Impossible"
},
{
"input": "|||+|||=||||",
"output": "||+|||=|||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "|||||+|||||=||||||||||||",
"output": "||||||+|||||=|||||||||||"
},
{
"input": "||+|||||||=|||||||",
"output": "|+|||||||=||||||||"
},
{
"input": "||||+||||=||||||",
"output": "|||+||||=|||||||"
},
{
"input": "||||+|||=|||||",
"output": "|||+|||=||||||"
},
{
"input": "||+|||=|||||||",
"output": "|||+|||=||||||"
},
{
"input": "||+|=|",
"output": "|+|=||"
},
{
"input": "|||+|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "||+|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+|=||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
}
] | 1,553,450,522
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 155
| 819,200
|
g = input()
import re
m = re.match("(\|+)\+(\|+)=(\|+)",g)
z = m.group
a = len(z(1))
b = len(z(2))
c = len(z(3))
if a + b == c - 2:
print('|' * (a + 1) + '+' + '|' * b + '=' + '|' * (c - 1))
elif a + b == c + 2:
if a == 1:
print('|' * a + '+' + '|' * (b - 1) + '=' + '|' * (c + 1))
else:
print('|' * (a - 1) + '+' + '|' * b + '=' + '|' * (c + 1))
elif a + b == c:
print(g)
else:
print("Impossible")
|
Title: Counting Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us?
Input Specification:
The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100.
Output Specification:
If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples.
Demo Input:
['||+|=|||||\n', '|||||+||=||\n', '|+|=||||||\n', '||||+||=||||||\n']
Demo Output:
['|||+|=||||\n', 'Impossible\n', 'Impossible\n', '||||+||=||||||\n']
Note:
In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks.
|
```python
g = input()
import re
m = re.match("(\|+)\+(\|+)=(\|+)",g)
z = m.group
a = len(z(1))
b = len(z(2))
c = len(z(3))
if a + b == c - 2:
print('|' * (a + 1) + '+' + '|' * b + '=' + '|' * (c - 1))
elif a + b == c + 2:
if a == 1:
print('|' * a + '+' + '|' * (b - 1) + '=' + '|' * (c + 1))
else:
print('|' * (a - 1) + '+' + '|' * b + '=' + '|' * (c + 1))
elif a + b == c:
print(g)
else:
print("Impossible")
```
| 3
|
|
20
|
C
|
Dijkstra?
|
PROGRAMMING
| 1,900
|
[
"graphs",
"shortest paths"
] |
C. Dijkstra?
|
1
|
64
|
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
|
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
|
[
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] |
[
"1 4 3 5 ",
"1 4 3 5 "
] |
none
| 1,500
|
[
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input": "3 3\n1 2 1\n1 3 2\n2 3 1",
"output": "1 3 "
},
{
"input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173",
"output": "1 5 8 7 3 10 "
},
{
"input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157",
"output": "1 8 10 "
},
{
"input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135",
"output": "1 9 5 8 10 "
},
{
"input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242",
"output": "1 4 6 10 "
}
] | 1,611,189,747
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 77
| 102,400
|
from collections import defaultdict
# Filling the graph.
V, E = map(int, input().split())
graph = defaultdict(list)
for _ in range(E):
a, b, w = map(int, input().split())
graph[a].append((b, w))
graph[b].append((a, w))
# Distance from the root to every other vertex in G.
# Distance from the root to itself is 1.
d = [float('inf')] * (V + 1)
d[1] = 0
# Visited vertices.
visited = [False] * (V + 1)
# Dijkstra algorithm.
for _ in range(1, V + 1):
mn = float('inf')
a = 1
for i in range(1, V + 1):
if d[i] < mn and not visited[i]:
mn = d[i]
a = i
visited[a] = True
neighbors = graph[a]
for u, w in neighbors:
d[u] = min(d[u], d[a] + w)
# Printing the answer.
ans = []
for i in range(2, V + 1):
if d[i] == float('inf'):
ans.append(-1)
else:
ans.append(d[i])
print(*ans)
# Testing.
# Answer should be 3 7 1 2
'''
5 7
1 2 6
1 4 1
2 3 5
2 4 2
2 5 2
3 5 5
4 5 1
'''
|
Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none
|
```python
from collections import defaultdict
# Filling the graph.
V, E = map(int, input().split())
graph = defaultdict(list)
for _ in range(E):
a, b, w = map(int, input().split())
graph[a].append((b, w))
graph[b].append((a, w))
# Distance from the root to every other vertex in G.
# Distance from the root to itself is 1.
d = [float('inf')] * (V + 1)
d[1] = 0
# Visited vertices.
visited = [False] * (V + 1)
# Dijkstra algorithm.
for _ in range(1, V + 1):
mn = float('inf')
a = 1
for i in range(1, V + 1):
if d[i] < mn and not visited[i]:
mn = d[i]
a = i
visited[a] = True
neighbors = graph[a]
for u, w in neighbors:
d[u] = min(d[u], d[a] + w)
# Printing the answer.
ans = []
for i in range(2, V + 1):
if d[i] == float('inf'):
ans.append(-1)
else:
ans.append(d[i])
print(*ans)
# Testing.
# Answer should be 3 7 1 2
'''
5 7
1 2 6
1 4 1
2 3 5
2 4 2
2 5 2
3 5 5
4 5 1
'''
```
| 0
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,653,465,157
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
from math import floor
n,m = map(int,input().split())
print(floor((n*m)/2))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
from math import floor
n,m = map(int,input().split())
print(floor((n*m)/2))
```
| 3.977
|
446
|
A
|
DZY Loves Sequences
|
PROGRAMMING
| 1,600
|
[
"dp",
"implementation",
"two pointers"
] | null | null |
DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
In a single line print the answer to the problem — the maximum length of the required subsegment.
|
[
"6\n7 2 3 1 5 6\n"
] |
[
"5\n"
] |
You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4.
