contestId int64 0 1.01k | index stringclasses 57 values | name stringlengths 2 58 | type stringclasses 2 values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522 values | time-limit stringclasses 8 values | memory-limit stringclasses 8 values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3 values | verdict stringclasses 14 values | testset stringclasses 12 values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 58697... | 1,693,592,507 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 124 | 0 | s = list(map(int,input().split()))
ss = set(s)
print(len(s)-len(ss)) | Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
s = list(map(int,input().split()))
ss = set(s)
print(len(s)-len(ss))
``` | 3 | |
432 | A | Choosing Teams | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? | The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. | Print a single number — the answer to the problem. | [
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | 500 | [
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0... | 1,687,900,316 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 46 | 0 | import sys
import math
INF = float('inf')
MOD = 1000000007
def read_int() -> int:
return int(sys.stdin.readline().strip())
def read_str() -> str:
return sys.stdin.readline().strip()
def read_list() -> list:
return list(map(int, sys.stdin.readline().split()))
def read_map() -> map:
return map(int, sys.stdin.readline().split())
def write(data) -> None:
sys.stdout.write(str(data) + "\n")
sys.stdout.flush()
def write_list(arr:list, sep=" ") -> None:
for i in arr:
sys.stdout.write(str(i) + sep)
sys.stdout.write("\n")
sys.stdout.flush()
n,k = read_map()
d = read_list()
ok=0
for i in d:
if i+k<=5:
ok+=1
write(str(ok//3)) | Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | ```python
import sys
import math
INF = float('inf')
MOD = 1000000007
def read_int() -> int:
return int(sys.stdin.readline().strip())
def read_str() -> str:
return sys.stdin.readline().strip()
def read_list() -> list:
return list(map(int, sys.stdin.readline().split()))
def read_map() -> map:
return map(int, sys.stdin.readline().split())
def write(data) -> None:
sys.stdout.write(str(data) + "\n")
sys.stdout.flush()
def write_list(arr:list, sep=" ") -> None:
for i in arr:
sys.stdout.write(str(i) + sep)
sys.stdout.write("\n")
sys.stdout.flush()
n,k = read_map()
d = read_list()
ok=0
for i in d:
if i+k<=5:
ok+=1
write(str(ok//3))
``` | 3 | |
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"... | 1,597,925,117 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 280 | 0 | def s(num):
if num == 0:
return 2
if num == 1:
return 1
i = 2
while i * i <= num:
if num % i == 0:
return i
else:
i += 1
return num
n = int(input())
arr = list(map(int, input().split()))
check = list()
for i in range(n):
check.append(s(arr[i]))
if check.count(check[0]) == 1:
print(1)
else:
for i in range(1, n):
if check[0] != check[i]:
print(i + 1) | 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
def s(num):
if num == 0:
return 2
if num == 1:
return 1
i = 2
while i * i <= num:
if num % i == 0:
return i
else:
i += 1
return num
n = int(input())
arr = list(map(int, input().split()))
check = list()
for i in range(n):
check.append(s(arr[i]))
if check.count(check[0]) == 1:
print(1)
else:
for i in range(1, n):
if check[0] != check[i]:
print(i + 1)
``` | 0 |
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"
},
... | 1,666,030,203 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 46 | 0 | n = int(input())
if n % 2:
print(n - 9, 9)
else:
print(n - 4, 4)
| 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
n = int(input())
if n % 2:
print(n - 9, 9)
else:
print(n - 4, 4)
``` | 3 | |
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"... | 1,673,620,305 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 92 | 0 | k=int(input())
a=[int(i) for i in input().split()]
b=[]
c=[]
count=0
sum=0
for i in range(k):
if(a[i]%2==0):
count+=1
b.append(i+1)
else:
sum+=1
c.append(i+1)
if(count>sum):
print(c[0])
else:
print(b[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
k=int(input())
a=[int(i) for i in input().split()]
b=[]
c=[]
count=0
sum=0
for i in range(k):
if(a[i]%2==0):
count+=1
b.append(i+1)
else:
sum+=1
c.append(i+1)
if(count>sum):
print(c[0])
else:
print(b[0])
``` | 3.977 |
911 | C | Three Garlands | PROGRAMMING | 1,400 | [
"brute force",
"constructive algorithms"
] | null | null | Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if *i*-th garland is switched on during *x*-th second, then it is lit only during seconds *x*, *x*<=+<=*k**i*, *x*<=+<=2*k**i*, *x*<=+<=3*k**i* and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers *x*1, *x*2 and *x*3 (not necessarily distinct) so that he will switch on the first garland during *x*1-th second, the second one — during *x*2-th second, and the third one — during *x*3-th second, respectively, and during each second starting from *max*(*x*1,<=*x*2,<=*x*3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this! | The first line contains three integers *k*1, *k*2 and *k*3 (1<=≤<=*k**i*<=≤<=1500) — time intervals of the garlands. | If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO. | [
"2 2 3\n",
"4 2 3\n"
] | [
"YES\n",
"NO\n"
] | In the first example Mishka can choose *x*<sub class="lower-index">1</sub> = 1, *x*<sub class="lower-index">2</sub> = 2, *x*<sub class="lower-index">3</sub> = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what *x*<sub class="lower-index">3</sub> is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit. | 0 | [
{
"input": "2 2 3",
"output": "YES"
},
{
"input": "4 2 3",
"output": "NO"
},
{
"input": "1499 1498 1500",
"output": "NO"
},
{
"input": "1500 1500 1500",
"output": "NO"
},
{
"input": "100 4 1",
"output": "YES"
},
{
"input": "4 2 4",
"output": "YES"
... | 1,635,181,855 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 4,505,600 | k1,k2,k3 = map(int,input().split())
if k1 == 1 or k2 == 1 or k3 == 1:
print("YES")
elif (k1 == 2 and k2 == 2) or (k2 == 2 and k3 == 2) or (k3 == 2 and k1 == 2):
print("YES")
elif k1 == 3 and k2 == 3 and k3 == 3:
print("YES")
else:
print("NO") | Title: Three Garlands
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if *i*-th garland is switched on during *x*-th second, then it is lit only during seconds *x*, *x*<=+<=*k**i*, *x*<=+<=2*k**i*, *x*<=+<=3*k**i* and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers *x*1, *x*2 and *x*3 (not necessarily distinct) so that he will switch on the first garland during *x*1-th second, the second one — during *x*2-th second, and the third one — during *x*3-th second, respectively, and during each second starting from *max*(*x*1,<=*x*2,<=*x*3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this!
Input Specification:
The first line contains three integers *k*1, *k*2 and *k*3 (1<=≤<=*k**i*<=≤<=1500) — time intervals of the garlands.
Output Specification:
If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO.
Demo Input:
['2 2 3\n', '4 2 3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Mishka can choose *x*<sub class="lower-index">1</sub> = 1, *x*<sub class="lower-index">2</sub> = 2, *x*<sub class="lower-index">3</sub> = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what *x*<sub class="lower-index">3</sub> is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit. | ```python
k1,k2,k3 = map(int,input().split())
if k1 == 1 or k2 == 1 or k3 == 1:
print("YES")
elif (k1 == 2 and k2 == 2) or (k2 == 2 and k3 == 2) or (k3 == 2 and k1 == 2):
print("YES")
elif k1 == 3 and k2 == 3 and k3 == 3:
print("YES")
else:
print("NO")
``` | 0 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,669,197,456 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | s=input()
n=len(s)
l=0
u=0
for i in range(n):
if s[i].islower():
l=l+1
elif s[i].isupper():
u=u+1
if u<=l:
s=s.lower()
print(s)
else:
s=s.upper()
print(s)
| Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
s=input()
n=len(s)
l=0
u=0
for i in range(n):
if s[i].islower():
l=l+1
elif s[i].isupper():
u=u+1
if u<=l:
s=s.lower()
print(s)
else:
s=s.upper()
print(s)
``` | 3.977 |
202 | A | LLPS | PROGRAMMING | 800 | [
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null | This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". | The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. | Print the lexicographically largest palindromic subsequence of string *s*. | [
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] | [
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] | Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | 500 | [
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
... | 1,690,977,127 | 2,147,483,647 | PyPy 3 | OK | TESTS | 54 | 280 | 2,150,400 | import math
import cmath
import string
result = list(string.ascii_lowercase)
result.sort(reverse = True)
def XOR(n): return [n, 1, n + 1, 0][n % 4]
def isprime(n): return (all([False for i in range(2,n) if n % i == 0 ]) and not n < 2)
def c(lis,l,r,e):
ans = 0
while l<= r :
m = (l+r)//2
if lis[m] > e :
r = m-1
else :
l = m+1
ans = l
return ans
aa = []
def f(l,i,j,n,m):
c = l[i][j]
i1 = i
j1 = j
i2 = i
j2 = j
i3 = i
j3 = j
i4 = i
j4 = j
while i1 > 0 and j1 > 0 :
i1-=1
j1-=1
c+=l[i1][j1]
while i2 < n-1 and j2 < m-1 :
i2+=1
j2+=1
c+=l[i2][j2]
while i3 > 0 and j3 < m-1 :
i3-=1
j3+=1
c+=l[i3][j3]
while i4 < n-1 and j4 > 0 :
i4+=1
j4-=1
c+=l[i4][j4]
return c
# for _ in range(int(input())):
ss = input()
for i in result :
if i in ss :
e = i
break
i = ss.count(e)
print(e*i) | Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | ```python
import math
import cmath
import string
result = list(string.ascii_lowercase)
result.sort(reverse = True)
def XOR(n): return [n, 1, n + 1, 0][n % 4]
def isprime(n): return (all([False for i in range(2,n) if n % i == 0 ]) and not n < 2)
def c(lis,l,r,e):
ans = 0
while l<= r :
m = (l+r)//2
if lis[m] > e :
r = m-1
else :
l = m+1
ans = l
return ans
aa = []
def f(l,i,j,n,m):
c = l[i][j]
i1 = i
j1 = j
i2 = i
j2 = j
i3 = i
j3 = j
i4 = i
j4 = j
while i1 > 0 and j1 > 0 :
i1-=1
j1-=1
c+=l[i1][j1]
while i2 < n-1 and j2 < m-1 :
i2+=1
j2+=1
c+=l[i2][j2]
while i3 > 0 and j3 < m-1 :
i3-=1
j3+=1
c+=l[i3][j3]
while i4 < n-1 and j4 > 0 :
i4+=1
j4-=1
c+=l[i4][j4]
return c
# for _ in range(int(input())):
ss = input()
for i in result :
if i in ss :
e = i
break
i = ss.count(e)
print(e*i)
``` | 3 | |
266 | B | Queue at the School | PROGRAMMING | 800 | [
"constructive algorithms",
"graph matchings",
"implementation",
"shortest paths"
] | null | null | During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds. | The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G". | Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G". | [
"5 1\nBGGBG\n",
"5 2\nBGGBG\n",
"4 1\nGGGB\n"
] | [
"GBGGB\n",
"GGBGB\n",
"GGGB\n"
] | none | 500 | [
{
"input": "5 1\nBGGBG",
"output": "GBGGB"
},
{
"input": "5 2\nBGGBG",
"output": "GGBGB"
},
{
"input": "4 1\nGGGB",
"output": "GGGB"
},
{
"input": "2 1\nBB",
"output": "BB"
},
{
"input": "2 1\nBG",
"output": "GB"
},
{
"input": "6 2\nBBGBBG",
"outpu... | 1,696,401,502 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | import numpy as np
n,t = map(int,input().split())
arr = list(input())
for i in range(0,n):
ans = np.zeros(n)
# print(ans)
for j in range(0,n-1):
k = j + 1
if arr[j] == 'B' and arr[k] == 'G' and ans[j] != t and ans[k] != t:
arr[j] = 'G'
arr[k] = 'B'
ans[j] = ans[j] + 1
ans[t] = ans[t] + 1
elif ans[j] != t and ans[k] != t:
ans[j] = ans[j] + 1
ans[t] = ans[t] + 1
str1 = "".join(arr)
print(str1) | Title: Queue at the School
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
Output Specification:
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
Demo Input:
['5 1\nBGGBG\n', '5 2\nBGGBG\n', '4 1\nGGGB\n']
Demo Output:
['GBGGB\n', 'GGBGB\n', 'GGGB\n']
Note:
none | ```python
import numpy as np
n,t = map(int,input().split())
arr = list(input())
for i in range(0,n):
ans = np.zeros(n)
# print(ans)
for j in range(0,n-1):
k = j + 1
if arr[j] == 'B' and arr[k] == 'G' and ans[j] != t and ans[k] != t:
arr[j] = 'G'
arr[k] = 'B'
ans[j] = ans[j] + 1
ans[t] = ans[t] + 1
elif ans[j] != t and ans[k] != t:
ans[j] = ans[j] + 1
ans[t] = ans[t] + 1
str1 = "".join(arr)
print(str1)
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Карта звёздного неба представляет собой прямоугольное поле, состоящее из *n* строк по *m* символов в каждой строке. Каждый символ — это либо «.» (означает пустой участок неба), либо «*» (означает то, что в этом месте на небе есть звезда).
Новое издание карты звёздного неба будет напечатано на квадратных листах, поэтому требуется найти минимально возможную сторону квадрата, в который могут поместиться все звезды. Границы искомого квадрата должны быть параллельны сторонам заданного прямоугольного поля. | В первой строке входных данных записаны два числа *n* и *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — количество строк и столбцов на карте звездного неба.
В следующих *n* строках задано по *m* символов. Каждый символ — это либо «.» (пустой участок неба), либо «*» (звезда).
Гарантируется, что на небе есть хотя бы одна звезда. | Выведите одно число — минимально возможную сторону квадрата, которым можно накрыть все звезды. | [
"4 4\n....\n..*.\n...*\n..**\n",
"1 3\n*.*\n",
"2 1\n.\n*\n"
] | [
"3\n",
"3\n",
"1\n"
] | Один из возможных ответов на первый тестовый пример:
Один из возможных ответов на второй тестовый пример (обратите внимание, что покрывающий квадрат выходит за пределы карты звездного неба):
Ответ на третий тестовый пример: | 0 | [
{
"input": "4 4\n....\n..*.\n...*\n..**",
"output": "3"
},
{
"input": "1 3\n*.*",
"output": "3"
},
{
"input": "2 1\n.\n*",
"output": "1"
},
{
"input": "1 1\n*",
"output": "1"
},
{
"input": "1 2\n.*",
"output": "1"
},
{
"input": "1 2\n*.",
"output":... | 1,458,932,089 | 132,889 | Python 3 | OK | TESTS | 208 | 545 | 6,144,000 | z = input().split()
a = []
for i in range(2):
a.append(int(z[i]))
n = a[0]
m = a[1]
b = []
x1 = 1001
x2 = 0
y1 = 1001
y2 = 0
for i in range(n):
b.append(str(input()))
for i in range(n):
for j in range(m):
if b[i][j] == '*':
if i < y1 :
y1 = i
if j < x1:
x1 = j
if i > y2:
y2 = i
if j > x2:
x2 = j
x = x2 - x1 + 1
y = y2 - y1 + 1
if x > y:
print(x)
else:
print(y)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Карта звёздного неба представляет собой прямоугольное поле, состоящее из *n* строк по *m* символов в каждой строке. Каждый символ — это либо «.» (означает пустой участок неба), либо «*» (означает то, что в этом месте на небе есть звезда).
Новое издание карты звёздного неба будет напечатано на квадратных листах, поэтому требуется найти минимально возможную сторону квадрата, в который могут поместиться все звезды. Границы искомого квадрата должны быть параллельны сторонам заданного прямоугольного поля.
Input Specification:
В первой строке входных данных записаны два числа *n* и *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — количество строк и столбцов на карте звездного неба.
В следующих *n* строках задано по *m* символов. Каждый символ — это либо «.» (пустой участок неба), либо «*» (звезда).
Гарантируется, что на небе есть хотя бы одна звезда.
Output Specification:
Выведите одно число — минимально возможную сторону квадрата, которым можно накрыть все звезды.
Demo Input:
['4 4\n....\n..*.\n...*\n..**\n', '1 3\n*.*\n', '2 1\n.\n*\n']
Demo Output:
['3\n', '3\n', '1\n']
Note:
Один из возможных ответов на первый тестовый пример:
Один из возможных ответов на второй тестовый пример (обратите внимание, что покрывающий квадрат выходит за пределы карты звездного неба):
Ответ на третий тестовый пример: | ```python
z = input().split()
a = []
for i in range(2):
a.append(int(z[i]))
n = a[0]
m = a[1]
b = []
x1 = 1001
x2 = 0
y1 = 1001
y2 = 0
for i in range(n):
b.append(str(input()))
for i in range(n):
for j in range(m):
if b[i][j] == '*':
if i < y1 :
y1 = i
if j < x1:
x1 = j
if i > y2:
y2 = i
if j > x2:
x2 = j
x = x2 - x1 + 1
y = y2 - y1 + 1
if x > y:
print(x)
else:
print(y)
``` | 3 | |
887 | A | Div. 64 | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system. | In the only line given a non-empty binary string *s* with length up to 100. | Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise. | [
"100010001\n",
"100\n"
] | [
"yes",
"no"
] | In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system) | 500 | [
{
"input": "100010001",
"output": "yes"
},
{
"input": "100",
"output": "no"
},
{
"input": "0000001000000",
"output": "yes"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "no"
},
{
"in... | 1,558,611,563 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 6 | 124 | 0 | s = input().strip()
print(['no', 'yes'][s.count('1') and s.count('0') >= 5]) | Title: Div. 64
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.
Input Specification:
In the only line given a non-empty binary string *s* with length up to 100.
Output Specification:
Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.
Demo Input:
['100010001\n', '100\n']
Demo Output:
['yes', 'no']
Note:
In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system) | ```python
s = input().strip()
print(['no', 'yes'][s.count('1') and s.count('0') >= 5])
``` | 0 | |
433 | A | Kitahara Haruki's Gift | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple. | In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes). | [
"3\n100 200 100\n",
"4\n100 100 100 200\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | 500 | [
{
"input": "3\n100 200 100",
"output": "YES"
},
{
"input": "4\n100 100 100 200",
"output": "NO"
},
{
"input": "1\n100",
"output": "NO"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "2\n100 100",
"output": "YES"
},
{
"input": "2\n200 200",
"o... | 1,688,536,806 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | n=int(input())
lis=list(map(int,input().strip().split()))
if n==1:
print("NO")
else:
if n%2!=0:
if lis.count(100)==0 or lis.count(200)==0:
print("NO")
else:
if (sum(lis)//100)%2==0:
print("YES")
else:
print("NO")
| Title: Kitahara Haruki's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
Output Specification:
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
Demo Input:
['3\n100 200 100\n', '4\n100 100 100 200\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | ```python
n=int(input())
lis=list(map(int,input().strip().split()))
if n==1:
print("NO")
else:
if n%2!=0:
if lis.count(100)==0 or lis.count(200)==0:
print("NO")
else:
if (sum(lis)//100)%2==0:
print("YES")
else:
print("NO")
``` | 0 | |
501 | A | Contest | PROGRAMMING | 900 | [
"implementation"
] | null | null | Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. | The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). | Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points. | [
"500 1000 20 30\n",
"1000 1000 1 1\n",
"1500 1000 176 177\n"
] | [
"Vasya\n",
"Tie\n",
"Misha\n"
] | none | 500 | [
{
"input": "500 1000 20 30",
"output": "Vasya"
},
{
"input": "1000 1000 1 1",
"output": "Tie"
},
{
"input": "1500 1000 176 177",
"output": "Misha"
},
{
"input": "1500 1000 74 177",
"output": "Misha"
},
{
"input": "750 2500 175 178",
"output": "Vasya"
},
{
... | 1,655,149,760 | 2,147,483,647 | PyPy 3 | OK | TESTS | 40 | 77 | 0 | #J - Contest
# misha point = a
# vasya point = b
# time misha = c
# time vasya = d
def cal_point(arr):
p = arr[0]
t = arr[1]
point1 = (3*p)/10
point2 = (p-(p*t)/250)
point = int(max(point1, point2))
return point
array = list(map(int, input().split()))
misha = [array[0], array[2]]
vasya = [array[1], array[3]]
m = cal_point(misha)
v = cal_point(vasya)
if m==v:
print("Tie")
elif m>v:
print("Misha")
else:
print("Vasya")
| Title: Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input Specification:
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output Specification:
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
Demo Input:
['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n']
Demo Output:
['Vasya\n', 'Tie\n', 'Misha\n']
Note:
none | ```python
#J - Contest
# misha point = a
# vasya point = b
# time misha = c
# time vasya = d
def cal_point(arr):
p = arr[0]
t = arr[1]
point1 = (3*p)/10
point2 = (p-(p*t)/250)
point = int(max(point1, point2))
return point
array = list(map(int, input().split()))
misha = [array[0], array[2]]
vasya = [array[1], array[3]]
m = cal_point(misha)
v = cal_point(vasya)
if m==v:
print("Tie")
elif m>v:
print("Misha")
else:
print("Vasya")
``` | 3 | |
520 | A | Pangram | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters. | Output "YES", if the string is a pangram and "NO" otherwise. | [
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXY... | 1,697,824,670 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 46 | 0 | def main():
n = int(input())
s = input().lower()
h = {}
for i in range(len(s)):
h[s[i]] = 1
if len(h.keys()) < 26:
print("NO")
else:
print("YES")
main()
| Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
def main():
n = int(input())
s = input().lower()
h = {}
for i in range(len(s)):
h[s[i]] = 1
if len(h.keys()) < 26:
print("NO")
else:
print("YES")
main()
``` | 3 | |
361 | A | Levko and Table | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*.
Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them. | The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000). | Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.
If there are multiple suitable tables, you are allowed to print any of them. | [
"2 4\n",
"4 7\n"
] | [
"1 3\n3 1\n",
"2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n"
] | In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.
In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements. | 500 | [
{
"input": "2 4",
"output": "4 0 \n0 4 "
},
{
"input": "4 7",
"output": "7 0 0 0 \n0 7 0 0 \n0 0 7 0 \n0 0 0 7 "
},
{
"input": "1 8",
"output": "8 "
},
{
"input": "9 3",
"output": "3 0 0 0 0 0 0 0 0 \n0 3 0 0 0 0 0 0 0 \n0 0 3 0 0 0 0 0 0 \n0 0 0 3 0 0 0 0 0 \n0 0 0 0 3 0... | 1,586,948,142 | 2,147,483,647 | Python 3 | OK | TESTS | 22 | 109 | 307,200 | n,k=map(int,input().split())
for i in range(n):
for j in range(n):
if (i==j):
print (k,end=' ')
else:
print (0,end=' ')
print ()
| Title: Levko and Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*.
Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them.
Input Specification:
The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000).
Output Specification:
Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.
If there are multiple suitable tables, you are allowed to print any of them.
Demo Input:
['2 4\n', '4 7\n']
Demo Output:
['1 3\n3 1\n', '2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n']
Note:
In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.
In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements. | ```python
n,k=map(int,input().split())
for i in range(n):
for j in range(n):
if (i==j):
print (k,end=' ')
else:
print (0,end=' ')
print ()
``` | 3 | |
294 | A | Shaass and Oskols | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. | The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. | On the *i*-th line of the output print the number of birds on the *i*-th wire. | [
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] | [
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] | none | 500 | [
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 8... | 1,616,454,291 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 154 | 307,200 | import math
n=int(input())
a=[]
a=list(map(int,input().strip().split()))
s=int(input())
for i in range(s):
x,y=map(int,input().split())
x-=1
if(len(a)==1 and s==1):
a[0]=0
break
if x==0:
a[x+1]+=a[x]-y
a[x]=0
elif x==n-1:
a[x-1]+=y-1
a[x]=0
else:
a[x-1]+=y-1
a[x+1]+=a[x]-y
a[x]=0
for i in a:
print(i) | Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none | ```python
import math
n=int(input())
a=[]
a=list(map(int,input().strip().split()))
s=int(input())
for i in range(s):
x,y=map(int,input().split())
x-=1
if(len(a)==1 and s==1):
a[0]=0
break
if x==0:
a[x+1]+=a[x]-y
a[x]=0
elif x==n-1:
a[x-1]+=y-1
a[x]=0
else:
a[x-1]+=y-1
a[x+1]+=a[x]-y
a[x]=0
for i in a:
print(i)
``` | 3 | |
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 ... | 1,688,636,187 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 15 | 31 | 0 | n=int(input())
p1,p2=0,0
for i in range(n):
m,c=map(int,input().split())
if m>c:
p1+=1
else:
p2+=1
if p1>p2:
print("Mishka")
elif p1<p2:
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())
p1,p2=0,0
for i in range(n):
m,c=map(int,input().split())
if m>c:
p1+=1
else:
p2+=1
if p1>p2:
print("Mishka")
elif p1<p2:
print("Chris")
else:
print("Friendship is magic!^^")
``` | 0 | |
990 | E | Post Lamps | PROGRAMMING | 2,100 | [
"brute force",
"greedy"
] | null | null | Adilbek's house is located on a street which can be represented as the OX axis. This street is really dark, so Adilbek wants to install some post lamps to illuminate it. Street has $n$ positions to install lamps, they correspond to the integer numbers from $0$ to $n - 1$ on the OX axis. However, some positions are blocked and no post lamp can be placed there.
There are post lamps of different types which differ only by their power. When placed in position $x$, post lamp of power $l$ illuminates the segment $[x; x + l]$. The power of each post lamp is always a positive integer number.
The post lamp shop provides an infinite amount of lamps of each type from power $1$ to power $k$. Though each customer is only allowed to order post lamps of exactly one type. Post lamps of power $l$ cost $a_l$ each.
What is the minimal total cost of the post lamps of exactly one type Adilbek can buy to illuminate the entire segment $[0; n]$ of the street? If some lamps illuminate any other segment of the street, Adilbek does not care, so, for example, he may place a lamp of power $3$ in position $n - 1$ (even though its illumination zone doesn't completely belong to segment $[0; n]$). | The first line contains three integer numbers $n$, $m$ and $k$ ($1 \le k \le n \le 10^6$, $0 \le m \le n$) — the length of the segment of the street Adilbek wants to illuminate, the number of the blocked positions and the maximum power of the post lamp available.
The second line contains $m$ integer numbers $s_1, s_2, \dots, s_m$ ($0 \le s_1 < s_2 < \dots s_m < n$) — the blocked positions.
The third line contains $k$ integer numbers $a_1, a_2, \dots, a_k$ ($1 \le a_i \le 10^6$) — the costs of the post lamps. | Print the minimal total cost of the post lamps of exactly one type Adilbek can buy to illuminate the entire segment $[0; n]$ of the street.
If illumintaing the entire segment $[0; n]$ is impossible, print -1. | [
"6 2 3\n1 3\n1 2 3\n",
"4 3 4\n1 2 3\n1 10 100 1000\n",
"5 1 5\n0\n3 3 3 3 3\n",
"7 4 3\n2 4 5 6\n3 14 15\n"
] | [
"6\n",
"1000\n",
"-1\n",
"-1\n"
] | none | 0 | [
{
"input": "6 2 3\n1 3\n1 2 3",
"output": "6"
},
{
"input": "4 3 4\n1 2 3\n1 10 100 1000",
"output": "1000"
},
{
"input": "5 1 5\n0\n3 3 3 3 3",
"output": "-1"
},
{
"input": "7 4 3\n2 4 5 6\n3 14 15",
"output": "-1"
},
{
"input": "1 0 1\n\n1000000",
"output": ... | 1,601,472,438 | 2,147,483,647 | PyPy 3 | OK | TESTS | 46 | 1,497 | 82,124,800 | import sys
from array import array
n, m, k = map(int, input().split())
block = list(map(int, input().split()))
a = [0] + list(map(int, input().split()))
if block and block[0] == 0:
print(-1)
exit()
prev = array('i', list(range(n)))
for x in block:
prev[x] = -1
for i in range(1, n):
if prev[i] == -1:
prev[i] = prev[i-1]
inf = ans = 10**18
for i in range(1, k+1):
s = 0
cost = 0
while True:
cost += a[i]
t = s+i
if t >= n:
break
if prev[t] == s:
cost = inf
break
s = prev[t]
ans = min(ans, cost)
print(ans if ans < inf else -1)
| Title: Post Lamps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Adilbek's house is located on a street which can be represented as the OX axis. This street is really dark, so Adilbek wants to install some post lamps to illuminate it. Street has $n$ positions to install lamps, they correspond to the integer numbers from $0$ to $n - 1$ on the OX axis. However, some positions are blocked and no post lamp can be placed there.
There are post lamps of different types which differ only by their power. When placed in position $x$, post lamp of power $l$ illuminates the segment $[x; x + l]$. The power of each post lamp is always a positive integer number.
The post lamp shop provides an infinite amount of lamps of each type from power $1$ to power $k$. Though each customer is only allowed to order post lamps of exactly one type. Post lamps of power $l$ cost $a_l$ each.
What is the minimal total cost of the post lamps of exactly one type Adilbek can buy to illuminate the entire segment $[0; n]$ of the street? If some lamps illuminate any other segment of the street, Adilbek does not care, so, for example, he may place a lamp of power $3$ in position $n - 1$ (even though its illumination zone doesn't completely belong to segment $[0; n]$).
Input Specification:
The first line contains three integer numbers $n$, $m$ and $k$ ($1 \le k \le n \le 10^6$, $0 \le m \le n$) — the length of the segment of the street Adilbek wants to illuminate, the number of the blocked positions and the maximum power of the post lamp available.
The second line contains $m$ integer numbers $s_1, s_2, \dots, s_m$ ($0 \le s_1 < s_2 < \dots s_m < n$) — the blocked positions.
The third line contains $k$ integer numbers $a_1, a_2, \dots, a_k$ ($1 \le a_i \le 10^6$) — the costs of the post lamps.
Output Specification:
Print the minimal total cost of the post lamps of exactly one type Adilbek can buy to illuminate the entire segment $[0; n]$ of the street.
If illumintaing the entire segment $[0; n]$ is impossible, print -1.
Demo Input:
['6 2 3\n1 3\n1 2 3\n', '4 3 4\n1 2 3\n1 10 100 1000\n', '5 1 5\n0\n3 3 3 3 3\n', '7 4 3\n2 4 5 6\n3 14 15\n']
Demo Output:
['6\n', '1000\n', '-1\n', '-1\n']
Note:
none | ```python
import sys
from array import array
n, m, k = map(int, input().split())
block = list(map(int, input().split()))
a = [0] + list(map(int, input().split()))
if block and block[0] == 0:
print(-1)
exit()
prev = array('i', list(range(n)))
for x in block:
prev[x] = -1
for i in range(1, n):
if prev[i] == -1:
prev[i] = prev[i-1]
inf = ans = 10**18
for i in range(1, k+1):
s = 0
cost = 0
while True:
cost += a[i]
t = s+i
if t >= n:
break
if prev[t] == s:
cost = inf
break
s = prev[t]
ans = min(ans, cost)
print(ans if ans < inf else -1)
``` | 3 | |
868 | C | Qualification Rounds | PROGRAMMING | 1,500 | [
"bitmasks",
"brute force",
"constructive algorithms",
"dp"
] | null | null | Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset! | The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise. | Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n",
"3 2\n1 0\n1 1\n0 1\n"
] | [
"NO\n",
"YES\n"
] | In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | 1,000 | [
{
"input": "5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0",
"output": "NO"
},
{
"input": "3 2\n1 0\n1 1\n0 1",
"output": "YES"
},
{
"input": "10 2\n1 0\n1 0\n0 0\n1 1\n0 0\n1 1\n0 0\n1 1\n0 1\n0 1",
"output": "YES"
},
{
"input": "10 3\n1 0 0\n0 1 1\n1 0 0\n0 1 0\n0 0 1\n1 0 1\n0 1 1... | 1,507,256,632 | 4,732 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 204,800 | def solve():
n, k = [int(st) for st in input().split(" ")]
# number of total problems = n, number of total teams = k
probknown = [[int(st) for st in input().split(" ")] for i in range(n)]
# probknown[i][j]: jth team knows ith problem
#knownbyteam = [sum([probknown[i][j] for i in range(n)]) for j in range(k)]
#print(probknown)
#print(knownbyteam)
summ = [0] * n
stack = [None] * n
def search(stacklevel, startindex):
ans = False
for i in range(startindex, n):
stack[stacklevel] = i
stacklevel += 1
for j in range(k):
summ[j] += probknown[i][j]
#print("Len: ", stacklevel)
#print("Selected Problems:", stack[:stacklevel])
#print("Known sum:", summ)
#print("-"*30)
if max(summ) <= (stacklevel)//2:
return True
elif max(summ) > n/2:
pass
else:
ans = ans or search(stacklevel, i+1)
stacklevel -= 1
stack[stacklevel] = None
for j in range(k):
summ[j] -= probknown[i][j]
if ans is True:
return True
return ans
if search(0, 0):
return "YES"
else:
return "NO"
print(solve())
# if 0000
# if 0001 0010 0100 1000
# if 0011 1100
# if 0101 1010
# 1100 0110 0011 1001
| Title: Qualification Rounds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset!
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise.
Output Specification:
Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n', '3 2\n1 0\n1 1\n0 1\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | ```python
def solve():
n, k = [int(st) for st in input().split(" ")]
# number of total problems = n, number of total teams = k
probknown = [[int(st) for st in input().split(" ")] for i in range(n)]
# probknown[i][j]: jth team knows ith problem
#knownbyteam = [sum([probknown[i][j] for i in range(n)]) for j in range(k)]
#print(probknown)
#print(knownbyteam)
summ = [0] * n
stack = [None] * n
def search(stacklevel, startindex):
ans = False
for i in range(startindex, n):
stack[stacklevel] = i
stacklevel += 1
for j in range(k):
summ[j] += probknown[i][j]
#print("Len: ", stacklevel)
#print("Selected Problems:", stack[:stacklevel])
#print("Known sum:", summ)
#print("-"*30)
if max(summ) <= (stacklevel)//2:
return True
elif max(summ) > n/2:
pass
else:
ans = ans or search(stacklevel, i+1)
stacklevel -= 1
stack[stacklevel] = None
for j in range(k):
summ[j] -= probknown[i][j]
if ans is True:
return True
return ans
if search(0, 0):
return "YES"
else:
return "NO"
print(solve())
# if 0000
# if 0001 0010 0100 1000
# if 0011 1100
# if 0101 1010
# 1100 0110 0011 1001
``` | 0 | |
834 | A | The Useless Toy | PROGRAMMING | 900 | [
"implementation"
] | null | null | Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this. | There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position. | Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise. | [
"^ >\n1\n",
"< ^\n3\n",
"^ v\n6\n"
] | [
"cw\n",
"ccw\n",
"undefined\n"
] | none | 500 | [
{
"input": "^ >\n1",
"output": "cw"
},
{
"input": "< ^\n3",
"output": "ccw"
},
{
"input": "^ v\n6",
"output": "undefined"
},
{
"input": "^ >\n999999999",
"output": "ccw"
},
{
"input": "> v\n1",
"output": "cw"
},
{
"input": "v <\n1",
"output": "cw"
... | 1,696,664,988 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 46 | 0 | spinner=['v','<','^','>']
start,stop=input().split()
n=int(input())
for i in range(4):
if spinner[i]==start:
start=i
if spinner[i]==stop:
stop=i
if (start+n)%4==stop:
can_cw=True
else:
can_cw=False
start=3-start
stop=3-stop
if (start+n)%4==stop:
can_ccw=True
else:
can_ccw=False
if can_cw and can_ccw:
print('undefined')
elif can_ccw:
print('ccw')
else:
print('cw')
| Title: The Useless Toy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this.
Input Specification:
There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position.
Output Specification:
Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise.
Demo Input:
['^ >\n1\n', '< ^\n3\n', '^ v\n6\n']
Demo Output:
['cw\n', 'ccw\n', 'undefined\n']
Note:
none | ```python
spinner=['v','<','^','>']
start,stop=input().split()
n=int(input())
for i in range(4):
if spinner[i]==start:
start=i
if spinner[i]==stop:
stop=i
if (start+n)%4==stop:
can_cw=True
else:
can_cw=False
start=3-start
stop=3-stop
if (start+n)%4==stop:
can_ccw=True
else:
can_ccw=False
if can_cw and can_ccw:
print('undefined')
elif can_ccw:
print('ccw')
else:
print('cw')
``` | 3 | |
747 | D | Winter Is Coming | PROGRAMMING | 1,800 | [
"dp",
"greedy",
"sortings"
] | null | null | The winter in Berland lasts *n* days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than *k* days at any average air temperature. After *k* days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these *k* days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires. | The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than *k* days in total.