| 500
|
[
{
"input": "6\n7 2 3 1 5 6",
"output": "5"
},
{
"input": "10\n424238336 649760493 681692778 714636916 719885387 804289384 846930887 957747794 596516650 189641422",
"output": "9"
},
{
"input": "50\n804289384 846930887 681692778 714636916 957747794 424238336 719885387 649760493 596516650 189641422 25202363 350490028 783368691 102520060 44897764 967513927 365180541 540383427 304089173 303455737 35005212 521595369 294702568 726956430 336465783 861021531 59961394 89018457 101513930 125898168 131176230 145174068 233665124 278722863 315634023 369133070 468703136 628175012 635723059 653377374 656478043 801979803 859484422 914544920 608413785 756898538 734575199 973594325 149798316 38664371",
"output": "19"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "2"
},
{
"input": "5\n1 2 3 4 1",
"output": "5"
},
{
"input": "10\n1 2 3 4 5 5 6 7 8 9",
"output": "6"
},
{
"input": "5\n1 1 1 1 1",
"output": "2"
},
{
"input": "5\n1 1 2 3 4",
"output": "5"
},
{
"input": "5\n1 2 3 1 6",
"output": "5"
},
{
"input": "1\n42",
"output": "1"
},
{
"input": "5\n1 2 42 3 4",
"output": "4"
},
{
"input": "5\n1 5 9 6 10",
"output": "4"
},
{
"input": "5\n5 2 3 4 5",
"output": "5"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "5\n1 2 3 3 4",
"output": "4"
},
{
"input": "8\n1 2 3 4 1 5 6 7",
"output": "5"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "3\n5 1 2",
"output": "3"
},
{
"input": "4\n1 4 3 4",
"output": "4"
},
{
"input": "6\n7 2 12 4 5 6",
"output": "5"
},
{
"input": "6\n7 2 3 1 4 5",
"output": "4"
},
{
"input": "6\n2 3 5 5 6 7",
"output": "6"
},
{
"input": "5\n2 4 7 6 8",
"output": "5"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n1 1 2",
"output": "3"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "5\n4 1 2 3 4",
"output": "5"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6",
"output": "7"
},
{
"input": "4\n1 2 1 3",
"output": "3"
},
{
"input": "4\n4 3 1 2",
"output": "3"
},
{
"input": "6\n1 2 2 3 4 5",
"output": "5"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "4\n5 1 2 3",
"output": "4"
},
{
"input": "5\n9 1 2 3 4",
"output": "5"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "5\n1 3 2 4 5",
"output": "4"
},
{
"input": "6\n1 2 1 2 4 5",
"output": "5"
},
{
"input": "10\n1 1 5 3 2 9 9 7 7 6",
"output": "3"
},
{
"input": "6\n1 2 3 100000 100 101",
"output": "6"
},
{
"input": "4\n3 3 3 4",
"output": "3"
},
{
"input": "3\n4 3 5",
"output": "3"
},
{
"input": "5\n1 3 2 3 4",
"output": "4"
},
{
"input": "10\n1 2 3 4 5 10 10 11 12 13",
"output": "10"
},
{
"input": "7\n11 2 1 2 13 4 14",
"output": "5"
},
{
"input": "3\n5 1 3",
"output": "3"
},
{
"input": "4\n1 5 3 4",
"output": "4"
},
{
"input": "10\n1 2 3 4 100 6 7 8 9 10",
"output": "10"
},
{
"input": "3\n5 3 5",
"output": "3"
},
{
"input": "5\n100 100 7 8 9",
"output": "4"
},
{
"input": "5\n1 2 3 4 5",
"output": "5"
},
{
"input": "5\n1 2 4 4 5",
"output": "5"
},
{
"input": "6\n7 4 5 6 7 8",
"output": "6"
},
{
"input": "9\n3 4 1 6 3 4 5 6 7",
"output": "7"
},
{
"input": "3\n1000 1 2",
"output": "3"
},
{
"input": "3\n20 1 9",
"output": "3"
},
{
"input": "6\n7 2 3 1 4 6",
"output": "4"
},
{
"input": "3\n100 5 10",
"output": "3"
},
{
"input": "4\n2 2 2 3",
"output": "3"
},
{
"input": "6\n4 2 8 1 2 5",
"output": "4"
},
{
"input": "3\n25 1 6",
"output": "3"
},
{
"input": "10\n17 99 23 72 78 36 5 43 95 9",
"output": "5"
},
{
"input": "7\n21 16 22 21 11 13 19",
"output": "4"
},
{
"input": "5\n1 2 5 3 4",
"output": "4"
},
{
"input": "6\n2 2 2 3 4 5",
"output": "5"
},
{
"input": "5\n1 3 1 2 3",
"output": "4"
},
{
"input": "3\n81 33 64",
"output": "3"
},
{
"input": "7\n14 3 3 19 13 19 15",
"output": "4"
},
{
"input": "9\n1 2 3 4 5 42 7 8 9",
"output": "9"
},
{
"input": "5\n2 3 7 5 6",
"output": "5"
},
{
"input": "5\n1 3 3 4 5",
"output": "5"
},
{
"input": "6\n1 5 4 3 4 5",
"output": "4"
}
] | 1,591,722,102
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 93
| 307,200
|
def sol():
n = int(input())
array = list(map(int, input().split(" ")))
array.append(array[-2] + array[-1])
array.insert(0,0)
count_a = 1
count_b = 0
count_c = 0
g = 0
w = 0
count_ans = 0
i = 2
while i <= n:
if array[i] > array[i-1]:
count_a += 1
i += 1
elif array[i] <= array[i-1] and count_c == 0:
if array[i] - array[i - 2] >= 2 or array[i+1] - array[i - 1] >= 2:
if array[i+1] - array[i - 1] >= 2:
array[i] = (array[i+1] + array[i - 1]) // 2
count_b = count_a+1
count_a = 0
i += 1
else:
array[i-1] = (array[i -2] + array[i]) // 2
count_b = count_a
count_a = 0
else:
count_ans = max(count_ans, count_a + 1, count_b+count_a)
array[i] = 1
array[i-1] = array[i+1]
count_a = 1
count_b = 0
i += 1
elif array[i] <= array[i-1] and count_c == 1:
count_ans = max(count_ans, count_a+count_b)
count_b = count_a
count_a = 1
count_c = 0
print(max(count_ans, count_b+count_a))
sol()
|
Title: DZY Loves Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
In a single line print the answer to the problem — the maximum length of the required subsegment.
Demo Input:
['6\n7 2 3 1 5 6\n']
Demo Output:
['5\n']
Note:
You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4.
|
```python
def sol():
n = int(input())
array = list(map(int, input().split(" ")))
array.append(array[-2] + array[-1])
array.insert(0,0)
count_a = 1
count_b = 0
count_c = 0
g = 0
w = 0
count_ans = 0
i = 2
while i <= n:
if array[i] > array[i-1]:
count_a += 1
i += 1
elif array[i] <= array[i-1] and count_c == 0:
if array[i] - array[i - 2] >= 2 or array[i+1] - array[i - 1] >= 2:
if array[i+1] - array[i - 1] >= 2:
array[i] = (array[i+1] + array[i - 1]) // 2
count_b = count_a+1
count_a = 0
i += 1
else:
array[i-1] = (array[i -2] + array[i]) // 2
count_b = count_a
count_a = 0
else:
count_ans = max(count_ans, count_a + 1, count_b+count_a)
array[i] = 1
array[i-1] = array[i+1]
count_a = 1
count_b = 0
i += 1
elif array[i] <= array[i-1] and count_c == 1:
count_ans = max(count_ans, count_a+count_b)
count_b = count_a
count_a = 1
count_c = 0
print(max(count_ans, count_b+count_a))
sol()
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,694,507,285
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 0
|
str_1=input()
str_2=input()
n=len(str_1)
a=''
for i in range(n):
if str_1[i]==str_2[i]:
a+='0'
else:
a+='1'
print(a)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
str_1=input()
str_2=input()
n=len(str_1)
a=''
for i in range(n):
if str_1[i]==str_2[i]:
a+='0'
else:
a+='1'
print(a)
```
| 3.9885
|
794
|
A
|
Bank Robbery
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes.
Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg.
The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather.
|
The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively.
The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes.
The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct.
|
Output a single integer: the maximum number of banknotes Oleg can take.
|
[
"5 3 7\n8\n4 7 5 5 3 6 2 8\n",
"6 5 7\n5\n1 5 7 92 3\n"
] |
[
"4\n",
"0\n"
] |
In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes.
For the second sample, Oleg can't take any banknotes without bumping into any of the security guards.
| 500
|
[
{
"input": "5 3 7\n8\n4 7 5 5 3 6 2 8",
"output": "4"
},
{
"input": "6 5 7\n5\n1 5 7 92 3",
"output": "0"
},
{
"input": "3 2 4\n1\n3",
"output": "1"
},
{
"input": "5 3 8\n12\n8 3 4 5 7 6 8 3 5 4 7 6",
"output": "8"
},
{
"input": "7 3 10\n5\n3 3 3 3 3",
"output": "0"
},
{
"input": "3 2 5\n4\n1 3 4 5",
"output": "2"
},
{
"input": "3 2 4\n1\n1",
"output": "0"
},
{
"input": "6 4 8\n1\n4",
"output": "0"
},
{
"input": "2 1 3\n1\n3",
"output": "0"
}
] | 1,621,525,828
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 124
| 8,704,000
|
a,b,c=map(int,input().split())
n=int(input())
l=list(map(int,input().split()))
x=0
for i in l:
# print(i>c,,i,c)
if i>b and i<c:
x+=1
print(x)
|
Title: Bank Robbery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes.
Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg.
The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather.
Input Specification:
The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively.
The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes.
The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct.
Output Specification:
Output a single integer: the maximum number of banknotes Oleg can take.
Demo Input:
['5 3 7\n8\n4 7 5 5 3 6 2 8\n', '6 5 7\n5\n1 5 7 92 3\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes.
For the second sample, Oleg can't take any banknotes without bumping into any of the security guards.
|
```python
a,b,c=map(int,input().split())
n=int(input())
l=list(map(int,input().split()))
x=0
for i in l:
# print(i>c,,i,c)
if i>b and i<c:
x+=1
print(x)
```
| 3
|
|
580
|
A
|
Kefa and First Steps
|
PROGRAMMING
| 900
|
[
"brute force",
"dp",
"implementation"
] | null | null |
Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.
Help Kefa cope with this task!
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*.
|
[
"6\n2 2 1 3 4 1\n",
"3\n2 2 9\n"
] |
[
"3",
"3"
] |
In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.
In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
| 750
|
[
{
"input": "6\n2 2 1 3 4 1",
"output": "3"
},
{
"input": "3\n2 2 9",
"output": "3"
},
{
"input": "5\n10 100 111 1 2",
"output": "3"
},
{
"input": "10\n1 2 3 4 1 2 3 4 5 6",
"output": "6"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "50"
},
{
"input": "100\n1 838 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 605 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 27 533 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 835 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 992",
"output": "42"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "6\n5 4 3 2 1 2",
"output": "2"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "9"
},
{
"input": "7\n99 100 1 2 3 4 5",
"output": "5"
},
{
"input": "5\n3 3 1 2 3",
"output": "3"
},
{
"input": "1\n100",
"output": "1"
},
{
"input": "1\n5",
"output": "1"
}
] | 1,687,567,868
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 93
| 13,926,400
|
n=int(input())
a=list(map(int,input().split()))
pre=cnt=m=0
for i in a:
cnt=cnt+1 if i>=pre else 1
m=max(m,cnt)
pre=i
print(m)
|
Title: Kefa and First Steps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kefa decided to make some money doing business on the Internet for exactly *n* days. He knows that on the *i*-th day (1<=≤<=*i*<=≤<=*n*) he makes *a**i* money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence *a**i*. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.
Help Kefa cope with this task!
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the length of the maximum non-decreasing subsegment of sequence *a*.
Demo Input:
['6\n2 2 1 3 4 1\n', '3\n2 2 9\n']
Demo Output:
['3', '3']
Note:
In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.