The second line contains a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=20<=≤<=*t**i*<=≤<=20) — the average air temperature in the *i*-th winter day. | Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1. | [
"4 3\n-5 20 -3 0\n",
"4 2\n-5 20 -3 0\n",
"10 6\n2 -5 1 3 0 0 -4 -3 1 0\n"
] | [
"2\n",
"4\n",
"3\n"
] | In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | 2,000 | [
{
"input": "4 3\n-5 20 -3 0",
"output": "2"
},
{
"input": "4 2\n-5 20 -3 0",
"output": "4"
},
{
"input": "10 6\n2 -5 1 3 0 0 -4 -3 1 0",
"output": "3"
},
{
"input": "4 4\n-5 20 -3 0",
"output": "1"
},
{
"input": "4 1\n-5 20 -3 0",
"output": "-1"
},
{
"... | 1,691,699,719 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1691699719.6118515")# 1691699719.611867 | Title: Winter Is Coming
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The winter in Berland lasts *n* days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than *k* days at any average air temperature. After *k* days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these *k* days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input Specification:
The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than *k* days in total.
The second line contains a sequence of *n* integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=20<=≤<=*t**i*<=≤<=20) — the average air temperature in the *i*-th winter day.
Output Specification:
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Demo Input:
['4 3\n-5 20 -3 0\n', '4 2\n-5 20 -3 0\n', '10 6\n2 -5 1 3 0 0 -4 -3 1 0\n']
Demo Output:
['2\n', '4\n', '3\n']
Note:
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | ```python
print("_RANDOM_GUESS_1691699719.6118515")# 1691699719.611867
``` | 0 | |
75 | A | Life Without Zeros | PROGRAMMING | 1,000 | [
"implementation"
] | A. Life Without Zeros | 2 | 256 | Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation. | The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*. | The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise. | [
"101\n102\n",
"105\n106\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
}... | 1,618,391,343 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 154 | 0 | def daoso (n):
tong = 0
while(n!=0):
le = n%10
tong = tong*10+le
n=n//10
return (tong)
def rmzero(n):
new = 0
count = 0
while (n!=0):
res = n%10
n = n//10
if res != 0:
new = new*10 + res
new = daoso(new)
return new
a = int(input())
b = int(input())
if rmzero(a+b) == (rmzero(a)+rmzero(b)):
print('YES')
else:
print('NO') | Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
def daoso (n):
tong = 0
while(n!=0):
le = n%10
tong = tong*10+le
n=n//10
return (tong)
def rmzero(n):
new = 0
count = 0
while (n!=0):
res = n%10
n = n//10
if res != 0:
new = new*10 + res
new = daoso(new)
return new
a = int(input())
b = int(input())
if rmzero(a+b) == (rmzero(a)+rmzero(b)):
print('YES')
else:
print('NO')
``` | 3.9615 |
84 | A | Toy Army | PROGRAMMING | 900 | [
"math",
"number theory"
] | A. Toy Army | 2 | 256 | The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows.
There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore.
The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends.
You are asked to calculate the maximum total number of soldiers that may be killed during the game. | The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields. | Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns. | [
"2\n",
"4\n"
] | [
"3\n",
"6\n"
] | The first sample test:
1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1.
2) Arcady's soldier 2 shoots at Valera's soldier 1.
3) Valera's soldier 1 shoots at Arcady's soldier 2.
There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2. | 500 | [
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "6"
},
{
"input": "6",
"output": "9"
},
{
"input": "8",
"output": "12"
},
{
"input": "10",
"output": "15"
},
{
"input": "140",
"output": "210"
},
{
"input": "500",
"output": "75... | 1,570,567,063 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 218 | 0 | n = int(input())
print(int(n*1.5))
| Title: Toy Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows.
There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore.
The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends.
You are asked to calculate the maximum total number of soldiers that may be killed during the game.
Input Specification:
The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields.
Output Specification:
Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns.
Demo Input:
['2\n', '4\n']
Demo Output:
['3\n', '6\n']
Note:
The first sample test:
1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1.
2) Arcady's soldier 2 shoots at Valera's soldier 1.
3) Valera's soldier 1 shoots at Arcady's soldier 2.
There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2. | ```python
n = int(input())
print(int(n*1.5))
``` | 3.9455 |
0 | none | none | none | 0 | [
"none"
] | null | null | From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of *n* lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation *n*<=-<=1 times:
- Remove any two elements *s* and *t* from the set, and add their concatenation *s*<=+<=*t* to the set.
The cost of such operation is defined to be , where *f*(*s*,<=*c*) denotes the number of times character *c* appears in string *s*.
Given a non-negative integer *k*, construct any valid non-empty set of no more than 100<=000 letters, such that the minimum accumulative cost of the whole process is exactly *k*. It can be shown that a solution always exists. | The first and only line of input contains a non-negative integer *k* (0<=≤<=*k*<=≤<=100<=000) — the required minimum cost. | Output a non-empty string of no more than 100<=000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters. | [
"12\n",
"3\n"
] | [
"abababab\n",
"codeforces\n"
] | For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
- {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0; - {"aba", "b", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "ab", "a", "b"}, with a cost of 0; - {"abab", "aba", "b"}, with a cost of 1; - {"abab", "abab"}, with a cost of 1; - {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | 0 | [
{
"input": "12",
"output": "abababab"
},
{
"input": "3",
"output": "codeforces"
},
{
"input": "0",
"output": "o"
},
{
"input": "2",
"output": "aabb"
},
{
"input": "5",
"output": "aaabbcc"
},
{
"input": "10",
"output": "aaaaa"
},
{
"input": ... | 1,504,276,821 | 3,921 | PyPy 3 | OK | TESTS | 25 | 93 | 307,200 | k = int(input())
ans = ""
cost = 0
for c in [chr(ord('a') + i) for i in range(26)]:
i = 0
while cost + i <= k:
ans += c
cost += i
i += 1
print(ans)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of *n* lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation *n*<=-<=1 times:
- Remove any two elements *s* and *t* from the set, and add their concatenation *s*<=+<=*t* to the set.
The cost of such operation is defined to be , where *f*(*s*,<=*c*) denotes the number of times character *c* appears in string *s*.
Given a non-negative integer *k*, construct any valid non-empty set of no more than 100<=000 letters, such that the minimum accumulative cost of the whole process is exactly *k*. It can be shown that a solution always exists.
Input Specification:
The first and only line of input contains a non-negative integer *k* (0<=≤<=*k*<=≤<=100<=000) — the required minimum cost.
Output Specification:
Output a non-empty string of no more than 100<=000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Demo Input:
['12\n', '3\n']
Demo Output:
['abababab\n', 'codeforces\n']
Note:
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
- {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0; - {"aba", "b", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "ab", "a", "b"}, with a cost of 0; - {"abab", "aba", "b"}, with a cost of 1; - {"abab", "abab"}, with a cost of 1; - {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | ```python
k = int(input())
ans = ""
cost = 0
for c in [chr(ord('a') + i) for i in range(26)]:
i = 0
while cost + i <= k:
ans += c
cost += i
i += 1
print(ans)
``` | 3 | |
14 | A | Letter | PROGRAMMING | 800 | [
"implementation"
] | A. Letter | 1 | 64 | A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides. | The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square. | Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better. | [
"6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n",
"3 3\n***\n*.*\n***\n"
] | [
"***\n*..\n***\n*..\n***\n",
"***\n*.*\n***\n"
] | none | 0 | [
{
"input": "6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..",
"output": "***\n*..\n***\n*..\n***"
},
{
"input": "3 3\n***\n*.*\n***",
"output": "***\n*.*\n***"
},
{
"input": "1 1\n*",
"output": "*"
},
{
"input": "2 1\n*\n*",
"output": "*\n*"
},
{
"input"... | 1,522,598,480 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 156 | 7,065,600 | n, m = map(int, input().split())
mat = [list(input().strip()) for i in range(n)]
left = -1
tmat = list(zip(*mat))
for i in range(m):
if '*' in tmat[i]:
left = i
break
for i in range(m - 1, -1, -1):
if '*' in tmat[i]:
right = i
break
for i in range(n):
if '*' in mat[i]:
top = i
break
for i in range(n - 1, -1, -1):
if '*' in mat[i]:
bottom = i
break
for i in range(top, bottom + 1):
ch = ''
for j in range(left, right + 1):
ch += mat[i][j]
print(ch)
| Title: Letter
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input Specification:
The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output Specification:
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Demo Input:
['6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n', '3 3\n***\n*.*\n***\n']
Demo Output:
['***\n*..\n***\n*..\n***\n', '***\n*.*\n***\n']
Note:
none | ```python
n, m = map(int, input().split())
mat = [list(input().strip()) for i in range(n)]
left = -1
tmat = list(zip(*mat))
for i in range(m):
if '*' in tmat[i]:
left = i
break
for i in range(m - 1, -1, -1):
if '*' in tmat[i]:
right = i
break
for i in range(n):
if '*' in mat[i]:
top = i
break
for i in range(n - 1, -1, -1):
if '*' in mat[i]:
bottom = i
break
for i in range(top, bottom + 1):
ch = ''
for j in range(left, right + 1):
ch += mat[i][j]
print(ch)
``` | 3.869357 |
940 | C | Phone Numbers | PROGRAMMING | 1,500 | [
"constructive algorithms",
"implementation",
"strings"
] | null | null | And where the are the phone numbers?
You are given a string *s* consisting of lowercase English letters and an integer *k*. Find the lexicographically smallest string *t* of length *k*, such that its set of letters is a subset of the set of letters of *s* and *s* is lexicographically smaller than *t*.
It's guaranteed that the answer exists.
Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {*a*,<=*b*,<=*d*}.
String *p* is lexicographically smaller than string *q*, if *p* is a prefix of *q*, is not equal to *q* or there exists *i*, such that *p**i*<=<<=*q**i* and for all *j*<=<<=*i* it is satisfied that *p**j*<==<=*q**j*. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a. | The first line of input contains two space separated integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100<=000) — the length of *s* and the required length of *t*.
The second line of input contains the string *s* consisting of *n* lowercase English letters. | Output the string *t* conforming to the requirements above.
It's guaranteed that the answer exists. | [
"3 3\nabc\n",
"3 2\nabc\n",
"3 3\nayy\n",
"2 3\nba\n"
] | [
"aca\n",
"ac\n",
"yaa\n",
"baa\n"
] | In the first example the list of strings *t* of length 3, such that the set of letters of *t* is a subset of letters of *s* is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca. | 1,250 | [
{
"input": "3 3\nabc",
"output": "aca"
},
{
"input": "3 2\nabc",
"output": "ac"
},
{
"input": "3 3\nayy",
"output": "yaa"
},
{
"input": "2 3\nba",
"output": "baa"
},
{
"input": "1 3\nf",
"output": "fff"
},
{
"input": "3 1\nazz",
"output": "z"
},
... | 1,519,889,060 | 2,147,483,647 | PyPy 3 | OK | TESTS | 54 | 93 | 25,088,000 | n, k = [int(x) for x in input().split(' ')]
s = input()
alph = sorted(list(set(s)))
if k <= n:
i = min(n, k) - 1
while s[i] == alph[-1]:
i -= 1
r = s[:i] + alph[alph.index(s[i])+1] + alph[0]*(k-i-1)
else:
r = s + alph[0]*(k-n)
print(r) | Title: Phone Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
And where the are the phone numbers?
You are given a string *s* consisting of lowercase English letters and an integer *k*. Find the lexicographically smallest string *t* of length *k*, such that its set of letters is a subset of the set of letters of *s* and *s* is lexicographically smaller than *t*.
It's guaranteed that the answer exists.
Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {*a*,<=*b*,<=*d*}.
String *p* is lexicographically smaller than string *q*, if *p* is a prefix of *q*, is not equal to *q* or there exists *i*, such that *p**i*<=<<=*q**i* and for all *j*<=<<=*i* it is satisfied that *p**j*<==<=*q**j*. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.
Input Specification:
The first line of input contains two space separated integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100<=000) — the length of *s* and the required length of *t*.
The second line of input contains the string *s* consisting of *n* lowercase English letters.
Output Specification:
Output the string *t* conforming to the requirements above.
It's guaranteed that the answer exists.
Demo Input:
['3 3\nabc\n', '3 2\nabc\n', '3 3\nayy\n', '2 3\nba\n']
Demo Output:
['aca\n', 'ac\n', 'yaa\n', 'baa\n']
Note:
In the first example the list of strings *t* of length 3, such that the set of letters of *t* is a subset of letters of *s* is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca. | ```python
n, k = [int(x) for x in input().split(' ')]
s = input()
alph = sorted(list(set(s)))
if k <= n:
i = min(n, k) - 1
while s[i] == alph[-1]:
i -= 1
r = s[:i] + alph[alph.index(s[i])+1] + alph[0]*(k-i-1)
else:
r = s + alph[0]*(k-n)
print(r)
``` | 3 | |
987 | B | High School: Become Human | PROGRAMMING | 1,100 | [
"math"
] | null | null | Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him. | On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$). | If $x^y < y^x$, then print '<' (without quotes). If $x^y > y^x$, then print '>' (without quotes). If $x^y = y^x$, then print '=' (without quotes). | [
"5 8\n",
"10 3\n",
"6 6\n"
] | [
">\n",
"<\n",
"=\n"
] | In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '>'.
In the second example $10^3 = 1000 < 3^{10} = 59049$.
In the third example $6^6 = 46656 = 6^6$. | 1,000 | [
{
"input": "5 8",
"output": ">"
},
{
"input": "10 3",
"output": "<"
},
{
"input": "6 6",
"output": "="
},
{
"input": "14 1",
"output": ">"
},
{
"input": "2 4",
"output": "="
},
{
"input": "987654321 123456987",
"output": "<"
},
{
"input": "... | 1,582,021,440 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 8,908,800 | def fast_pow(x, y):
if y == 0:
return 1
p = fast_pow(x, y // 2)
p *= p
if y % 2:
p *= x
return p
x, y = list(map(int, input().split()))
xy = fast_pow(x, y)
yx = fast_pow(y, x)
if xy > yx:
print(">")
elif xy == yx:
print("=")
else:
print("<")
| Title: High School: Become Human
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him.
Input Specification:
On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$).
Output Specification:
If $x^y < y^x$, then print '<' (without quotes). If $x^y > y^x$, then print '>' (without quotes). If $x^y = y^x$, then print '=' (without quotes).
Demo Input:
['5 8\n', '10 3\n', '6 6\n']
Demo Output:
['>\n', '<\n', '=\n']
Note:
In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '>'.
In the second example $10^3 = 1000 < 3^{10} = 59049$.
In the third example $6^6 = 46656 = 6^6$. | ```python
def fast_pow(x, y):
if y == 0:
return 1
p = fast_pow(x, y // 2)
p *= p
if y % 2:
p *= x
return p
x, y = list(map(int, input().split()))
xy = fast_pow(x, y)
yx = fast_pow(y, x)
if xy > yx:
print(">")
elif xy == yx:
print("=")
else:
print("<")
``` | 0 | |
690 | C1 | Brain Network (easy) | PROGRAMMING | 1,300 | [] | null | null | One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know why their movements are so sluggish, the problem of laggy thinking has been recently resolved. It turns out that the reason is not (as previously suspected) any kind of brain defect – it's the opposite! Independent researchers confirmed that the nervous system of a zombie is highly complicated – it consists of *n* brains (much like a cow has several stomachs). They are interconnected by brain connectors, which are veins capable of transmitting thoughts between brains. There are two important properties such a brain network should have to function properly:
1. It should be possible to exchange thoughts between any two pairs of brains (perhaps indirectly, through other brains). 1. There should be no redundant brain connectors, that is, removing any brain connector would make property 1 false.
If both properties are satisfied, we say that the nervous system is valid. Unfortunately (?), if the system is not valid, the zombie stops thinking and becomes (even more) dead. Your task is to analyze a given nervous system of a zombie and find out whether it is valid. | The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) denoting the number of brains (which are conveniently numbered from 1 to *n*) and the number of brain connectors in the nervous system, respectively. In the next *m* lines, descriptions of brain connectors follow. Every connector is given as a pair of brains *a* *b* it connects (1<=≤<=*a*,<=*b*<=≤<=*n*, *a*<=≠<=*b*). | The output consists of one line, containing either yes or no depending on whether the nervous system is valid. | [
"4 4\n1 2\n2 3\n3 1\n4 1\n",
"6 5\n1 2\n2 3\n3 4\n4 5\n3 6\n"
] | [
"no\n",
"yes\n"
] | none | 0 | [
{
"input": "4 4\n1 2\n2 3\n3 1\n4 1",
"output": "no"
},
{
"input": "6 5\n1 2\n2 3\n3 4\n4 5\n3 6",
"output": "yes"
},
{
"input": "2 1\n1 2",
"output": "yes"
},
{
"input": "3 3\n2 1\n1 3\n3 2",
"output": "no"
},
{
"input": "3 2\n1 2\n2 3",
"output": "yes"
},
... | 1,654,718,345 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 |
def DFS(node,visited, graph,count):
visited[node] = True
count[0] += 1
for neighbour in graph[node]:
if(visited[neighbour] == False):
DFS(neighbour,visited, graph,count)
def main():
inp = input().split()
n,m = [int(i) for i in inp]
if(m != n-1):
print("NO")
return
graph = []
for _ in range(n):
graph.append([])
for _ in range(m):
inp = input().split()
u,v = [int(i) for i in inp]
graph[u].append(v)
graph[v].append(u)
visited = []
count = [0]
for _ in range(n):
visited.append(False)
DFS(0,visited, graph, count)
if(count[0] == n):
print("YES")
else:
print("NO")
if __name__ == "__main__":
main() | Title: Brain Network (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know why their movements are so sluggish, the problem of laggy thinking has been recently resolved. It turns out that the reason is not (as previously suspected) any kind of brain defect – it's the opposite! Independent researchers confirmed that the nervous system of a zombie is highly complicated – it consists of *n* brains (much like a cow has several stomachs). They are interconnected by brain connectors, which are veins capable of transmitting thoughts between brains. There are two important properties such a brain network should have to function properly:
1. It should be possible to exchange thoughts between any two pairs of brains (perhaps indirectly, through other brains). 1. There should be no redundant brain connectors, that is, removing any brain connector would make property 1 false.