In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
|
```python
n=int(input())
a=list(map(int,input().split()))
pre=cnt=m=0
for i in a:
cnt=cnt+1 if i>=pre else 1
m=max(m,cnt)
pre=i
print(m)
```
| 3
|
|
439
|
A
|
Devu, the Singer and Churu, the Joker
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
|
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
|
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
|
[
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] |
[
"5\n",
"-1\n"
] |
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
| 500
|
[
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 20 9 14 14 10 13 6 13 14 17 6 8 20 12 10 15 13 17 5 12 13 11 7 5 5 2 3 15 13 7 14 14 19 2 13 14 5 15 3 19 15 16 4 1",
"output": "1891"
},
{
"input": "100 9000\n5 2 3 1 1 3 4 9 9 6 7 10 10 10 2 10 6 8 8 6 7 9 9 5 6 2 1 10 10 9 4 5 9 2 4 3 8 5 6 1 1 5 3 6 2 6 6 6 5 8 3 6 7 3 1 10 9 1 8 3 10 9 5 6 3 4 1 1 10 10 2 3 4 8 10 10 5 1 5 3 6 8 10 6 10 2 1 8 10 1 7 6 9 10 5 2 3 5 3 2",
"output": "1688"
},
{
"input": "100 8007\n5 19 14 18 9 6 15 8 1 14 11 20 3 17 7 12 2 6 3 17 7 20 1 14 20 17 2 10 13 7 18 18 9 10 16 8 1 11 11 9 13 18 9 20 12 12 7 15 12 17 11 5 11 15 9 2 15 1 18 3 18 16 15 4 10 5 18 13 13 12 3 8 17 2 12 2 13 3 1 13 2 4 9 10 18 10 14 4 4 17 12 19 2 9 6 5 5 20 18 12",
"output": "1391"
},
{
"input": "39 2412\n1 1 1 1 1 1 26 1 1 1 99 1 1 1 1 1 1 1 1 1 1 88 7 1 1 1 1 76 1 1 1 93 40 1 13 1 68 1 32",
"output": "368"
},
{
"input": "39 2617\n47 1 1 1 63 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 70 1 99 63 1 1 1 1 1 1 1 1 64 1 1",
"output": "435"
},
{
"input": "39 3681\n83 77 1 94 85 47 1 98 29 16 1 1 1 71 96 85 31 97 96 93 40 50 98 1 60 51 1 96 100 72 1 1 1 89 1 93 1 92 100",
"output": "326"
},
{
"input": "45 894\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 28 28 1 1 1 1 1 1 1 1 1 1 1 1 1 1 99 3 1 1",
"output": "139"
},
{
"input": "45 4534\n1 99 65 99 4 46 54 80 51 30 96 1 28 30 44 70 78 1 1 100 1 62 1 1 1 85 1 1 1 61 1 46 75 1 61 77 97 26 67 1 1 63 81 85 86",
"output": "514"
},
{
"input": "72 3538\n52 1 8 1 1 1 7 1 1 1 1 48 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 40 1 1 38 1 1 1 1 1 1 1 1 1 1 1 35 1 93 79 1 1 1 1 1 1 1 1 1 51 1 1 1 1 1 1 1 1 1 1 1 1 96 1",
"output": "586"
},
{
"input": "81 2200\n1 59 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 93 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 50 1 1 1 1 1 1 1 1 1 1 1",
"output": "384"
},
{
"input": "81 2577\n85 91 1 1 2 1 1 100 1 80 1 1 17 86 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 37 1 66 24 1 1 96 49 1 66 1 44 1 1 1 1 98 1 1 1 1 35 1 37 3 35 1 1 87 64 1 24 1 58 1 1 42 83 5 1 1 1 1 1 95 1 94 1 50 1 1",
"output": "174"
},
{
"input": "81 4131\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "807"
},
{
"input": "81 6315\n1 1 67 100 1 99 36 1 92 5 1 96 42 12 1 57 91 1 1 66 41 30 74 95 1 37 1 39 91 69 1 52 77 47 65 1 1 93 96 74 90 35 85 76 71 92 92 1 1 67 92 74 1 1 86 76 35 1 56 16 27 57 37 95 1 40 20 100 51 1 80 60 45 79 95 1 46 1 25 100 96",
"output": "490"
},
{
"input": "96 1688\n1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 71 1 1 1 30 1 1 1",
"output": "284"
},
{
"input": "96 8889\n1 1 18 1 1 1 1 1 1 1 1 1 99 1 1 1 1 88 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 96 1 1 1 1 21 1 1 1 1 1 1 1 73 1 1 1 1 1 10 1 1 1 1 1 1 1 46 43 1 1 1 1 1 98 1 1 1 1 1 1 6 1 1 1 1 1 74 1 25 1 55 1 1 1 13 1 1 54 1 1 1",
"output": "1589"
},
{
"input": "10 100\n1 1 1 1 1 1 1 1 1 1",
"output": "18"
},
{
"input": "100 10000\n54 46 72 94 79 83 91 54 73 3 24 55 54 31 28 20 19 6 25 19 47 23 1 70 15 87 51 39 54 77 55 5 60 3 15 99 56 88 22 78 79 21 38 27 28 86 7 88 12 59 55 70 25 1 70 49 1 45 69 72 50 17 4 56 8 100 90 34 35 20 61 76 88 79 4 74 65 68 75 26 40 72 59 94 10 67 96 85 29 90 47 24 44 1 66 93 55 36 1 99",
"output": "1017"
},
{
"input": "100 6000\n41 31 23 17 24 78 26 96 93 48 46 2 49 33 35 9 73 100 34 48 83 36 33 69 43 24 3 74 8 81 27 33 94 38 77 9 76 90 62 90 21 67 22 22 12 2 17 27 61 18 72 85 59 65 71 38 90 75 74 66 60 47 58 50 90 95 75 10 5 100 97 29 83 88 65 26 93 90 22 98 36 55 70 38 50 92 88 72 99 96 25 14 74 16 25 92 67 94 77 96",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 6\n1",
"output": "1"
},
{
"input": "1 5\n1",
"output": "0"
},
{
"input": "1 3\n4",
"output": "-1"
},
{
"input": "3 24\n2 1 2",
"output": "-1"
}
] | 1,600,326,898
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 0
|
x, y=map(int, input().split())
p=list(map(int, input().split()))
jokes=(d-sum(p)-10*(n-1))//5
if time<=d:
print(jokes+2*(n-1))
else:
print(-1)
|
Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
|
```python
x, y=map(int, input().split())
p=list(map(int, input().split()))
jokes=(d-sum(p)-10*(n-1))//5
if time<=d:
print(jokes+2*(n-1))
else:
print(-1)
```
| -1
|
|
389
|
B
|
Fox and Cross
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Fox Ciel has a board with *n* rows and *n* columns. So, the board consists of *n*<=×<=*n* cells. Each cell contains either a symbol '.', or a symbol '#'.