If both properties are satisfied, we say that the nervous system is valid. Unfortunately (?), if the system is not valid, the zombie stops thinking and becomes (even more) dead. Your task is to analyze a given nervous system of a zombie and find out whether it is valid.
Input Specification:
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) denoting the number of brains (which are conveniently numbered from 1 to *n*) and the number of brain connectors in the nervous system, respectively. In the next *m* lines, descriptions of brain connectors follow. Every connector is given as a pair of brains *a* *b* it connects (1<=≤<=*a*,<=*b*<=≤<=*n*, *a*<=≠<=*b*).
Output Specification:
The output consists of one line, containing either yes or no depending on whether the nervous system is valid.
Demo Input:
['4 4\n1 2\n2 3\n3 1\n4 1\n', '6 5\n1 2\n2 3\n3 4\n4 5\n3 6\n']
Demo Output:
['no\n', 'yes\n']
Note:
none | ```python
def DFS(node,visited, graph,count):
visited[node] = True
count[0] += 1
for neighbour in graph[node]:
if(visited[neighbour] == False):
DFS(neighbour,visited, graph,count)
def main():
inp = input().split()
n,m = [int(i) for i in inp]
if(m != n-1):
print("NO")
return
graph = []
for _ in range(n):
graph.append([])
for _ in range(m):
inp = input().split()
u,v = [int(i) for i in inp]
graph[u].append(v)
graph[v].append(u)
visited = []
count = [0]
for _ in range(n):
visited.append(False)
DFS(0,visited, graph, count)
if(count[0] == n):
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
``` | 0 | |
194 | A | Exams | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | One day the Codeforces round author sat exams. He had *n* exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than *k*. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than *k*, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all *n* exams equal exactly *k*. | The single input line contains space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=250) — the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass *n* exams in the way that makes the sum of marks equal exactly *k*. | Print the single number — the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal *k*. | [
"4 8\n",
"4 10\n",
"1 3\n"
] | [
"4\n",
"2\n",
"0\n"
] | In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam. | 500 | [
{
"input": "4 8",
"output": "4"
},
{
"input": "4 10",
"output": "2"
},
{
"input": "1 3",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "50 234",
"output": "0"
},
{
"input": "50 100",
"... | 1,610,275,227 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 124 | 0 | n,k=input().split()
n=int(n)
k=int(k)
print(max(n*3-k,0)) | Title: Exams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day the Codeforces round author sat exams. He had *n* exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than *k*. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than *k*, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all *n* exams equal exactly *k*.
Input Specification:
The single input line contains space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=250) — the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass *n* exams in the way that makes the sum of marks equal exactly *k*.
Output Specification:
Print the single number — the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal *k*.
Demo Input:
['4 8\n', '4 10\n', '1 3\n']
Demo Output:
['4\n', '2\n', '0\n']
Note:
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam. | ```python
n,k=input().split()
n=int(n)
k=int(k)
print(max(n*3-k,0))
``` | 3 | |
230 | B | T-primes | PROGRAMMING | 1,300 | [
"binary search",
"implementation",
"math",
"number theory"
] | null | null | We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not. | The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier. | Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't. | [
"3\n4 5 6\n"
] | [
"YES\nNO\nNO\n"
] | The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO". | 500 | [
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"... | 1,697,371,073 | 2,147,483,647 | Python 3 | OK | TESTS | 68 | 1,496 | 23,040,000 | import math
n=input()
prime=[1 for i in range(1000001)]
for i in range(2,1000001):
if prime[i]:
for k in range(i*2,1000001,i):
prime[k]=0
def is_Tprime(a):
tmp=math.sqrt(a)
if int(tmp)==tmp:
if prime[int(tmp)]:
return 1
return 0
arr=list(map(int,input().split()))
for i in arr:
if i==1:
print("NO")
continue
if is_Tprime(i):
print("YES")
else:
print("NO") | Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO". | ```python
import math
n=input()
prime=[1 for i in range(1000001)]
for i in range(2,1000001):
if prime[i]:
for k in range(i*2,1000001,i):
prime[k]=0
def is_Tprime(a):
tmp=math.sqrt(a)
if int(tmp)==tmp:
if prime[int(tmp)]:
return 1
return 0
arr=list(map(int,input().split()))
for i in arr:
if i==1:
print("NO")
continue
if is_Tprime(i):
print("YES")
else:
print("NO")
``` | 3 | |
112 | A | Petya and Strings | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Petya and Strings | 2 | 256 | Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison. | Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters. | If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared. | [
"aaaa\naaaA\n",
"abs\nAbz\n",
"abcdefg\nAbCdEfF\n"
] | [
"0\n",
"-1\n",
"1\n"
] | If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | 500 | [
{
"input": "aaaa\naaaA",
"output": "0"
},
{
"input": "abs\nAbz",
"output": "-1"
},
{
"input": "abcdefg\nAbCdEfF",
"output": "1"
},
{
"input": "asadasdasd\nasdwasdawd",
"output": "-1"
},
{
"input": "aslkjlkasdd\nasdlkjdajwi",
"output": "1"
},
{
"input":... | 1,695,301,934 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | import sys
import math
import bisect
def input(): return sys.stdin.readline().rstrip("\r\n")
def main():
length = int(input())
#string = input()import sys
import math
import bisect
def input(): return sys.stdin.readline().rstrip("\r\n")
def main():
#length = int(input())
string = input()
string_2 = input()
ans = 0
#numbers = [*map(int, input().split())]
a=sum(ord(i) for i in string.lower())
b=sum(ord(i) for i in string_2.lower())
if a==b:print(0)
else:
print(1 if a>b else -1)
if __name__ == "__main__":
main()
ans = 0
#numbers = [*map(int, input().split())]
for i in range(length):
if "+" in input():
ans += 1
else:
ans -= 1
print(ans)
if __name__ == "__main__":
main()
| Title: Petya and Strings
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input Specification:
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output Specification:
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Demo Input:
['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | ```python
import sys
import math
import bisect
def input(): return sys.stdin.readline().rstrip("\r\n")
def main():
length = int(input())
#string = input()import sys
import math
import bisect
def input(): return sys.stdin.readline().rstrip("\r\n")
def main():
#length = int(input())
string = input()
string_2 = input()
ans = 0
#numbers = [*map(int, input().split())]
a=sum(ord(i) for i in string.lower())
b=sum(ord(i) for i in string_2.lower())
if a==b:print(0)
else:
print(1 if a>b else -1)
if __name__ == "__main__":
main()
ans = 0
#numbers = [*map(int, input().split())]
for i in range(length):
if "+" in input():
ans += 1
else:
ans -= 1
print(ans)
if __name__ == "__main__":
main()
``` | -1 |
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"
... | 1,691,180,550 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 92 | 0 | input()
c = 0
a = list(map(int, input().split()))
a.sort()
a.reverse()
t = 0
for elem in a:
t += elem
if t % 2 == 0:
g = t // 2 + 1
else:
g = t // 2
t = 0
for elem in a:
c += 1
t += elem
if t > g:
print(c)
break | 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
input()
c = 0
a = list(map(int, input().split()))
a.sort()
a.reverse()
t = 0
for elem in a:
t += elem
if t % 2 == 0:
g = t // 2 + 1
else:
g = t // 2
t = 0
for elem in a:
c += 1
t += elem
if t > g:
print(c)
break
``` | 0 | |
867 | A | Between the Offices | PROGRAMMING | 800 | [
"implementation"
] | null | null | As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not. | The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence. | Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower). | [
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] | [
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] | In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | 500 | [
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFF... | 1,574,434,644 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 124 | 0 | n = input()
word = input()
SF = 0
FS = 0
for i in range (0, int(n)-1) :
if word[i] + word[i+1] == "SF" :
SF += 1
elif word[i] + word[i+1] == "FS" :
FS += 1
if SF > FS :
print("yes")
else :
print("no")
| Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | ```python
n = input()
word = input()
SF = 0
FS = 0
for i in range (0, int(n)-1) :
if word[i] + word[i+1] == "SF" :
SF += 1
elif word[i] + word[i+1] == "FS" :
FS += 1
if SF > FS :
print("yes")
else :
print("no")
``` | 3 | |
35 | A | Shell Game | PROGRAMMING | 1,000 | [
"implementation"
] | A. Shell Game | 2 | 64 | Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too? | The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3. | In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles. | [
"1\n1 2\n2 1\n2 1\n",
"1\n2 1\n3 1\n1 3\n"
] | [
"2\n",
"2\n"
] | none | 500 | [
{
"input": "1\n1 2\n2 1\n2 1",
"output": "2"
},
{
"input": "1\n2 1\n3 1\n1 3",
"output": "2"
},
{
"input": "3\n3 1\n2 1\n1 2",
"output": "1"
},
{
"input": "1\n1 3\n1 2\n2 3",
"output": "2"
},
{
"input": "3\n3 2\n3 1\n3 1",
"output": "2"
},
{
"input": "... | 1,595,962,300 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 154 | 6,656,000 | # -*- coding: utf-8 -*-
lista = [0,0,0]
lista[int(input())-1] = 1
for i in range(3):
x,y = map(int,input().split())
lista[x-1],lista[y-1] = lista[y-1],lista[x-1]
print(lista.index(1)+1)
| Title: Shell Game
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too?
Input Specification:
The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3.
Output Specification:
In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles.
Demo Input:
['1\n1 2\n2 1\n2 1\n', '1\n2 1\n3 1\n1 3\n']
Demo Output:
['2\n', '2\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
lista = [0,0,0]
lista[int(input())-1] = 1
for i in range(3):
x,y = map(int,input().split())
lista[x-1],lista[y-1] = lista[y-1],lista[x-1]
print(lista.index(1)+1)
``` | -1 |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,695,896,918 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 31 | 0 | n, m, a = map(int, input().split())
x = 0
if a == 1:
x = (m*n)//a
else:
while n > 0:
n -= a
x +=1
while m > 0:
m -= a
x +=1
print(x) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n, m, a = map(int, input().split())
x = 0
if a == 1:
x = (m*n)//a
else:
while n > 0:
n -= a
x +=1
while m > 0:
m -= a
x +=1
print(x)
``` | 0 |
946 | B | Weird Subtraction Process | PROGRAMMING | 1,100 | [
"math",
"number theory"
] | null | null | You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends. | The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*. | Print two integers — the values of *a* and *b* after the end of the process. | [
"12 5\n",
"31 12\n"
] | [
"0 1\n",
"7 12\n"
] | Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12. | 0 | [
{
"input": "12 5",
"output": "0 1"
},
{
"input": "31 12",
"output": "7 12"
},
{
"input": "1000000000000000000 7",
"output": "8 7"
},
{
"input": "31960284556200 8515664064180",
"output": "14928956427840 8515664064180"
},
{
"input": "1000000000000000000 100000000000... | 1,678,548,222 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 30 | 512,000 | n, m = map(int, input().split())
def solution(a: int, b: int):
if a == 0 or b == 0:
return (a, b)
else:
if a >= 2 * b:
a -= 2 * b
return solution(a, b)
else:
if b >= 2 * a:
b -= 2*a
return solution(a, b)
else:
return (a, b)
result = solution(n, m)
print(result[0], result[1]) | Title: Weird Subtraction Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*.
Output Specification:
Print two integers — the values of *a* and *b* after the end of the process.
Demo Input:
['12 5\n', '31 12\n']
Demo Output:
['0 1\n', '7 12\n']
Note:
Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12. | ```python
n, m = map(int, input().split())
def solution(a: int, b: int):
if a == 0 or b == 0:
return (a, b)
else:
if a >= 2 * b:
a -= 2 * b
return solution(a, b)
else:
if b >= 2 * a:
b -= 2*a
return solution(a, b)
else:
return (a, b)
result = solution(n, m)
print(result[0], result[1])
``` | -1 | |
771 | A | Bear and Friendship Condition | PROGRAMMING | 1,500 | [
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes. | The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input. | If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes). | [
"4 3\n1 3\n3 4\n1 4\n",
"4 4\n3 1\n2 3\n3 4\n1 2\n",
"10 4\n4 3\n5 10\n8 9\n1 2\n",
"3 2\n1 2\n2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] | The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | 250 | [
{
"input": "4 3\n1 3\n3 4\n1 4",
"output": "YES"
},
{
"input": "4 4\n3 1\n2 3\n3 4\n1 2",
"output": "NO"
},
{
"input": "10 4\n4 3\n5 10\n8 9\n1 2",
"output": "YES"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 0",
"output": "YES"
},
{
... | 1,489,857,845 | 6,545 | Python 3 | WRONG_ANSWER | TESTS | 27 | 904 | 21,606,400 | """ Created by Henrikh Kantuni on 3/18/17 """
if __name__ == "__main__":
n, m = [int(x) for x in input().split()]
mc = m
friends = {}
if m == 0 or m == 1:
print("YES")
exit()
while mc > 0:
x, y = [int(x) for x in input().split()]
if x not in friends:
friends[x] = []
friends[x].append(y)
if y not in friends:
friends[y] = []
friends[y].append(x)
mc -= 1
reasonable = True
for k in friends.keys():
for v in friends[k]:
if len(friends[k]) != len(friends[v]):
reasonable = False
break
if reasonable:
print("YES")
else:
print("NO")
| Title: Bear and Friendship Condition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input Specification:
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
Output Specification:
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Demo Input:
['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | ```python
""" Created by Henrikh Kantuni on 3/18/17 """
if __name__ == "__main__":
n, m = [int(x) for x in input().split()]
mc = m
friends = {}
if m == 0 or m == 1:
print("YES")
exit()
while mc > 0:
x, y = [int(x) for x in input().split()]
if x not in friends:
friends[x] = []
friends[x].append(y)
if y not in friends:
friends[y] = []
friends[y].append(x)
mc -= 1
reasonable = True
for k in friends.keys():
for v in friends[k]:
if len(friends[k]) != len(friends[v]):
reasonable = False
break
if reasonable:
print("YES")
else:
print("NO")
``` | 0 | |
854 | A | Fraction | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"math"
] | null | null | Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem. | In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction. | Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum. | [
"3\n",
"4\n",
"12\n"
] | [
"1 2\n",
"1 3\n",
"5 7\n"
] | none | 500 | [
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"ou... | 1,640,920,622 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 108 | 0 | # test = int(input())
# for tt in range(0, test, 1):
n=int(input())
if n&1:
print(int(n/2),n-int(n/2))
else:
print(int(n/2-1),int(n/2+1)) | Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none | ```python
# test = int(input())
# for tt in range(0, test, 1):
n=int(input())
if n&1:
print(int(n/2),n-int(n/2))
else:
print(int(n/2-1),int(n/2+1))
``` | 0 | |
527 | C | Glass Carving | PROGRAMMING | 1,500 | [
"binary search",
"data structures",
"implementation"
] | null | null | Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree? | The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts. | After each cut print on a single line the area of the maximum available glass fragment in mm2. | [
"4 3 4\nH 2\nV 2\nV 3\nV 1\n",
"7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n"
] | [
"8\n4\n4\n2\n",
"28\n16\n12\n6\n4\n"
] | Picture for the first sample test: | 1,500 | [
{
"input": "4 3 4\nH 2\nV 2\nV 3\nV 1",
"output": "8\n4\n4\n2"
},
{
"input": "7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1",
"output": "28\n16\n12\n6\n4"
},
{
"input": "2 2 1\nV 1",
"output": "2"
},
{
"input": "2 2 1\nH 1",
"output": "2"
},
{
"input": "2 2 2\nV 1\nH 1",
"ou... | 1,658,488,711 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int N=4e5+100;
const int MOD = 2147483647;
struct Node
{
int left;
int right;
int val;
int key;
int size;
}tree1[N],tree2[N],tree3[N],tree4[N];
int root[10],tot[10];
int add(int val, int num, Node *tree){
tot[num]++;
tree[tot[num]].left=0;
tree[tot[num]].right=0;
tree[tot[num]].val=val;
tree[tot[num]].key=rand()%MOD;
tree[tot[num]].size=1;
return tot[num];
}
//
void update_root(int now, int num, Node *tree){
int left,right;
left=tree[now].left;
right=tree[now].right;
tree[now].size=tree[left].size+tree[right].size+1;
}
//
void spilt_new(int now, int &a, int &b, int val, int num, Node *tree){
if(now==0){
a=0;
b=0;
return;
}
if(tree[now].val<=val){
a=now;
spilt_new(tree[now].right, tree[a].right, b, val, num, tree);
}else{
b=now;
spilt_new(tree[now].left, a, tree[b].left, val, num, tree);
}
update_root(now, num, tree);
}
//
void merge_new(int &now, int a, int b, int num, Node *tree){
if(a==0 || b==0){
now=a+b;
return;
}
if(tree[a].key<tree[b].key){
now=a;
merge_new(tree[now].right, tree[a].right, b, num, tree);
}else{
now=b;
merge_new(tree[now].left, a, tree[b].left, num, tree);
}
update_root(now, num, tree);
}
//
void insert_new(int val, int num, Node *tree){
int x,y,z;
x=0;
y=0;
z=add(val,num,tree);
spilt_new(root[num],x,y,val,num,tree);
merge_new(x,x,z,num,tree);
merge_new(root[num],x,y,num,tree);
}
//
void del_new(int val, int num, Node *tree){
int x,y,z;
x=y=z=0;
spilt_new(root[num],x,y,val,num,tree);
spilt_new(x,x,z,val-1,num,tree);
merge_new(z,tree[z].left,tree[z].right,num,tree);
merge_new(x,x,z,num,tree);
merge_new(root[num],x,y,num,tree);
}
//
int find_new(int now, int rank, int num, Node *tree){
while(tree[tree[now].left].size+1!=rank){
if(tree[tree[now].left].size>=rank){
now=tree[now].left;
}else{
rank-=tree[tree[now].left].size+1;
now=tree[now].right;
}
}
return tree[now].val;
}
//
int get_val(int rank, int num, Node *tree){
return find_new(root[num], rank, num, tree);
}
int get_pre(int val, int num, Node *tree){
int x,y;
int ans;
x=y=0;
spilt_new(root[num], x, y, val-1, num, tree);
ans=find_new(x, tree[x].size, num, tree);
merge_new(root[num], x, y, num, tree);
return ans;
}
//
int get_next(int val, int num, Node *tree){
int x,y;
int ans;
x=y=0;
spilt_new(root[num], x, y, val, num, tree);
ans=find_new(y,1,num,tree);
merge_new(root[num], x, y, num, tree);
return ans;
}
int main(){
int wid,hig,left,right;
int m,x,w,h;
char op;
ll ans;
scanf("%d%d%d", &wid, &hig, &m);
memset(tree1,0,sizeof(tree1));
memset(tree2,0,sizeof(tree2));
memset(tree3,0,sizeof(tree3));
memset(tree4,0,sizeof(tree4));
root[1]=add(0,1,tree1);
insert_new(wid,1,tree1);
root[3]=add(0,3,tree3);
insert_new(hig,3,tree3);
root[2]=add(wid,2,tree2);
root[4]=add(hig,4,tree4);
for(int i=1;i<=m;i++){
scanf(" %c%d", &op, &x);
if(op=='H'){
left=get_pre(x, 3, tree3);
right=get_next(x, 3, tree3);
insert_new(x,3,tree3);
insert_new(x-left, 4, tree4);
insert_new(right-x, 4, tree4);
del_new(right-left, 4, tree4);
}else{
left=get_pre(x, 1, tree1);
right=get_next(x, 1, tree1);
insert_new(x, 1, tree1);
insert_new(x-left, 2, tree2);
insert_new(right-x, 2, tree2);
del_new(right-left, 2, tree2);
}
w=get_val(tree2[root[2]].size, 2, tree2);
h=get_val(tree4[root[4]].size, 4, tree4);
ans=(ll)h*w;
printf("%lld\n", ans);
}
return 0;
}
| Title: Glass Carving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
Input Specification:
The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
Output Specification:
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Demo Input:
['4 3 4\nH 2\nV 2\nV 3\nV 1\n', '7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n']
Demo Output:
['8\n4\n4\n2\n', '28\n16\n12\n6\n4\n']
Note:
Picture for the first sample test: | ```python
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int N=4e5+100;
const int MOD = 2147483647;
struct Node
{
int left;
int right;
int val;
int key;
int size;
}tree1[N],tree2[N],tree3[N],tree4[N];
int root[10],tot[10];
int add(int val, int num, Node *tree){
tot[num]++;
tree[tot[num]].left=0;
tree[tot[num]].right=0;
tree[tot[num]].val=val;
tree[tot[num]].key=rand()%MOD;
tree[tot[num]].size=1;
return tot[num];
}
//
void update_root(int now, int num, Node *tree){
int left,right;
left=tree[now].left;
right=tree[now].right;
tree[now].size=tree[left].size+tree[right].size+1;
}
//
void spilt_new(int now, int &a, int &b, int val, int num, Node *tree){
if(now==0){
a=0;
b=0;
return;
}
if(tree[now].val<=val){
a=now;
spilt_new(tree[now].right, tree[a].right, b, val, num, tree);
}else{
b=now;
spilt_new(tree[now].left, a, tree[b].left, val, num, tree);
}
update_root(now, num, tree);
}
//
void merge_new(int &now, int a, int b, int num, Node *tree){
if(a==0 || b==0){
now=a+b;
return;
}
if(tree[a].key<tree[b].key){
now=a;
merge_new(tree[now].right, tree[a].right, b, num, tree);
}else{
now=b;
merge_new(tree[now].left, a, tree[b].left, num, tree);
}
update_root(now, num, tree);
}
//
void insert_new(int val, int num, Node *tree){
int x,y,z;
x=0;
y=0;
z=add(val,num,tree);
spilt_new(root[num],x,y,val,num,tree);
merge_new(x,x,z,num,tree);
merge_new(root[num],x,y,num,tree);
}
//
void del_new(int val, int num, Node *tree){
int x,y,z;
x=y=z=0;
spilt_new(root[num],x,y,val,num,tree);
spilt_new(x,x,z,val-1,num,tree);
merge_new(z,tree[z].left,tree[z].right,num,tree);
merge_new(x,x,z,num,tree);
merge_new(root[num],x,y,num,tree);
}
//
int find_new(int now, int rank, int num, Node *tree){
while(tree[tree[now].left].size+1!=rank){
if(tree[tree[now].left].size>=rank){
now=tree[now].left;
}else{
rank-=tree[tree[now].left].size+1;
now=tree[now].right;
}
}
return tree[now].val;
}
//
int get_val(int rank, int num, Node *tree){
return find_new(root[num], rank, num, tree);
}
int get_pre(int val, int num, Node *tree){
int x,y;
int ans;
x=y=0;
spilt_new(root[num], x, y, val-1, num, tree);
ans=find_new(x, tree[x].size, num, tree);
merge_new(root[num], x, y, num, tree);
return ans;
}
//
int get_next(int val, int num, Node *tree){
int x,y;
int ans;
x=y=0;
spilt_new(root[num], x, y, val, num, tree);
ans=find_new(y,1,num,tree);
merge_new(root[num], x, y, num, tree);
return ans;
}
int main(){
int wid,hig,left,right;
int m,x,w,h;
char op;
ll ans;
scanf("%d%d%d", &wid, &hig, &m);
memset(tree1,0,sizeof(tree1));
memset(tree2,0,sizeof(tree2));
memset(tree3,0,sizeof(tree3));
memset(tree4,0,sizeof(tree4));
root[1]=add(0,1,tree1);
insert_new(wid,1,tree1);
root[3]=add(0,3,tree3);
insert_new(hig,3,tree3);
root[2]=add(wid,2,tree2);
root[4]=add(hig,4,tree4);
for(int i=1;i<=m;i++){
scanf(" %c%d", &op, &x);
if(op=='H'){
left=get_pre(x, 3, tree3);
right=get_next(x, 3, tree3);
insert_new(x,3,tree3);
insert_new(x-left, 4, tree4);
insert_new(right-x, 4, tree4);
del_new(right-left, 4, tree4);
}else{
left=get_pre(x, 1, tree1);
right=get_next(x, 1, tree1);
insert_new(x, 1, tree1);
insert_new(x-left, 2, tree2);
insert_new(right-x, 2, tree2);
del_new(right-left, 2, tree2);
}
w=get_val(tree2[root[2]].size, 2, tree2);
h=get_val(tree4[root[4]].size, 4, tree4);
ans=(ll)h*w;
printf("%lld\n", ans);
}
return 0;
}
``` | -1 | |
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": "... | 1,559,836,909 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 6 | 108 | 0 | s = input()
i = 0
while (i < len(s) and s[i] == 'a'):
i += 1
if i == len(s):
s = s[-1:] + 'z'
else:
j = i
p = ''
while (j < len(s) and s[j] != 'a'):
p = p + chr(ord(s[j]) - 1)
j += 1
print(s[:i] + p + s[j:])
| 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()
i = 0
while (i < len(s) and s[i] == 'a'):
i += 1
if i == len(s):
s = s[-1:] + 'z'
else:
j = i
p = ''
while (j < len(s) and s[j] != 'a'):
p = p + chr(ord(s[j]) - 1)
j += 1
print(s[:i] + p + s[j:])
``` | -1 | |
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... | 1,694,120,550 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <algorithm>
using namespace std;
int main(){
string r , s;
cin >> r >> s;
reverse(s.begin(), s.end());
if ( s == r){
cout << "YES" <<'\n';
}
else{
cout <<"NO"<< '\n' ;
}
return 0;
}
| 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
#include <iostream>
#include <algorithm>
using namespace std;
int main(){
string r , s;
cin >> r >> s;
reverse(s.begin(), s.end());
if ( s == r){
cout << "YES" <<'\n';
}
else{
cout <<"NO"<< '\n' ;
}
return 0;
}
``` | -1 |
199 | A | Hexadecimal's theorem | PROGRAMMING | 900 | [
"brute force",
"constructive algorithms",
"implementation",
"number theory"
] | null | null | Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible. | The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number. | Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them. | [
"3\n",
"13\n"
] | [
"1 1 1\n",
"2 3 8\n"
] | none | 500 | [
{
"input": "3",
"output": "1 1 1"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input... | 1,613,323,029 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 124 | 409,600 | import functools
n = int(input())
c = 0
@functools.lru_cache(maxsize=128)
def check_num_of_divisors(num):
counter = 0
for divisor in range(2,num//2+1):
if num % divisor == 0:
counter +=1
return counter
for n1 in range(2,n+1):
temp_counter = 0
for n2 in range(2,(n1//2)+1):
if check_num_of_divisors(n2) > 0:
continue
elif n1 % n2 ==0:
temp_counter+=1
else:
if temp_counter == 2:
c+=1
print(c)
| Title: Hexadecimal's theorem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
Input Specification:
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output Specification:
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
Demo Input:
['3\n', '13\n']
Demo Output:
['1 1 1\n', '2 3 8\n']
Note:
none | ```python
import functools
n = int(input())
c = 0
@functools.lru_cache(maxsize=128)
def check_num_of_divisors(num):
counter = 0
for divisor in range(2,num//2+1):
if num % divisor == 0:
counter +=1
return counter
for n1 in range(2,n+1):
temp_counter = 0
for n2 in range(2,(n1//2)+1):
if check_num_of_divisors(n2) > 0:
continue
elif n1 % n2 ==0:
temp_counter+=1
else:
if temp_counter == 2:
c+=1
print(c)
``` | 0 | |
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... | 1,505,590,645 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 307,200 | def to_list(s):
return list(map(int, s.split(" ")))
def shortest_path(graph, start, end):
queue = [(1,0)] # (vertex, distance)
visited = {} # vertex : True or False
history = {} # vertex : [ (prev_vertex, distance), ... ]
while len(queue) > 0:
# pop the next vertex off the priority queue
vertex, distance = queue[0]
queue.pop(0)
# mark it as visited
visited[vertex] = True
# if we've reached our destination, find the shortest path by back tracking
if vertex == end:
cv = end
path = [cv]
while cv != 1:
prev = min(history[cv], key=lambda x: x[1])[0]
path.append(prev)
cv = prev
path = list(reversed(path))
print(" ".join(str(v) for v in path))
return
# otherwise, add the vertex's neighbors to the queue
for v in graph[vertex]:
# if the neighbor has not been visited
if not visited.get(v[0], False):
# add the neighbor to the queue with updated distance
queue.append((v[0], v[1] + distance))
# file the neighbor away in the history hash (prev_vertex, distance_through_prev_vertex)
history[v[0]] = history.get(v[0], []) + [(vertex, v[1] + distance)]
# sort the queue in ascending order by weight
list.sort(queue, key=lambda x: x[1])
# parse the input and create the graph (adjacency list)
vertices, edges = to_list(input())
graph = {}
for _ in range(0, edges):
v1, v2, w = to_list(input())
graph[v1] = graph.get(v1, []) + [(v2, w)]
graph[v2] = graph.get(v2, []) + [(v1, w)]
# print out the shortest path
shortest_path(graph, 1, vertices) | 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
def to_list(s):
return list(map(int, s.split(" ")))
def shortest_path(graph, start, end):
queue = [(1,0)] # (vertex, distance)
visited = {} # vertex : True or False
history = {} # vertex : [ (prev_vertex, distance), ... ]
while len(queue) > 0:
# pop the next vertex off the priority queue
vertex, distance = queue[0]
queue.pop(0)
# mark it as visited
visited[vertex] = True
# if we've reached our destination, find the shortest path by back tracking
if vertex == end:
cv = end
path = [cv]
while cv != 1:
prev = min(history[cv], key=lambda x: x[1])[0]
path.append(prev)
cv = prev
path = list(reversed(path))
print(" ".join(str(v) for v in path))
return
# otherwise, add the vertex's neighbors to the queue
for v in graph[vertex]:
# if the neighbor has not been visited
if not visited.get(v[0], False):
# add the neighbor to the queue with updated distance
queue.append((v[0], v[1] + distance))
# file the neighbor away in the history hash (prev_vertex, distance_through_prev_vertex)
history[v[0]] = history.get(v[0], []) + [(vertex, v[1] + distance)]
# sort the queue in ascending order by weight
list.sort(queue, key=lambda x: x[1])
# parse the input and create the graph (adjacency list)
vertices, edges = to_list(input())
graph = {}
for _ in range(0, edges):
v1, v2, w = to_list(input())
graph[v1] = graph.get(v1, []) + [(v2, w)]
graph[v2] = graph.get(v2, []) + [(v1, w)]
# print out the shortest path
shortest_path(graph, 1, vertices)
``` | 0 |
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 ... | 1,496,698,306 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 10 | 92 | 0 | x = int(input())
if x == 1:
print('NO')
else:
y = set(list(map(int, input().split())))
y = sorted(y)
print(y[1]) | Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
x = int(input())
if x == 1:
print('NO')
else:
y = set(list(map(int, input().split())))
y = sorted(y)
print(y[1])
``` | -1 |
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... | 1,670,738,760 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | texta= input()
text = texta[::-1]
textb= input()
if text== textb:
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
texta= input()
text = texta[::-1]
textb= input()
if text== textb:
print('yes')
else:
print('no')
``` | 0 |
996 | A | Hit the Lottery | PROGRAMMING | 800 | [
"dp",
"greedy"
] | null | null | Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance? | The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$). | Output the minimum number of bills that Allen could receive. | [
"125\n",
"43\n",
"1000000000\n"
] | [
"3\n",
"5\n",
"10000000\n"
] | In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | 500 | [
{
"input": "125",
"output": "3"
},
{
"input": "43",
"output": "5"
},
{
"input": "1000000000",
"output": "10000000"
},
{
"input": "4",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "74",
"o... | 1,694,841,456 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 46 | 0 | t= int(input())
dem=[100,20,10,5,1]
cnt=0
for d in dem:
cnt+=t//d
t%=d
print(cnt) | Title: Hit the Lottery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input Specification:
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output Specification:
Output the minimum number of bills that Allen could receive.
Demo Input:
['125\n', '43\n', '1000000000\n']
Demo Output:
['3\n', '5\n', '10000000\n']
Note:
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | ```python
t= int(input())
dem=[100,20,10,5,1]
cnt=0
for d in dem:
cnt+=t//d
t%=d
print(cnt)
``` | 3 | |
146 | A | Lucky Ticket | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky. | The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros. | On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes). | [
"2\n47\n",
"4\n4738\n",
"4\n4774\n"
] | [
"NO\n",
"NO\n",
"YES\n"
] | In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | 500 | [
{
"input": "2\n47",
"output": "NO"
},
{
"input": "4\n4738",
"output": "NO"
},
{
"input": "4\n4774",
"output": "YES"
},
{
"input": "4\n4570",
"output": "NO"
},
{
"input": "6\n477477",
"output": "YES"
},
{
"input": "6\n777777",
"output": "YES"
},
... | 1,624,009,509 | 69 | Python 3 | OK | TESTS | 46 | 124 | 0 | def lucky(l,n):
for i in l:
if i!=4 and i!=7:
return "NO"
if sum(l[0:n])==sum(l[n:]):
return "YES"
return "NO"
n=int(input())//2
s=input()
l=list(s)
l=[int(i) for i in l]
print(lucky(l,n)) | Title: Lucky Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
Input Specification:
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
Output Specification:
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['2\n47\n', '4\n4738\n', '4\n4774\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n']
Note:
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | ```python
def lucky(l,n):
for i in l:
if i!=4 and i!=7:
return "NO"
if sum(l[0:n])==sum(l[n:]):
return "YES"
return "NO"
n=int(input())//2
s=input()
l=list(s)
l=[int(i) for i in l]
print(lucky(l,n))
``` | 3 | |
710 | C | Magic Odd Square | PROGRAMMING | 1,500 | [
"constructive algorithms",
"math"
] | null | null | Find an *n*<=×<=*n* matrix with different numbers from 1 to *n*2, so the sum in each row, column and both main diagonals are odd. | The only line contains odd integer *n* (1<=≤<=*n*<=≤<=49). | Print *n* lines with *n* integers. All the integers should be different and from 1 to *n*2. The sum in each row, column and both main diagonals should be odd. | [
"1\n",
"3\n"
] | [
"1\n",
"2 1 4\n3 5 7\n6 9 8\n"
] | none | 0 | [
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "2 1 4\n3 5 7\n6 9 8"
},
{
"input": "5",
"output": "2 4 1 6 8\n10 3 5 7 12\n9 11 13 15 17\n14 19 21 23 16\n18 20 25 22 24"
},
{
"input": "7",
"output": "2 4 6 1 8 10 12\n14 16 3 5 7 18 20\n22 9 11 13 15 17 24\n19 ... | 1,599,887,993 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 93 | 307,200 | n=int(input())
ans=[[0 for i in range(n)] for j in range(n)]
# for i in ans:
# print(i)
start=0
od=1
do=n
for i in range(start,n):
for j in range(start, start+do):
ans[i][j]=od
od+=2
start+=1
do-=2
start=n//2
do=1
for i in range(0,n//2):
for j in range(start, start+do):
ans[i][j]=od
od+=2
start-=1
do+=2
ev=2
for i in ans:
for j in i:
if j:
print(j,end=' ')
else:
print(ev,end=' ')
ev+=2
print('')
| Title: Magic Odd Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Find an *n*<=×<=*n* matrix with different numbers from 1 to *n*2, so the sum in each row, column and both main diagonals are odd.
Input Specification:
The only line contains odd integer *n* (1<=≤<=*n*<=≤<=49).
Output Specification:
Print *n* lines with *n* integers. All the integers should be different and from 1 to *n*2. The sum in each row, column and both main diagonals should be odd.