A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.
Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.
Please, tell Ciel if she can draw the crosses in the described way.
|
The first line contains an integer *n* (3<=≤<=*n*<=≤<=100) — the size of the board.
Each of the next *n* lines describes one row of the board. The *i*-th line describes the *i*-th row of the board and consists of *n* characters. Each character is either a symbol '.', or a symbol '#'.
|
Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".
|
[
"5\n.#...\n####.\n.####\n...#.\n.....\n",
"4\n####\n####\n####\n####\n",
"6\n.#....\n####..\n.####.\n.#.##.\n######\n.#..#.\n",
"6\n.#..#.\n######\n.####.\n.####.\n######\n.#..#.\n",
"3\n...\n...\n...\n"
] |
[
"YES\n",
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] |
In example 1, you can draw two crosses. The picture below shows what they look like.
In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.
| 1,000
|
[
{
"input": "4\n####\n####\n####\n####",
"output": "NO"
},
{
"input": "6\n.#....\n####..\n.####.\n.#.##.\n######\n.#..#.",
"output": "YES"
},
{
"input": "6\n.#..#.\n######\n.####.\n.####.\n######\n.#..#.",
"output": "NO"
},
{
"input": "5\n.....\n.#...\n####.\n.####\n...#.",
"output": "YES"
},
{
"input": "5\n#....\n###..\n.###.\n..##.\n..##.",
"output": "NO"
},
{
"input": "10\n....#...#.\n.#.###.###\n#####.#.#.\n.#######..\n.#.##.#.#.\n######.###\n.#..#.#.#.\n.#..#####.\n##########\n.#..#...#.",
"output": "YES"
},
{
"input": "8\n.#...##.\n##.##.#.\n#....##.\n#.###.##\n...##..#\n#..#.##.\n#..#.#..\n#....###",
"output": "NO"
},
{
"input": "15\n.#...##.##.....\n.###.#.#.##..#.\n##.#####......#\n#....###.#..#..\n#.##.#.###.###.\n.###....#.###.#\n##.....##..#...\n.#.########...#\n##..#..##..#.##\n#...#....##....\n.......#.......\n.#..#.#.####...\n.#..#...#..###.\n#..#.#..#......\n###.....##....#",
"output": "NO"
},
{
"input": "7\n.......\n.......\n.......\n.......\n.....#.\n....###\n.....#.",
"output": "YES"
},
{
"input": "9\n.#.#....#\n#.#.....#\n.#..#..#.\n#.#....##\n...#...#.\n#..###.#.\n....#..##\n.#.###...\n....##..#",
"output": "NO"
},
{
"input": "5\n#...#\n.....\n..#..\n.....\n#...#",
"output": "NO"
},
{
"input": "3\n.#.\n#.#\n.#.",
"output": "NO"
},
{
"input": "3\n...\n...\n..#",
"output": "NO"
}
] | 1,565,107,533
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 42
| 140
| 1,843,200
|
def cross(i, j, v, n):
v[i*n+j] = '.'
if v[(i+1)*n+j] == '#':
v[(i+1)*n+j] = '.'
if v[(i+2)*n+j] == '#':
v[(i+2)*n+j] = '.'
if v[(i+1)*n+j+1] == '#':
v[(i+1)*n+j+1] = '.'
if v[(i+1)*n+j-1] == '#':
v[(i+1)*n+j-1] = '.'
return True
else:
return False
else:
return False
else:
return False
else:
return False
n = int(input())
v = ['.'] * n * n
ans = True
for i in range(n):
s = input()
for j in range(n):
v[i*n+j] = s[j]
for i in range(n):
if ans:
for j in range(n):
if v[i*n+j] == '#':
if j == 0 or j == (n - 1) or i > (n - 3):
ans = False
break
else:
ans = cross(i, j, v, n)
else:
break
if ans:
print('YES')
else:
print('NO')
|
Title: Fox and Cross
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has a board with *n* rows and *n* columns. So, the board consists of *n*<=×<=*n* cells. Each cell contains either a symbol '.', or a symbol '#'.
A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.
Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.
Please, tell Ciel if she can draw the crosses in the described way.
Input Specification:
The first line contains an integer *n* (3<=≤<=*n*<=≤<=100) — the size of the board.
Each of the next *n* lines describes one row of the board. The *i*-th line describes the *i*-th row of the board and consists of *n* characters. Each character is either a symbol '.', or a symbol '#'.
Output Specification:
Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".
Demo Input:
['5\n.#...\n####.\n.####\n...#.\n.....\n', '4\n####\n####\n####\n####\n', '6\n.#....\n####..\n.####.\n.#.##.\n######\n.#..#.\n', '6\n.#..#.\n######\n.####.\n.####.\n######\n.#..#.\n', '3\n...\n...\n...\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In example 1, you can draw two crosses. The picture below shows what they look like.
In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.
|
```python
def cross(i, j, v, n):
v[i*n+j] = '.'
if v[(i+1)*n+j] == '#':
v[(i+1)*n+j] = '.'
if v[(i+2)*n+j] == '#':
v[(i+2)*n+j] = '.'
if v[(i+1)*n+j+1] == '#':
v[(i+1)*n+j+1] = '.'
if v[(i+1)*n+j-1] == '#':
v[(i+1)*n+j-1] = '.'
return True
else:
return False
else:
return False
else:
return False
else:
return False
n = int(input())
v = ['.'] * n * n
ans = True
for i in range(n):
s = input()
for j in range(n):
v[i*n+j] = s[j]
for i in range(n):
if ans:
for j in range(n):
if v[i*n+j] == '#':
if j == 0 or j == (n - 1) or i > (n - 3):
ans = False
break
else:
ans = cross(i, j, v, n)
else:
break
if ans:
print('YES')
else:
print('NO')
```
| 3
|
|
551
|
A
|
GukiZ and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
|
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
|
[
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] |
[
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] |
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
| 500
|
[
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4 4 4 2"
},
{
"input": "1\n2000",
"output": "1"
},
{
"input": "2\n2000 2000",
"output": "1 1"
},
{
"input": "3\n500 501 502",
"output": "3 2 1"
},
{
"input": "10\n105 106 1 1 1 11 1000 999 1000 999",
"output": "6 5 8 8 8 7 1 3 1 3"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "6 5 4 3 2 1"
},
{
"input": "7\n6 5 4 3 2 1 1",
"output": "1 2 3 4 5 6 6"
},
{
"input": "8\n153 100 87 14 10 8 6 5",
"output": "1 2 3 4 5 6 7 8"
},
{
"input": "70\n11 54 37 62 1 46 13 17 38 47 28 15 63 5 61 34 49 66 32 59 3 41 58 28 23 62 41 64 20 5 14 41 10 37 51 32 65 46 61 8 15 19 16 44 31 42 19 46 66 25 26 58 60 5 19 18 69 53 20 40 45 27 24 41 32 23 57 56 62 10",
"output": "62 18 35 7 70 23 61 56 34 22 42 58 6 66 10 37 21 2 38 13 69 29 14 42 48 7 29 5 50 66 60 29 63 35 20 38 4 23 10 65 58 52 57 27 41 28 52 23 2 46 45 14 12 66 52 55 1 19 50 33 26 44 47 29 38 48 16 17 7 63"
},
{
"input": "5\n1 2000 1 1 2000",
"output": "3 1 3 3 1"
}
] | 1,434,128,836
| 1,336
|
Python 3
|
RUNTIME_ERROR
|
PRETESTS
| 0
| 46
| 0
|
a=input()
b=list(map(int, input().split()))
b2=sorted(b, reverse=True)
b3=[]
for i in b:
b3.append(b2.index(i)+1)
#print(b3)
sys.stdout.write(b3)
sys.stdout.flush()
|
Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
|
```python
a=input()
b=list(map(int, input().split()))
b2=sorted(b, reverse=True)
b3=[]
for i in b:
b3.append(b2.index(i)+1)
#print(b3)
sys.stdout.write(b3)
sys.stdout.flush()
```
| -1
|
|
934
|
A
|
A Compatible Pair
|
PROGRAMMING
| 1,400
|
[
"brute force",
"games"
] | null | null |
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109.
|
Print a single integer — the brightness of the chosen pair.
|
[
"2 2\n20 18\n2 14\n",
"5 3\n-1 0 1 2 3\n-1 0 1\n"
] |
[
"252\n",
"2\n"
] |
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
| 500
|
[
{
"input": "2 2\n20 18\n2 14",
"output": "252"
},
{
"input": "5 3\n-1 0 1 2 3\n-1 0 1",
"output": "2"
},
{
"input": "10 2\n1 6 2 10 2 3 2 10 6 4\n5 7",
"output": "70"
},
{
"input": "50 50\n1 6 2 10 2 3 2 10 6 4 5 0 3 1 7 3 2 4 4 2 1 5 0 6 10 1 8 0 10 9 0 4 10 5 5 7 4 9 9 5 5 2 6 7 9 4 3 7 2 0\n0 5 9 4 4 6 1 8 2 1 6 6 8 6 4 4 7 2 1 8 6 7 4 9 8 3 0 2 0 10 7 1 4 9 4 4 2 5 3 5 1 3 2 4 1 6 5 3 8 6",
"output": "100"
},
{
"input": "5 7\n-130464232 -73113866 -542094710 -53118823 -63528720\n-775179088 631683023 -974858199 -157471745 -629658630 71825477 -6235611",
"output": "127184126241438168"
},
{
"input": "16 15\n-94580188 -713689767 -559972014 -632609438 -930348091 -567718487 -611395744 -819913097 -924009672 -427913920 -812510647 -546415480 -982072775 -693369647 -693004777 -714181162\n-772924706 -202246100 -165871667 -991426281 -490838183 209351416 134956137 -36128588 -754413937 -616596290 696201705 -201191199 967464971 -244181984 -729907974",
"output": "922371547895579571"
},
{
"input": "12 22\n-102896616 -311161241 -67541276 -402842686 -830595520 -813834033 -44046671 -584806552 -598620444 -968935604 -303048547 -545969410\n545786451 262898403 442511997 -441241260 -479587986 -752123290 720443264 500646237 737842681 -571966572 -798463881 -477248830 89875164 410339460 -359022689 -251280099 -441455542 -538431186 -406793869 374561004 -108755237 -440143410",
"output": "663200522440413120"
},
{
"input": "33 14\n-576562007 -218618150 -471719380 -583840778 -256368365 -68451917 -405045344 -775538133 -896830082 -439261765 -947070124 -716577019 -456110999 -689862512 -132480131 -10805271 -518903339 -196240188 -222292638 -828546042 -43887962 -161359263 -281422097 -484060534 963147664 -492377073 -154570101 -52145116 187803553 858844161 66540410 418777176 434025748\n-78301978 -319393213 -12393024 542953412 786804661 845642067 754996432 -985617475 -487171947 56142664 203173079 -268261708 -817080591 -511720682",
"output": "883931400924882950"
},
{
"input": "15 8\n-966400308 -992207261 -302395973 -837980754 -516443826 -492405613 -378127629 -762650324 -519519776 -36132939 -286460372 -351445284 -407653342 -604960925 -523442015\n610042288 27129580 -103108347 -942517864 842060508 -588904868 614786155 37455106",
"output": "910849554065102112"
},
{
"input": "6 30\n-524297819 -947277203 -444186475 -182837689 -385379656 -453917269\n834529938 35245081 663687669 585422565 164412867 850052113 796429008 -307345676 -127653313 426960600 211854713 -733687358 251466836 -33491050 -882811238 455544614 774581544 768447941 -241033484 441104324 -493975870 308277556 275268265 935941507 -152292053 -961509996 -740482111 -954176110 -924254634 -518710544",