Demo Input:
['1\n', '3\n']
Demo Output:
['1\n', '2 1 4\n3 5 7\n6 9 8\n']
Note:
none | ```python
n=int(input())
ans=[[0 for i in range(n)] for j in range(n)]
# for i in ans:
# print(i)
start=0
od=1
do=n
for i in range(start,n):
for j in range(start, start+do):
ans[i][j]=od
od+=2
start+=1
do-=2
start=n//2
do=1
for i in range(0,n//2):
for j in range(start, start+do):
ans[i][j]=od
od+=2
start-=1
do+=2
ev=2
for i in ans:
for j in i:
if j:
print(j,end=' ')
else:
print(ev,end=' ')
ev+=2
print('')
``` | 0 | |
862 | B | Mahmoud and Ehab and the bipartiteness | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same . | The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree. | Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions. | [
"3\n1 2\n1 3\n",
"5\n1 2\n2 3\n3 4\n4 5\n"
] | [
"0\n",
"2\n"
] | Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | 1,000 | [
{
"input": "3\n1 2\n1 3",
"output": "0"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5",
"output": "2"
},
{
"input": "10\n3 8\n6 2\n9 7\n10 1\n3 5\n1 3\n6 7\n5 4\n3 6",
"output": "16"
},
{
"input": "10\n7 6\n2 7\n4 1\n8 5\n9 4\n5 3\n8 7\n10 8\n10 4",
"output": "16"
},
{
"in... | 1,656,866,302 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 296 | 14,745,600 | import sys
input = sys.stdin.readline
n = int(input())
d = [[] for _ in range(n)]
for _ in range(n-1):
a, b = map(int, input().split())
d[a-1].append(b)
d[b-1].append(a)
q = [1]
e = [1] + [0]*(n-1)
while len(q) > 0:
x = q.pop()
for i in d[x-1]:
if e[i-1] == 0:
q.append(i)
e[i-1] = -e[x-1]
x = e.count(1)
print(x*(n-x)-n+1) | Title: Mahmoud and Ehab and the bipartiteness
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
Input Specification:
The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree.
Output Specification:
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
Demo Input:
['3\n1 2\n1 3\n', '5\n1 2\n2 3\n3 4\n4 5\n']
Demo Output:
['0\n', '2\n']
Note:
Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | ```python
import sys
input = sys.stdin.readline
n = int(input())
d = [[] for _ in range(n)]
for _ in range(n-1):
a, b = map(int, input().split())
d[a-1].append(b)
d[b-1].append(a)
q = [1]
e = [1] + [0]*(n-1)
while len(q) > 0:
x = q.pop()
for i in d[x-1]:
if e[i-1] == 0:
q.append(i)
e[i-1] = -e[x-1]
x = e.count(1)
print(x*(n-x)-n+1)
``` | 3 | |
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": "1000022... | 1,643,958,221 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 404 | 10,752,000 | from cmath import inf
from re import T, X
import math
from tkinter import N, Y
def swap(a,b):
a = a^b
b = a^b
a = a^b
return a,b
# call using a,b = swap(a,b)
def even(a):
if a&1:
return False
return True
def gcd(x, y):
while(y):
x, y = y, x % y
return x
def lcm(x, y):
lcm = (x*y) // gcd(x,y)
return lcm
# it will return index of "key" in "sorted list".
def BS_elemPosi(low, high, key, li):
while low<high:
mid = (low+high)//2
if li[mid]<=key:
low = mid+1
else:
high = mid
return low
###########################(CODE)##################################
# 0 -> .
# 1 -> -.
# 2 -> --
s = input()
i = 0
ans = ""
while i<len(s):
if s[i]=='.':
ans+="0"
elif s[i]=="-":
if (i+1)<len(s):
if s[i+1]==".":
ans+="1"
i+=1
elif s[i+1]=="-":
ans+="2"
i+=1
i+=1
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
from cmath import inf
from re import T, X
import math
from tkinter import N, Y
def swap(a,b):
a = a^b
b = a^b
a = a^b
return a,b
# call using a,b = swap(a,b)
def even(a):
if a&1:
return False
return True
def gcd(x, y):
while(y):
x, y = y, x % y
return x
def lcm(x, y):
lcm = (x*y) // gcd(x,y)
return lcm
# it will return index of "key" in "sorted list".
def BS_elemPosi(low, high, key, li):
while low<high:
mid = (low+high)//2
if li[mid]<=key:
low = mid+1
else:
high = mid
return low
###########################(CODE)##################################
# 0 -> .
# 1 -> -.
# 2 -> --
s = input()
i = 0
ans = ""
while i<len(s):
if s[i]=='.':
ans+="0"
elif s[i]=="-":
if (i+1)<len(s):
if s[i+1]==".":
ans+="1"
i+=1
elif s[i+1]=="-":
ans+="2"
i+=1
i+=1
print(ans)
#####################################################################
``` | 3.878973 |
822 | A | I'm bored with life | PROGRAMMING | 800 | [
"implementation",
"math",
"number theory"
] | null | null | Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you? | The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12). | Print a single integer denoting the greatest common divisor of integers *A*! and *B*!. | [
"4 3\n"
] | [
"6\n"
] | Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | 500 | [
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"ou... | 1,656,318,396 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 46 | 4,300,800 | a,b=map(int,input().split())
mini=min(a, b)
minifact=1
for i in range(1,mini+1):
minifact*=i
print(minifact) | Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | ```python
a,b=map(int,input().split())
mini=min(a, b)
minifact=1
for i in range(1,mini+1):
minifact*=i
print(minifact)
``` | 3 | |
841 | A | Generous Kefa | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons. | Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary. | [
"4 2\naabb\n",
"6 3\naacaab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO». | 500 | [
{
"input": "4 2\naabb",
"output": "YES"
},
{
"input": "6 3\naacaab",
"output": "NO"
},
{
"input": "2 2\nlu",
"output": "YES"
},
{
"input": "5 3\novvoo",
"output": "YES"
},
{
"input": "36 13\nbzbzcffczzcbcbzzfzbbfzfzzbfbbcbfccbf",
"output": "YES"
},
{
"... | 1,504,655,740 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | x,y = map(int,input().split())
p =input()
m = 0
for i in p:
if p.count(i)>x:
m=1
print("NO" if m==1 else "YES") | Title: Generous Kefa
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons.
Output Specification:
Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary.
Demo Input:
['4 2\naabb\n', '6 3\naacaab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO». | ```python
x,y = map(int,input().split())
p =input()
m = 0
for i in p:
if p.count(i)>x:
m=1
print("NO" if m==1 else "YES")
``` | 0 | |
513 | A | Game | PROGRAMMING | 800 | [
"constructive algorithms",
"math"
] | null | null | Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally. | The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission. | Output "First" if the first player wins and "Second" otherwise. | [
"2 2 1 2\n",
"2 1 1 1\n"
] | [
"Second\n",
"First\n"
] | Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | 3 | [
{
"input": "2 2 1 2",
"output": "Second"
},
{
"input": "2 1 1 1",
"output": "First"
},
{
"input": "5 7 4 1",
"output": "Second"
},
{
"input": "5 7 1 4",
"output": "Second"
},
{
"input": "5 7 10 10",
"output": "Second"
},
{
"input": "5 7 1 10",
"out... | 1,584,856,559 | 2,147,483,647 | PyPy 3 | OK | TESTS | 20 | 155 | 0 | import sys
import math
import bisect
def solve(n1, n2):
return n1 > n2
def main():
n1, n2, k1, k2 = map(int, input().split())
if solve(n1, n2):
print('First')
else:
print('Second')
if __name__ == "__main__":
main()
| Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
Input Specification:
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
Output "First" if the first player wins and "Second" otherwise.
Demo Input:
['2 2 1 2\n', '2 1 1 1\n']
Demo Output:
['Second\n', 'First\n']
Note:
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | ```python
import sys
import math
import bisect
def solve(n1, n2):
return n1 > n2
def main():
n1, n2, k1, k2 = map(int, input().split())
if solve(n1, n2):
print('First')
else:
print('Second')
if __name__ == "__main__":
main()
``` | 3 | |
292 | B | Network Topology | PROGRAMMING | 1,200 | [
"graphs",
"implementation"
] | null | null | This problem uses a simplified network topology model, please read the problem statement carefully and use it as a formal document as you develop the solution.
Polycarpus continues working as a system administrator in a large corporation. The computer network of this corporation consists of *n* computers, some of them are connected by a cable. The computers are indexed by integers from 1 to *n*. It's known that any two computers connected by cable directly or through other computers
Polycarpus decided to find out the network's topology. A network topology is the way of describing the network configuration, the scheme that shows the location and the connections of network devices.
Polycarpus knows three main network topologies: bus, ring and star. A bus is the topology that represents a shared cable with all computers connected with it. In the ring topology the cable connects each computer only with two other ones. A star is the topology where all computers of a network are connected to the single central node.
Let's represent each of these network topologies as a connected non-directed graph. A bus is a connected graph that is the only path, that is, the graph where all nodes are connected with two other ones except for some two nodes that are the beginning and the end of the path. A ring is a connected graph, where all nodes are connected with two other ones. A star is a connected graph, where a single central node is singled out and connected with all other nodes. For clarifications, see the picture.
You've got a connected non-directed graph that characterizes the computer network in Polycarpus' corporation. Help him find out, which topology type the given network is. If that is impossible to do, say that the network's topology is unknown. | The first line contains two space-separated integers *n* and *m* (4<=≤<=*n*<=≤<=105; 3<=≤<=*m*<=≤<=105) — the number of nodes and edges in the graph, correspondingly. Next *m* lines contain the description of the graph's edges. The *i*-th line contains a space-separated pair of integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the numbers of nodes that are connected by the *i*-the edge.
It is guaranteed that the given graph is connected. There is at most one edge between any two nodes. No edge connects a node with itself. | In a single line print the network topology name of the given graph. If the answer is the bus, print "bus topology" (without the quotes), if the answer is the ring, print "ring topology" (without the quotes), if the answer is the star, print "star topology" (without the quotes). If no answer fits, print "unknown topology" (without the quotes). | [
"4 3\n1 2\n2 3\n3 4\n",
"4 4\n1 2\n2 3\n3 4\n4 1\n",
"4 3\n1 2\n1 3\n1 4\n",
"4 4\n1 2\n2 3\n3 1\n1 4\n"
] | [
"bus topology\n",
"ring topology\n",
"star topology\n",
"unknown topology\n"
] | none | 1,000 | [
{
"input": "4 3\n1 2\n2 3\n3 4",
"output": "bus topology"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n4 1",
"output": "ring topology"
},
{
"input": "4 3\n1 2\n1 3\n1 4",
"output": "star topology"
},
{
"input": "4 4\n1 2\n2 3\n3 1\n1 4",
"output": "unknown topology"
},
{
"i... | 1,691,154,032 | 2,147,483,647 | PyPy 3 | OK | TESTS | 45 | 1,932 | 28,672,000 | def build_graph(n, m, edges):
graph = {}
for i in range(1, n + 1):
graph[i] = []
for u, v in edges:
graph[u].append(v)
graph[v].append(u) # Since the graph is undirected, we add both directions
return graph
from collections import deque
def bfs(graph, start):
visited = set()
queue = deque([start])
while queue:
node = queue.popleft()
visited.add(node)
for neighbor in graph[node]:
if neighbor not in visited:
queue.append(neighbor)
return visited
def is_disconnected(graph):
visited = set()
for node in graph:
if node not in visited:
connected_nodes = bfs(graph, node)
visited.update(connected_nodes)
return len(visited) != len(graph)
n, m = map(int, input().split())
edges = [tuple(map(int, input().split())) for _ in range(m)]
graph = build_graph(n, m, edges)
if is_disconnected(graph):
print("unknown topology")
else:
list=[]
for neighbors in graph.items():
# print(neighbors)
list.append(len(neighbors[1]))
# print(list)
x=list.count(n-1)
y=list.count(1)
z=list.count(2)
# print(x,y,z)
# print(n)
if x==1 and y==n-1:
print("star topology")
elif y==2 and z==n-2:
print("bus topology")
elif z==n:
print("ring topology")
else:
print("unknown topology")
| Title: Network Topology
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem uses a simplified network topology model, please read the problem statement carefully and use it as a formal document as you develop the solution.
Polycarpus continues working as a system administrator in a large corporation. The computer network of this corporation consists of *n* computers, some of them are connected by a cable. The computers are indexed by integers from 1 to *n*. It's known that any two computers connected by cable directly or through other computers
Polycarpus decided to find out the network's topology. A network topology is the way of describing the network configuration, the scheme that shows the location and the connections of network devices.
Polycarpus knows three main network topologies: bus, ring and star. A bus is the topology that represents a shared cable with all computers connected with it. In the ring topology the cable connects each computer only with two other ones. A star is the topology where all computers of a network are connected to the single central node.
Let's represent each of these network topologies as a connected non-directed graph. A bus is a connected graph that is the only path, that is, the graph where all nodes are connected with two other ones except for some two nodes that are the beginning and the end of the path. A ring is a connected graph, where all nodes are connected with two other ones. A star is a connected graph, where a single central node is singled out and connected with all other nodes. For clarifications, see the picture.
You've got a connected non-directed graph that characterizes the computer network in Polycarpus' corporation. Help him find out, which topology type the given network is. If that is impossible to do, say that the network's topology is unknown.
Input Specification:
The first line contains two space-separated integers *n* and *m* (4<=≤<=*n*<=≤<=105; 3<=≤<=*m*<=≤<=105) — the number of nodes and edges in the graph, correspondingly. Next *m* lines contain the description of the graph's edges. The *i*-th line contains a space-separated pair of integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the numbers of nodes that are connected by the *i*-the edge.
It is guaranteed that the given graph is connected. There is at most one edge between any two nodes. No edge connects a node with itself.
Output Specification:
In a single line print the network topology name of the given graph. If the answer is the bus, print "bus topology" (without the quotes), if the answer is the ring, print "ring topology" (without the quotes), if the answer is the star, print "star topology" (without the quotes). If no answer fits, print "unknown topology" (without the quotes).
Demo Input:
['4 3\n1 2\n2 3\n3 4\n', '4 4\n1 2\n2 3\n3 4\n4 1\n', '4 3\n1 2\n1 3\n1 4\n', '4 4\n1 2\n2 3\n3 1\n1 4\n']
Demo Output:
['bus topology\n', 'ring topology\n', 'star topology\n', 'unknown topology\n']
Note:
none | ```python
def build_graph(n, m, edges):
graph = {}
for i in range(1, n + 1):
graph[i] = []
for u, v in edges:
graph[u].append(v)
graph[v].append(u) # Since the graph is undirected, we add both directions
return graph
from collections import deque
def bfs(graph, start):
visited = set()
queue = deque([start])
while queue:
node = queue.popleft()
visited.add(node)
for neighbor in graph[node]:
if neighbor not in visited:
queue.append(neighbor)
return visited
def is_disconnected(graph):
visited = set()
for node in graph:
if node not in visited:
connected_nodes = bfs(graph, node)
visited.update(connected_nodes)
return len(visited) != len(graph)
n, m = map(int, input().split())
edges = [tuple(map(int, input().split())) for _ in range(m)]
graph = build_graph(n, m, edges)
if is_disconnected(graph):
print("unknown topology")
else:
list=[]
for neighbors in graph.items():
# print(neighbors)
list.append(len(neighbors[1]))
# print(list)
x=list.count(n-1)
y=list.count(1)
z=list.count(2)
# print(x,y,z)
# print(n)
if x==1 and y==n-1:
print("star topology")
elif y==2 and z==n-2:
print("bus topology")
elif z==n:
print("ring topology")
else:
print("unknown topology")
``` | 3 | |
493 | B | Vasya and Wrestling | PROGRAMMING | 1,400 | [
"implementation"
] | null | null | Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won. | The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105).
The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points.
The techniques are given in chronological order. | If the first wrestler wins, print string "first", otherwise print "second" | [
"5\n1\n2\n-3\n-4\n3\n",
"3\n-1\n-2\n3\n",
"2\n4\n-4\n"
] | [
"second\n",
"first\n",
"second\n"
] | Sequence *x* = *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y* = *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*| > |*y*| and *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">|*y*|</sub> = *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r* < |*x*|, *r* < |*y*|), that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> > *y*<sub class="lower-index">*r* + 1</sub>.
We use notation |*a*| to denote length of sequence *a*. | 1,000 | [
{
"input": "5\n1\n2\n-3\n-4\n3",
"output": "second"
},
{
"input": "3\n-1\n-2\n3",
"output": "first"
},
{
"input": "2\n4\n-4",
"output": "second"
},
{
"input": "7\n1\n2\n-3\n4\n5\n-6\n7",
"output": "first"
},
{
"input": "14\n1\n2\n3\n4\n5\n6\n7\n-8\n-9\n-10\n-11\n-... | 1,434,733,740 | 2,640 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 0 | fin = open('input.txt','r')
fout = open('output.txt','w')
n = int(fin.readline())
a = []
b = []
s1 = 0
s2 = 0
for i in range(n):
m = int(fin.readline())
if (m >= 0):
a.append(m)
s1 = s1 + m
else:
b.append(m)
s2 = s2 + m
if (s1 > s2):
print('first',file = fout)
elif (s1 < s2):
print('second',file = fout)
else:
if (len(a) > len(b)):
print('first',file = fout)
elif (len(a) < len(b)):
print('second',file = fout)
else:
for i in range(len(a)):
if (a[i] > b[i]):
print('first',file = fout)
break
else:
print('second',file = fout)
break
fin.close()
fout.close() | Title: Vasya and Wrestling
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input Specification:
The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105).
The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points.
The techniques are given in chronological order.
Output Specification:
If the first wrestler wins, print string "first", otherwise print "second"
Demo Input:
['5\n1\n2\n-3\n-4\n3\n', '3\n-1\n-2\n3\n', '2\n4\n-4\n']
Demo Output:
['second\n', 'first\n', 'second\n']
Note:
Sequence *x* = *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y* = *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*| > |*y*| and *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">|*y*|</sub> = *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r* < |*x*|, *r* < |*y*|), that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> > *y*<sub class="lower-index">*r* + 1</sub>.