"output": "504117593849498724"
},
{
"input": "5 32\n-540510995 -841481393 -94342377 -74818927 -93445356\n686714668 -82581175 736472406 502016312 575563638 -899308712 503504178 -644271272 -437408397 385778869 -746757839 306275973 -663503743 -431116516 -418708278 -515261493 -988182324 900230931 218258353 -714420102 -241118202 294802602 -937785552 -857537498 -723195312 -690515139 -214508504 -44086454 -231621215 -418360090 -810003786 -675944617",
"output": "534123411186652380"
},
{
"input": "32 13\n-999451897 -96946179 -524159869 -906101658 -63367320 -629803888 -968586834 -658416130 -874232857 -926556428 -749908220 -517073321 -659752288 -910152878 -786916085 -607633039 -191428642 -867952926 -873793977 -584331784 -733245792 -779809700 -554228536 -464503499 561577340 258991071 -569805979 -372655165 -106685554 -619607960 188856473 -268960803\n886429660 -587284372 911396803 -462990289 -228681210 -876239914 -822830527 -750131315 -401234943 116991909 -582713480 979631847 813552478",
"output": "848714444125692276"
},
{
"input": "12 25\n-464030345 -914672073 -483242132 -856226270 -925135169 -353124606 -294027092 -619650850 -490724485 -240424784 -483066792 -921640365\n279850608 726838739 -431610610 242749870 -244020223 -396865433 129534799 182767854 -939698671 342579400 330027106 893561388 -263513962 643369418 276245179 -99206565 -473767261 -168908664 -853755837 -270920164 -661186118 199341055 765543053 908211534 -93363867",
"output": "866064226130454915"
},
{
"input": "10 13\n-749120991 -186261632 -335412349 -231354880 -195919225 -808736065 -481883825 -263383991 -664780611 -605377134\n718174936 -140362196 -669193674 -598621021 -464130929 450701419 -331183926 107203430 946959233 -565825915 -558199897 246556991 -666216081",
"output": "501307028237810934"
},
{
"input": "17 13\n-483786205 -947257449 -125949195 -294711143 -420288876 -812462057 -250049555 -911026413 -188146919 -129501682 -869006661 -649643966 -26976411 -275761039 -869067490 -272248209 -342067346\n445539900 529728842 -808170728 673157826 -70778491 642872105 299298867 -76674218 -902394063 377664752 723887448 -121522827 906464625",
"output": "822104826327386019"
},
{
"input": "15 29\n-716525085 -464205793 -577203110 -979997115 -491032521 -70793687 -770595947 -817983495 -767886763 -223333719 -971913221 -944656683 -200397825 -295615495 -945544540\n-877638425 -146878165 523758517 -158778747 -49535534 597311016 77325385 494128313 12111658 -4196724 295706874 477139483 375083042 726254399 -439255703 662913604 -481588088 673747948 -345999555 -723334478 -656721905 276267528 628773156 851420802 -585029291 -643535709 -968999740 -384418713 -510285542",
"output": "941783658451562540"
},
{
"input": "5 7\n-130464232 -73113866 -542094710 -53118823 -63528720\n449942926 482853427 861095072 316710734 194604468 20277633 668816604",
"output": "-1288212069119760"
},
{
"input": "24 24\n-700068683 -418791905 -24650102 -167277317 -182309202 -517748507 -663050677 -854097070 -426998982 -197009558 -101944229 -746589957 -849018439 -774208211 -946709040 -594578249 -276703474 -434567489 -743600446 -625029074 -977300284 -895608684 -878936220 -850670748\n704881272 169877679 705460701 94083210 403943695 987978311 786162506 658067668 697640875 186287 295558596 286470276 251313879 353071193 755450449 173370603 805550377 192465301 168935494 110161743 285139426 985238736 723221868 520679017",
"output": "-18990884587723"
},
{
"input": "39 9\n44558618 981372779 318891054 283079237 285093436 907256321 414759796 652683534 79042330 249010687 7020063 309415438 788425492 138577429 714835649 954204512 795507844 389962019 507308352 408180613 194676444 44962879 922688019 101163040 327953325 560462120 183657590 273616448 226876035 233697890 720185285 689340674 372938362 15088928 283418109 796807778 149989495 694808087 276385512\n-681609072 -210918688 -757170622 -205635977 -597872997 -496188744 -97031207 -311654366 -389141528",
"output": "-1464096896176096"
},
{
"input": "5 7\n869535768 926886134 457905290 946881177 936471280\n-550057074 -517146573 -138904928 -683289266 -805395532 -979722367 -331183396",
"output": "-120782803247464704"
},
{
"input": "24 24\n299931317 581208095 975349898 832722683 817690798 482251493 336949323 145902930 573001018 802990442 898055771 253410043 150981561 225791789 53290960 405421751 723296526 565432511 256399554 374970926 22699716 104391316 121063780 149329252\n-295118728 -830122321 -294539299 -905916790 -596056305 -12021689 -213837494 -341932332 -302359125 -999813713 -704441404 -713529724 -748686121 -646928807 -244549551 -826629397 -194449623 -807534699 -831064506 -889838257 -714860574 -14761264 -276778132 -479320983",
"output": "-640647347631440"
},
{
"input": "14 8\n-1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "-1000000000000000000"
}
] | 1,545,310,260
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 87
| 109
| 0
|
n=input().split()
a=list(map(int,input().split(' ')))
b=list(map(int,input().split(' ')))
t=[]
for i in range(len(a)):
p=a[0]
del a[0]
t.append(max(max(a)*max(b),min(a)*min(b),min(a)*max(b),max(a)*min(b)))
a.append(p)
print(min(t))
|
Title: A Compatible Pair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109.
Output Specification:
Print a single integer — the brightness of the chosen pair.
Demo Input:
['2 2\n20 18\n2 14\n', '5 3\n-1 0 1 2 3\n-1 0 1\n']
Demo Output:
['252\n', '2\n']
Note:
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
|
```python
n=input().split()
a=list(map(int,input().split(' ')))
b=list(map(int,input().split(' ')))
t=[]
for i in range(len(a)):
p=a[0]
del a[0]
t.append(max(max(a)*max(b),min(a)*min(b),min(a)*max(b),max(a)*min(b)))
a.append(p)
print(min(t))
```
| 3
|
|
614
|
A
|
Link/Cut Tree
|
PROGRAMMING
| 1,500
|
[
"brute force",
"implementation"
] | null | null |
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
|
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
|
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
|
[
"1 10 2\n",
"2 4 5\n"
] |
[
"1 2 4 8 ",
"-1"
] |
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed.
| 500
|
[
{
"input": "1 10 2",
"output": "1 2 4 8 "
},
{
"input": "2 4 5",
"output": "-1"
},
{
"input": "18102 43332383920 28554",
"output": "28554 815330916 "
},
{
"input": "19562 31702689720 17701",
"output": "313325401 "
},
{
"input": "11729 55221128400 313",
"output": "97969 30664297 9597924961 "
},
{
"input": "5482 100347128000 342",
"output": "116964 40001688 13680577296 "
},
{
"input": "3680 37745933600 10",
"output": "10000 100000 1000000 10000000 100000000 1000000000 10000000000 "
},
{
"input": "17098 191120104800 43",
"output": "79507 3418801 147008443 6321363049 "
},
{
"input": "10462 418807699200 2",
"output": "16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 "
},
{
"input": "30061 641846400000 3",
"output": "59049 177147 531441 1594323 4782969 14348907 43046721 129140163 387420489 1162261467 3486784401 10460353203 31381059609 94143178827 282429536481 "
},
{
"input": "1 1000000000000000000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 4398046511104 8796093022208 17592186044416 35184372088832 70368744177664 140737488355328 281474976710656 562949953421312 1125899906842624 2251799813685248 4503599627370496 900719925474099..."
},
{
"input": "32 2498039712000 4",
"output": "64 256 1024 4096 16384 65536 262144 1048576 4194304 16777216 67108864 268435456 1073741824 4294967296 17179869184 68719476736 274877906944 1099511627776 "
},
{
"input": "1 2576683920000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 "
},
{
"input": "5 25 5",
"output": "5 25 "
},
{
"input": "1 90 90",
"output": "1 90 "
},
{
"input": "95 2200128528000 68",
"output": "4624 314432 21381376 1453933568 98867482624 "
},
{
"input": "64 426314644000 53",
"output": "2809 148877 7890481 418195493 22164361129 "
},
{
"input": "198765 198765 198765",
"output": "198765 "
},
{
"input": "42 2845016496000 12",
"output": "144 1728 20736 248832 2985984 35831808 429981696 5159780352 61917364224 743008370688 "
},
{
"input": "6 6 3",
"output": "-1"
},
{
"input": "1 10 11",
"output": "1 "
},
{
"input": "2 10 11",
"output": "-1"
},
{
"input": "87 160 41",
"output": "-1"
},
{
"input": "237171123124584251 923523399718980912 7150",
"output": "-1"
},
{
"input": "101021572000739548 453766043506276015 8898",
"output": "-1"
},
{
"input": "366070689449360724 928290634811046396 8230",
"output": "-1"
},
{
"input": "438133886369772308 942612870269666780 7193",
"output": "-1"
},
{
"input": "10 10 10",
"output": "10 "
},
{
"input": "16 16 256",
"output": "-1"
},
{
"input": "1 1000000000000000000 1000000000",
"output": "1 1000000000 1000000000000000000 "
},
{
"input": "1000000000000000000 1000000000000000000 1000000000",
"output": "1000000000000000000 "
},
{
"input": "1000000000 1000000000000000000 1000000000",
"output": "1000000000 1000000000000000000 "
},
{
"input": "1 1 4",
"output": "1 "
},
{
"input": "1 999999999999999999 1000000000",
"output": "1 1000000000 "
},
{
"input": "1 1000000000000000000 999999990",
"output": "1 999999990 999999980000000100 "
},
{
"input": "1 1000000000000000000 999999984",
"output": "1 999999984 999999968000000256 "
},
{
"input": "1 1000000000000000000 324325",
"output": "1 324325 105186705625 34114678301828125 "
},
{
"input": "1 1000000000000000000 999999523",
"output": "1 999999523 999999046000227529 "
},
{
"input": "1 243 3",
"output": "1 3 9 27 81 243 "
},
{
"input": "62769392426654367 567152589733560993 688813",
"output": "326816522793383797 "
},
{
"input": "1 1000000000000000000 690852001",
"output": "1 690852001 477276487285704001 "
},
{
"input": "1 1000000000000000000 918745157",
"output": "1 918745157 844092663510954649 "
},
{
"input": "1 1000000000000000000 131299843",
"output": "1 131299843 17239648771824649 "
},
{
"input": "2861381721051425 2861381721051425 1234",
"output": "-1"
}
] | 1,454,430,348
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 77
| 0
|
l, r, k = input().split()
l = int(l)
r = int(r)
k = int(k)
ar = []
i = 0
while (True):
p = k**i
if p > r:
break
if p >= l and p <= r:
ar.append(p)
i += 1
flag = True
for i in ar:
print(i, end = " ")
flag = False
if(flag):
print(-1)
|
Title: Link/Cut Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
Input Specification:
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
Output Specification:
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
Demo Input:
['1 10 2\n', '2 4 5\n']
Demo Output:
['1 2 4 8 ', '-1']
Note:
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed.
|
```python
l, r, k = input().split()
l = int(l)
r = int(r)
k = int(k)
ar = []
i = 0
while (True):
p = k**i
if p > r:
break
if p >= l and p <= r:
ar.append(p)
i += 1
flag = True
for i in ar:
print(i, end = " ")
flag = False
if(flag):
print(-1)
```
| 3
|
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