We use notation |*a*| to denote length of sequence *a*. | ```python
fin = open('input.txt','r')
fout = open('output.txt','w')
n = int(fin.readline())
a = []
b = []
s1 = 0
s2 = 0
for i in range(n):
m = int(fin.readline())
if (m >= 0):
a.append(m)
s1 = s1 + m
else:
b.append(m)
s2 = s2 + m
if (s1 > s2):
print('first',file = fout)
elif (s1 < s2):
print('second',file = fout)
else:
if (len(a) > len(b)):
print('first',file = fout)
elif (len(a) < len(b)):
print('second',file = fout)
else:
for i in range(len(a)):
if (a[i] > b[i]):
print('first',file = fout)
break
else:
print('second',file = fout)
break
fin.close()
fout.close()
``` | -1 | |
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
... | 1,555,333,827 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 248 | 0 | n = int(input())
n -= 10
if (n < 10 or n==1 or n == 11) and n != 0:
print (4)
elif n == 0 or n > 11:
print (0)
else:
print (15) | Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
n = int(input())
n -= 10
if (n < 10 or n==1 or n == 11) and n != 0:
print (4)
elif n == 0 or n > 11:
print (0)
else:
print (15)
``` | 0 |
990 | B | Micro-World | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope. | The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have. | Print the only integer — minimal possible number of bacteria can remain. | [
"7 1\n101 53 42 102 101 55 54\n",
"6 5\n20 15 10 15 20 25\n",
"7 1000000\n1 1 1 1 1 1 1\n"
] | [
"3\n",
"1\n",
"7\n"
] | The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria. | 0 | [
{
"input": "7 1\n101 53 42 102 101 55 54",
"output": "3"
},
{
"input": "6 5\n20 15 10 15 20 25",
"output": "1"
},
{
"input": "7 1000000\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 4\n8",
"output": "1"
},
{
"inp... | 1,528,876,596 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 14,336,000 | n, k = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
s = len(a)
i = 0
while i < len(a)-1:
j = i + 1
canSwallow = False
while j < len(a) and not canSwallow:
diff = a[j] - a[i]
j += 1
if diff == 0:
continue
if diff > k:
break
else:
canSwallow = True
if canSwallow:
s -= 1
i += 1
print(s)
| Title: Micro-World
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.
Input Specification:
The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) — number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — sizes of bacteria you have.
Output Specification:
Print the only integer — minimal possible number of bacteria can remain.
Demo Input:
['7 1\n101 53 42 102 101 55 54\n', '6 5\n20 15 10 15 20 25\n', '7 1000000\n1 1 1 1 1 1 1\n']
Demo Output:
['3\n', '1\n', '7\n']
Note:
The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria. | ```python
n, k = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
s = len(a)
i = 0
while i < len(a)-1:
j = i + 1
canSwallow = False
while j < len(a) and not canSwallow:
diff = a[j] - a[i]
j += 1
if diff == 0:
continue
if diff > k:
break
else:
canSwallow = True
if canSwallow:
s -= 1
i += 1
print(s)
``` | 0 | |
764 | A | Taymyr is calling you | PROGRAMMING | 800 | [
"brute force",
"implementation",
"math"
] | null | null | Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. | The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). | Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. | [
"1 1 10\n",
"1 2 5\n",
"2 3 9\n"
] | [
"10\n",
"2\n",
"1\n"
] | Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | 500 | [
{
"input": "1 1 10",
"output": "10"
},
{
"input": "1 2 5",
"output": "2"
},
{
"input": "2 3 9",
"output": "1"
},
{
"input": "4 8 9",
"output": "1"
},
{
"input": "7 9 2",
"output": "0"
},
{
"input": "10000 10000 10000",
"output": "1"
},
{
"i... | 1,560,196,650 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 109 | 0 | # import sys
# sys.stdin=open("input.in","r")
# sys.stdout=open("output.out","w")
n,m,z=map(int,input().split())
def gcd(n,m):
if m==0:
return n
else:
return gcd(m,n%m)
lcm=int((m*n)/gcd(m,n))
print(int(z/lcm)) | Title: Taymyr is calling you
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input Specification:
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
Output Specification:
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Demo Input:
['1 1 10\n', '1 2 5\n', '2 3 9\n']
Demo Output:
['10\n', '2\n', '1\n']
Note:
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | ```python
# import sys
# sys.stdin=open("input.in","r")
# sys.stdout=open("output.out","w")
n,m,z=map(int,input().split())
def gcd(n,m):
if m==0:
return n
else:
return gcd(m,n%m)
lcm=int((m*n)/gcd(m,n))
print(int(z/lcm))
``` | 3 | |
991 | C | Candies | PROGRAMMING | 1,500 | [
"binary search",
"implementation"
] | null | null | After passing a test, Vasya got himself a box of $n$ candies. He decided to eat an equal amount of candies each morning until there are no more candies. However, Petya also noticed the box and decided to get some candies for himself.
This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$, same for all days. After that, in the morning he eats $k$ candies from the box (if there are less than $k$ candies in the box, he eats them all), then in the evening Petya eats $10\%$ of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats $k$ candies again, and Petya — $10\%$ of the candies left in a box, and so on.
If the amount of candies in the box is not divisible by $10$, Petya rounds the amount he takes from the box down. For example, if there were $97$ candies in the box, Petya would eat only $9$ of them. In particular, if there are less than $10$ candies in a box, Petya won't eat any at all.
Your task is to find out the minimal amount of $k$ that can be chosen by Vasya so that he would eat at least half of the $n$ candies he initially got. Note that the number $k$ must be integer. | The first line contains a single integer $n$ ($1 \leq n \leq 10^{18}$) — the initial amount of candies in the box. | Output a single integer — the minimal amount of $k$ that would allow Vasya to eat at least half of candies he got. | [
"68\n"
] | [
"3\n"
] | In the sample, the amount of candies, with $k=3$, would change in the following way (Vasya eats first):
$68 \to 65 \to 59 \to 56 \to 51 \to 48 \to 44 \to 41 \\ \to 37 \to 34 \to 31 \to 28 \to 26 \to 23 \to 21 \to 18 \to 17 \to 14 \\ \to 13 \to 10 \to 9 \to 6 \to 6 \to 3 \to 3 \to 0$.
In total, Vasya would eat $39$ candies, while Petya — $29$. | 1,250 | [
{
"input": "68",
"output": "3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "42",
"output": "1"
},
{
"input": "43",
"output": "2"
},
{
"input": "756",
"output": "29"
},
{
"input": "999999972",
"output"... | 1,631,543,231 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 61 | 6,963,200 | import math
from sys import *
# stdin = open('input.txt', 'r')
def can_eat(k, n):
v_eat = 0
p_eat = 0
while n > 0:
v_eat += min(n, k)
n = n - k
p_eat += max(0, n/10)
n -= n/10
return v_eat >= p_eat
def main():
n = int(stdin.readline())
l = 1
r = n/2
while l < r:
k = math.floor((l + r)/2)
if can_eat(k, n):
r = k
else:
l = k + 1
print(l)
main()
| Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After passing a test, Vasya got himself a box of $n$ candies. He decided to eat an equal amount of candies each morning until there are no more candies. However, Petya also noticed the box and decided to get some candies for himself.
This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$, same for all days. After that, in the morning he eats $k$ candies from the box (if there are less than $k$ candies in the box, he eats them all), then in the evening Petya eats $10\%$ of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats $k$ candies again, and Petya — $10\%$ of the candies left in a box, and so on.
If the amount of candies in the box is not divisible by $10$, Petya rounds the amount he takes from the box down. For example, if there were $97$ candies in the box, Petya would eat only $9$ of them. In particular, if there are less than $10$ candies in a box, Petya won't eat any at all.
Your task is to find out the minimal amount of $k$ that can be chosen by Vasya so that he would eat at least half of the $n$ candies he initially got. Note that the number $k$ must be integer.
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 10^{18}$) — the initial amount of candies in the box.
Output Specification:
Output a single integer — the minimal amount of $k$ that would allow Vasya to eat at least half of candies he got.
Demo Input:
['68\n']
Demo Output:
['3\n']
Note:
In the sample, the amount of candies, with $k=3$, would change in the following way (Vasya eats first):
$68 \to 65 \to 59 \to 56 \to 51 \to 48 \to 44 \to 41 \\ \to 37 \to 34 \to 31 \to 28 \to 26 \to 23 \to 21 \to 18 \to 17 \to 14 \\ \to 13 \to 10 \to 9 \to 6 \to 6 \to 3 \to 3 \to 0$.
In total, Vasya would eat $39$ candies, while Petya — $29$. | ```python
import math
from sys import *
# stdin = open('input.txt', 'r')
def can_eat(k, n):
v_eat = 0
p_eat = 0
while n > 0:
v_eat += min(n, k)
n = n - k
p_eat += max(0, n/10)
n -= n/10
return v_eat >= p_eat
def main():
n = int(stdin.readline())
l = 1
r = n/2
while l < r:
k = math.floor((l + r)/2)
if can_eat(k, n):
r = k
else:
l = k + 1
print(l)
main()
``` | 0 | |
371 | C | Hamburgers | PROGRAMMING | 1,600 | [
"binary search",
"brute force"
] | null | null | Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient. | The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
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. | Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0. | [
"BBBSSC\n6 4 1\n1 2 3\n4\n",
"BBC\n1 10 1\n1 10 1\n21\n",
"BSC\n1 1 1\n1 1 3\n1000000000000\n"
] | [
"2\n",
"7\n",
"200000000001\n"
] | none | 1,500 | [
{
"input": "BBBSSC\n6 4 1\n1 2 3\n4",
"output": "2"
},
{
"input": "BBC\n1 10 1\n1 10 1\n21",
"output": "7"
},
{
"input": "BSC\n1 1 1\n1 1 3\n1000000000000",
"output": "200000000001"
},
{
"input": "B\n1 1 1\n1 1 1\n381",
"output": "382"
},
{
"input": "BSC\n3 5 6\n7... | 1,594,828,603 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 140 | 20,172,800 | strs = input()
rb = 0
rs = 0
rc = 0
for i in strs:
if i == "B":
rb += 1
if i == "S":
rs += 1
if i == "C":
rc += 1
temparr = input()
temparr = temparr.split()
kb = int(temparr[0])
ks = int(temparr[1])
kc = int (temparr[2])
temparr = input()
temparr = temparr.split()
cb = int(temparr[0])
cs = int(temparr[1])
cc = int(temparr[2])
costofburger = cb * rb + rs * cs + rc * cc
ruble = int(input())
ans = 0
while True:
if kb >= rb and kc >= rc and ks >= rs:
ans += 1
kb -= rb
kc -= rc
ks -= rs
continue
lackb = 0
lacks = 0
lackc = 0
if rb > kb:
lackb = rb - kb
if rc > kc:
lackc = rc - kc
if rs > ks:
lacks = rs - ks
costruble = lackb * rb + lacks * cs + lackc * cc
if ruble >= costruble:
ans += 1
ruble -= costruble
break
ans += ruble // costofburger
print(ans)
| Title: Hamburgers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.
Input Specification:
The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
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.
Output Specification:
Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.
Demo Input:
['BBBSSC\n6 4 1\n1 2 3\n4\n', 'BBC\n1 10 1\n1 10 1\n21\n', 'BSC\n1 1 1\n1 1 3\n1000000000000\n']
Demo Output:
['2\n', '7\n', '200000000001\n']
Note:
none | ```python
strs = input()
rb = 0
rs = 0
rc = 0
for i in strs:
if i == "B":
rb += 1
if i == "S":
rs += 1
if i == "C":
rc += 1
temparr = input()
temparr = temparr.split()
kb = int(temparr[0])
ks = int(temparr[1])
kc = int (temparr[2])
temparr = input()
temparr = temparr.split()
cb = int(temparr[0])
cs = int(temparr[1])
cc = int(temparr[2])
costofburger = cb * rb + rs * cs + rc * cc
ruble = int(input())
ans = 0
while True:
if kb >= rb and kc >= rc and ks >= rs:
ans += 1
kb -= rb
kc -= rc
ks -= rs
continue
lackb = 0
lacks = 0
lackc = 0
if rb > kb:
lackb = rb - kb
if rc > kc:
lackc = rc - kc
if rs > ks:
lacks = rs - ks
costruble = lackb * rb + lacks * cs + lackc * cc
if ruble >= costruble:
ans += 1
ruble -= costruble
break
ans += ruble // costofburger
print(ans)
``` | 0 | |
366 | A | Dima and Guards | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Nothing has changed since the last round. Dima and Inna still love each other and want to be together. They've made a deal with Seryozha and now they need to make a deal with the dorm guards...
There are four guardposts in Dima's dorm. Each post contains two guards (in Russia they are usually elderly women). You can bribe a guard by a chocolate bar or a box of juice. For each guard you know the minimum price of the chocolate bar she can accept as a gift and the minimum price of the box of juice she can accept as a gift. If a chocolate bar for some guard costs less than the minimum chocolate bar price for this guard is, or if a box of juice for some guard costs less than the minimum box of juice price for this guard is, then the guard doesn't accept such a gift.
In order to pass through a guardpost, one needs to bribe both guards.
The shop has an unlimited amount of juice and chocolate of any price starting with 1. Dima wants to choose some guardpost, buy one gift for each guard from the guardpost and spend exactly *n* rubles on it.
Help him choose a post through which he can safely sneak Inna or otherwise say that this is impossible. Mind you, Inna would be very sorry to hear that! | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=105) — the money Dima wants to spend. Then follow four lines describing the guardposts. Each line contains four integers *a*,<=*b*,<=*c*,<=*d* (1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=105) — the minimum price of the chocolate and the minimum price of the juice for the first guard and the minimum price of the chocolate and the minimum price of the juice for the second guard, correspondingly. | In a single line of the output print three space-separated integers: the number of the guardpost, the cost of the first present and the cost of the second present. If there is no guardpost Dima can sneak Inna through at such conditions, print -1 in a single line.
The guardposts are numbered from 1 to 4 according to the order given in the input.
If there are multiple solutions, you can print any of them. | [
"10\n5 6 5 6\n6 6 7 7\n5 8 6 6\n9 9 9 9\n",
"10\n6 6 6 6\n7 7 7 7\n4 4 4 4\n8 8 8 8\n",
"5\n3 3 3 3\n3 3 3 3\n3 3 3 3\n3 3 3 3\n"
] | [
"1 5 5\n",
"3 4 6\n",
"-1\n"
] | Explanation of the first example.
The only way to spend 10 rubles to buy the gifts that won't be less than the minimum prices is to buy two 5 ruble chocolates to both guards from the first guardpost.
Explanation of the second example.
Dima needs 12 rubles for the first guardpost, 14 for the second one, 16 for the fourth one. So the only guardpost we can sneak through is the third one. So, Dima can buy 4 ruble chocolate for the first guard and 6 ruble juice of the second guard. | 500 | [
{
"input": "10\n5 6 5 6\n6 6 7 7\n5 8 6 6\n9 9 9 9",
"output": "1 5 5"
},
{
"input": "10\n6 6 6 6\n7 7 7 7\n4 4 4 4\n8 8 8 8",
"output": "3 4 6"
},
{
"input": "5\n3 3 3 3\n3 3 3 3\n3 3 3 3\n3 3 3 3",
"output": "-1"
},
{
"input": "100000\n100000 100000 100000 100000\n100000 10... | 1,602,612,599 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 109 | 307,200 | n = int(input())
w = False
arr = []
for i in range(4):
arr.append([int(x) for x in input().split()])
for i in range(4):
a,b,c,d = arr[i]
m1 = min(a,b)
m2 = min(c,d)
if m1+m2<=n:
print(str(i+1) + " " + str(m1) + " " + str(n-m1))
w = True
break
if not w:
print(-1) | Title: Dima and Guards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nothing has changed since the last round. Dima and Inna still love each other and want to be together. They've made a deal with Seryozha and now they need to make a deal with the dorm guards...
There are four guardposts in Dima's dorm. Each post contains two guards (in Russia they are usually elderly women). You can bribe a guard by a chocolate bar or a box of juice. For each guard you know the minimum price of the chocolate bar she can accept as a gift and the minimum price of the box of juice she can accept as a gift. If a chocolate bar for some guard costs less than the minimum chocolate bar price for this guard is, or if a box of juice for some guard costs less than the minimum box of juice price for this guard is, then the guard doesn't accept such a gift.
In order to pass through a guardpost, one needs to bribe both guards.
The shop has an unlimited amount of juice and chocolate of any price starting with 1. Dima wants to choose some guardpost, buy one gift for each guard from the guardpost and spend exactly *n* rubles on it.
Help him choose a post through which he can safely sneak Inna or otherwise say that this is impossible. Mind you, Inna would be very sorry to hear that!
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=105) — the money Dima wants to spend. Then follow four lines describing the guardposts. Each line contains four integers *a*,<=*b*,<=*c*,<=*d* (1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=105) — the minimum price of the chocolate and the minimum price of the juice for the first guard and the minimum price of the chocolate and the minimum price of the juice for the second guard, correspondingly.
Output Specification:
In a single line of the output print three space-separated integers: the number of the guardpost, the cost of the first present and the cost of the second present. If there is no guardpost Dima can sneak Inna through at such conditions, print -1 in a single line.
The guardposts are numbered from 1 to 4 according to the order given in the input.
If there are multiple solutions, you can print any of them.
Demo Input:
['10\n5 6 5 6\n6 6 7 7\n5 8 6 6\n9 9 9 9\n', '10\n6 6 6 6\n7 7 7 7\n4 4 4 4\n8 8 8 8\n', '5\n3 3 3 3\n3 3 3 3\n3 3 3 3\n3 3 3 3\n']
Demo Output:
['1 5 5\n', '3 4 6\n', '-1\n']
Note:
Explanation of the first example.
The only way to spend 10 rubles to buy the gifts that won't be less than the minimum prices is to buy two 5 ruble chocolates to both guards from the first guardpost.
Explanation of the second example.
Dima needs 12 rubles for the first guardpost, 14 for the second one, 16 for the fourth one. So the only guardpost we can sneak through is the third one. So, Dima can buy 4 ruble chocolate for the first guard and 6 ruble juice of the second guard. | ```python
n = int(input())
w = False
arr = []
for i in range(4):
arr.append([int(x) for x in input().split()])
for i in range(4):
a,b,c,d = arr[i]
m1 = min(a,b)
m2 = min(c,d)
if m1+m2<=n:
print(str(i+1) + " " + str(m1) + " " + str(n-m1))
w = True
break
if not w:
print(-1)
``` | 3 | |
895 | A | Pizza Separation | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360. | Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya. | [
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] | [
"0\n",
"40\n",
"360\n",
"0\n"
] | In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | 500 | [
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 4... | 1,511,877,360 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 49 | 62 | 5,632,000 |
x = int(input())
inS = input()
S = [int(m) for m in inS.split()]
v = []
p = []
for i in range(x):
if sum(v) > sum(p):
p.append(max(S))
S.remove(max(S))
else:
v.append(max(S))
S.remove(max(S))
print(abs(sum(v) - sum(p)))
| Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | ```python
x = int(input())
inS = input()
S = [int(m) for m in inS.split()]
v = []
p = []
for i in range(x):
if sum(v) > sum(p):
p.append(max(S))
S.remove(max(S))
else:
v.append(max(S))
S.remove(max(S))
print(abs(sum(v) - sum(p)))
``` | 0 | |
78 | E | Evacuation | PROGRAMMING | 2,300 | [
"flows",
"graphs",
"shortest paths"
] | E. Evacuation | 1 | 256 | They've screwed something up yet again... In one nuclear reactor of a research station an uncontrolled reaction is in progress and explosion which will destroy the whole station will happen soon.
The station is represented by a square *n*<=×<=*n* divided into 1<=×<=1 blocks. Each block is either a reactor or a laboratory. There can be several reactors and exactly one of them will explode soon. The reactors can be considered impassable blocks, but one can move through laboratories. Between any two laboratories, which are in adjacent blocks, there is a corridor. Blocks are considered adjacent if they have a common edge.
In each laboratory there is some number of scientists and some number of rescue capsules. Once the scientist climbs into a capsule, he is considered to be saved. Each capsule has room for not more than one scientist.
The reactor, which is about to explode, is damaged and a toxic coolant trickles from it into the neighboring blocks. The block, which contains the reactor, is considered infected. Every minute the coolant spreads over the laboratories through corridors. If at some moment one of the blocks is infected, then the next minute all the neighboring laboratories also become infected. Once a lab is infected, all the scientists there that are not in rescue capsules die. The coolant does not spread through reactor blocks.
There are exactly *t* minutes to the explosion. Any scientist in a minute can move down the corridor to the next lab, if it is not infected. On any corridor an unlimited number of scientists can simultaneously move in both directions. It is believed that the scientists inside a lab moves without consuming time. Moreover, any scientist could get into the rescue capsule instantly. It is also believed that any scientist at any given moment always has the time to perform their actions (move from the given laboratory into the next one, or climb into the rescue capsule) before the laboratory will be infected.
Find the maximum number of scientists who will be able to escape. | The first line contains two integers *n* and *t* (2<=≤<=*n*<=≤<=10, 1<=≤<=*t*<=≤<=60). Each of the next *n* lines contains *n* characters. These lines describe the scientists' locations. Then exactly one empty line follows. Each of the next *n* more lines contains *n* characters. These lines describe the rescue capsules' locations.
In the description of the scientists' and the rescue capsules' locations the character "Y" stands for a properly functioning reactor, "Z" stands for the malfunctioning reactor. The reactors' positions in both descriptions coincide. There is exactly one malfunctioning reactor on the station. The digits "0" - "9" stand for the laboratories. In the description of the scientists' locations those numbers stand for the number of scientists in the corresponding laboratories. In the rescue capsules' descriptions they stand for the number of such capsules in each laboratory. | Print a single number — the maximum number of scientists who will manage to save themselves. | [
"3 3\n1YZ\n1YY\n100\n\n0YZ\n0YY\n003\n",
"4 4\nY110\n1Y1Z\n1Y0Y\n0100\n\nY001\n0Y0Z\n0Y0Y\n0005\n"
] | [
"2",
"3"
] | In the second sample the events could take place as follows: | 2,500 | [] | 1,692,648,814 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1692648814.3426898")# 1692648814.342706 | Title: Evacuation
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
They've screwed something up yet again... In one nuclear reactor of a research station an uncontrolled reaction is in progress and explosion which will destroy the whole station will happen soon.
The station is represented by a square *n*<=×<=*n* divided into 1<=×<=1 blocks. Each block is either a reactor or a laboratory. There can be several reactors and exactly one of them will explode soon. The reactors can be considered impassable blocks, but one can move through laboratories. Between any two laboratories, which are in adjacent blocks, there is a corridor. Blocks are considered adjacent if they have a common edge.
In each laboratory there is some number of scientists and some number of rescue capsules. Once the scientist climbs into a capsule, he is considered to be saved. Each capsule has room for not more than one scientist.
The reactor, which is about to explode, is damaged and a toxic coolant trickles from it into the neighboring blocks. The block, which contains the reactor, is considered infected. Every minute the coolant spreads over the laboratories through corridors. If at some moment one of the blocks is infected, then the next minute all the neighboring laboratories also become infected. Once a lab is infected, all the scientists there that are not in rescue capsules die. The coolant does not spread through reactor blocks.
There are exactly *t* minutes to the explosion. Any scientist in a minute can move down the corridor to the next lab, if it is not infected. On any corridor an unlimited number of scientists can simultaneously move in both directions. It is believed that the scientists inside a lab moves without consuming time. Moreover, any scientist could get into the rescue capsule instantly. It is also believed that any scientist at any given moment always has the time to perform their actions (move from the given laboratory into the next one, or climb into the rescue capsule) before the laboratory will be infected.
Find the maximum number of scientists who will be able to escape.
Input Specification:
The first line contains two integers *n* and *t* (2<=≤<=*n*<=≤<=10, 1<=≤<=*t*<=≤<=60). Each of the next *n* lines contains *n* characters. These lines describe the scientists' locations. Then exactly one empty line follows. Each of the next *n* more lines contains *n* characters. These lines describe the rescue capsules' locations.
In the description of the scientists' and the rescue capsules' locations the character "Y" stands for a properly functioning reactor, "Z" stands for the malfunctioning reactor. The reactors' positions in both descriptions coincide. There is exactly one malfunctioning reactor on the station. The digits "0" - "9" stand for the laboratories. In the description of the scientists' locations those numbers stand for the number of scientists in the corresponding laboratories. In the rescue capsules' descriptions they stand for the number of such capsules in each laboratory.
Output Specification:
Print a single number — the maximum number of scientists who will manage to save themselves.
Demo Input:
['3 3\n1YZ\n1YY\n100\n\n0YZ\n0YY\n003\n', '4 4\nY110\n1Y1Z\n1Y0Y\n0100\n\nY001\n0Y0Z\n0Y0Y\n0005\n']
Demo Output:
['2', '3']
Note:
In the second sample the events could take place as follows: | ```python
print("_RANDOM_GUESS_1692648814.3426898")# 1692648814.342706
``` | 0 |
641 | A | Little Artem and Grasshopper | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Little Artem found a grasshopper. He brought it to his house and constructed a jumping area for him.
The area looks like a strip of cells 1<=×<=*n*. Each cell contains the direction for the next jump and the length of that jump. Grasshopper starts in the first cell and follows the instructions written on the cells. Grasshopper stops immediately if it jumps out of the strip. Now Artem wants to find out if this will ever happen. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — length of the strip.
Next line contains a string of length *n* which consists of characters "<" and ">" only, that provide the direction of the jump from the corresponding cell. Next line contains *n* integers *d**i* (1<=≤<=*d**i*<=≤<=109) — the length of the jump from the *i*-th cell. | Print "INFINITE" (without quotes) if grasshopper will continue his jumps forever. Otherwise print "FINITE" (without quotes). | [
"2\n><\n1 2\n",
"3\n>><\n2 1 1\n"
] | [
"FINITE\n",
"INFINITE"
] | In the first sample grasshopper starts from the first cell and jumps to the right on the next cell. When he is in the second cell he needs to jump two cells left so he will jump out of the strip.
Second sample grasshopper path is 1 - 3 - 2 - 3 - 2 - 3 and so on. The path is infinite. | 500 | [
{
"input": "2\n><\n1 2",
"output": "FINITE"
},
{
"input": "3\n>><\n2 1 1",
"output": "INFINITE"
},
{
"input": "1\n>\n1000000000",
"output": "FINITE"
},
{
"input": "1\n<\n1000000000",
"output": "FINITE"
},
{
"input": "2\n>>\n1 1",
"output": "FINITE"
},
{
... | 1,562,259,896 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 0 |
n=int(input())
mystr=input()
m=list(map(int,input().split()))
index=1
flag=False
x=[]
while index<=n:
if mystr[index-1]=='>':
index+=m[index-1]
else:
index-=m[index-1]
if index in x:
flag=True
break
x.append(index)
if flag==True:
print("INFINITE")
else:
print("FINITE")
| Title: Little Artem and Grasshopper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem found a grasshopper. He brought it to his house and constructed a jumping area for him.
The area looks like a strip of cells 1<=×<=*n*. Each cell contains the direction for the next jump and the length of that jump. Grasshopper starts in the first cell and follows the instructions written on the cells. Grasshopper stops immediately if it jumps out of the strip. Now Artem wants to find out if this will ever happen.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — length of the strip.
Next line contains a string of length *n* which consists of characters "<" and ">" only, that provide the direction of the jump from the corresponding cell. Next line contains *n* integers *d**i* (1<=≤<=*d**i*<=≤<=109) — the length of the jump from the *i*-th cell.
Output Specification:
Print "INFINITE" (without quotes) if grasshopper will continue his jumps forever. Otherwise print "FINITE" (without quotes).
Demo Input:
['2\n><\n1 2\n', '3\n>><\n2 1 1\n']
Demo Output:
['FINITE\n', 'INFINITE']
Note:
In the first sample grasshopper starts from the first cell and jumps to the right on the next cell. When he is in the second cell he needs to jump two cells left so he will jump out of the strip.
Second sample grasshopper path is 1 - 3 - 2 - 3 - 2 - 3 and so on. The path is infinite. | ```python
n=int(input())
mystr=input()
m=list(map(int,input().split()))
index=1
flag=False
x=[]
while index<=n:
if mystr[index-1]=='>':
index+=m[index-1]
else:
index-=m[index-1]
if index in x:
flag=True
break
x.append(index)
if flag==True:
print("INFINITE")
else:
print("FINITE")
``` | -1 | |
57 | A | Square Earth? | PROGRAMMING | 1,300 | [
"dfs and similar",
"greedy",
"implementation"
] | A. Square Earth? | 2 | 256 | Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side *n*. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0,<=0), (*n*,<=0), (0,<=*n*) and (*n*,<=*n*). | The single line contains 5 space-separated integers: *n*,<=*x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*x*1,<=*y*1,<=*x*2,<=*y*2<=≤<=*n*) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square. | You must print on a single line the shortest distance between the points. | [
"2 0 0 1 0\n",
"2 0 1 2 1\n",
"100 0 0 100 100\n"
] | [
"1\n",
"4\n",
"200\n"
] | none | 500 | [
{
"input": "2 0 0 1 0",
"output": "1"
},
{
"input": "2 0 1 2 1",
"output": "4"
},
{
"input": "100 0 0 100 100",
"output": "200"
},
{
"input": "4 0 3 1 4",
"output": "2"
},
{
"input": "10 8 10 10 0",
"output": "12"
},
{
"input": "26 21 0 26 14",
"ou... | 1,643,969,122 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 92 | 0 | I = input().split()
n = int(I[0])
P1 = (int(I[1]), int(I[2]))
P2 = (int(I[3]), int(I[4]))
l = [0, n]
adjacent = False
opposite = False
for i in range(2):
if P1[i] in l:
if P2[int(not i)] in l:
adjacent = True
if P2[i] in l:
if P1[i] == P2[i]:
adjacent = True
else:
opposite = True
if adjacent:
shortest_distance = abs(P1[0] - P2[0]) + abs(P1[1] - P2[1])
print(shortest_distance)
if opposite and not adjacent:
distance = []
if P1[0] in l and P2[0] in l:
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2*(min(P1[1], P2[1])))
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2*(n - max(P1[1], P2[1])))
else:
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2 * (min(P1[0], P2[0])))
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2 * (n - max(P1[0], P2[0])))
shortest_distance = min(distance)
print(shortest_distance) | Title: Square Earth?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side *n*. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0,<=0), (*n*,<=0), (0,<=*n*) and (*n*,<=*n*).
Input Specification:
The single line contains 5 space-separated integers: *n*,<=*x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*x*1,<=*y*1,<=*x*2,<=*y*2<=≤<=*n*) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square.
Output Specification:
You must print on a single line the shortest distance between the points.
Demo Input:
['2 0 0 1 0\n', '2 0 1 2 1\n', '100 0 0 100 100\n']
Demo Output:
['1\n', '4\n', '200\n']
Note:
none | ```python
I = input().split()
n = int(I[0])
P1 = (int(I[1]), int(I[2]))
P2 = (int(I[3]), int(I[4]))
l = [0, n]
adjacent = False
opposite = False
for i in range(2):
if P1[i] in l:
if P2[int(not i)] in l:
adjacent = True
if P2[i] in l:
if P1[i] == P2[i]:
adjacent = True
else:
opposite = True
if adjacent:
shortest_distance = abs(P1[0] - P2[0]) + abs(P1[1] - P2[1])
print(shortest_distance)
if opposite and not adjacent:
distance = []
if P1[0] in l and P2[0] in l:
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2*(min(P1[1], P2[1])))
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2*(n - max(P1[1], P2[1])))
else:
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2 * (min(P1[0], P2[0])))
distance.append(abs(P1[0] - P2[0]) + abs(P1[1] - P2[1]) + 2 * (n - max(P1[0], P2[0])))
shortest_distance = min(distance)
print(shortest_distance)
``` | 3.977 |
579 | A | Raising Bacteria | PROGRAMMING | 1,000 | [
"bitmasks"
] | null | null | You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days? | The only line containing one integer *x* (1<=≤<=*x*<=≤<=109). | The only line containing one integer: the answer. | [
"5\n",
"8\n"
] | [
"2\n",
"1\n"
] | For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | 250 | [
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697... | 1,686,154,969 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | import math
n=int(input())
m=int(math.log2(n))
m=1+(n-(2**m))
print(m) | Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | ```python
import math
n=int(input())
m=int(math.log2(n))
m=1+(n-(2**m))
print(m)
``` | 0 | |
443 | A | Anton and Letters | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set. | The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space. | Print a single number — the number of distinct letters in Anton's set. | [
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] | [
"3\n",
"2\n",
"0\n"
] | none | 500 | [
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"o... | 1,682,932,826 | 2,147,483,647 | PyPy 3 | OK | TESTS | 27 | 93 | 0 | s = set(input()[1:-1].split(", "))
if s == {''}:
print(0)
else:
print(len(s)) | Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none | ```python
s = set(input()[1:-1].split(", "))
if s == {''}:
print(0)
else:
print(len(s))
``` | 3 | |
334 | A | Candy Bags | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies. | The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers. | Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits. | [
"2\n"
] | [
"1 4\n2 3\n"
] | The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother. | 500 | [
{
"input": "2",
"output": "1 4\n2 3"
},
{
"input": "4",
"output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9"
},
{
"input": "6",
"output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19"
},
{
"input": "8",
"output"... | 1,528,472,804 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 186 | 102,400 | brotherNo = int(input())
candyPerBro = int((brotherNo**3 + brotherNo)/2)
for i in range(1, brotherNo + 1):
sum = 0
j = i
while sum < candyPerBro:
print(str(j) + " " + str(brotherNo*brotherNo + 1 - j ) + " ", end = "")
sum += j
sum += brotherNo*brotherNo - j + 1
j += brotherNo
print() | Title: Candy Bags
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
Input Specification:
The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers.
Output Specification:
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
Demo Input:
['2\n']
Demo Output:
['1 4\n2 3\n']
Note:
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother. | ```python
brotherNo = int(input())
candyPerBro = int((brotherNo**3 + brotherNo)/2)
for i in range(1, brotherNo + 1):
sum = 0
j = i
while sum < candyPerBro:
print(str(j) + " " + str(brotherNo*brotherNo + 1 - j ) + " ", end = "")
sum += j
sum += brotherNo*brotherNo - j + 1
j += brotherNo
print()
``` | 3 | |
295 | B | Greg and Graph | PROGRAMMING | 1,700 | [
"dp",
"graphs",
"shortest paths"
] | null | null | Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step. | The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes. | Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier. | [
"1\n0\n1\n",
"2\n0 5\n4 0\n1 2\n",
"4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n"
] | [
"0 ",
"9 0 ",
"17 23 404 0 "
] | none | 1,000 | [
{
"input": "1\n0\n1",
"output": "0 "
},
{
"input": "2\n0 5\n4 0\n1 2",
"output": "9 0 "
},
{
"input": "4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3",
"output": "17 23 404 0 "
},
{
"input": "4\n0 57148 51001 13357\n71125 0 98369 67226\n49388 90852 0 66291\n39573 38165 9700... | 1,575,418,985 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 13 | 3,000 | 7,987,200 | n = int(input())
matrix = []
for i in range(n):
elem = list(map(int, input().split()))
matrix.append(elem)
delete = list(map(int, input().split()))
ans = [0 for i in range(n)]
v = []
for i in range(n-1, -1, -1):
di = delete[i] - 1
v.append(di)
for j in range(n):
for k in range(n):
matrix[j][k] = min(matrix[j][k], matrix[j][di] + matrix[di][k])
for j in range(len(v)):
for k in range(len(v)):
ans[i] += matrix[v[j]][v[k]]
s = ""
ans = [str(a) for a in ans]
print(" ".join(ans)) | Title: Greg and Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes.
Output Specification:
Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Demo Input:
['1\n0\n1\n', '2\n0 5\n4 0\n1 2\n', '4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n']
Demo Output:
['0 ', '9 0 ', '17 23 404 0 ']
Note:
none | ```python
n = int(input())
matrix = []
for i in range(n):
elem = list(map(int, input().split()))
matrix.append(elem)
delete = list(map(int, input().split()))
ans = [0 for i in range(n)]
v = []
for i in range(n-1, -1, -1):
di = delete[i] - 1
v.append(di)
for j in range(n):
for k in range(n):
matrix[j][k] = min(matrix[j][k], matrix[j][di] + matrix[di][k])
for j in range(len(v)):
for k in range(len(v)):
ans[i] += matrix[v[j]][v[k]]
s = ""
ans = [str(a) for a in ans]
print(" ".join(ans))
``` | 0 | |
218 | B | Airport | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total. | Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly. | [
"4 3\n2 1 1\n",
"4 3\n2 2 2\n"
] | [
"5 5\n",
"7 6\n"
] | In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "5 5"
},
{
"input": "4 3\n2 2 2",
"output": "7 6"
},
{
"input": "10 5\n10 3 3 1 2",
"output": "58 26"
},
{
"input": "10 1\n10",
"output": "55 55"
},
{
"input": "10 1\n100",
"output": "955 955"
},
{
"input": "10 2\n4 7... | 1,613,664,531 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 307,200 | n,m=map(int,input().split())
arr=list(map(int,input().split()))
if sum(arr)==n:
ans=0
for i in arr:
ans+=sum(range(i+1))
print(ans)
else:
arr2=arr.copy()
temp=n
ans=0
while temp>0:
k=max(arr)
ind=arr.index(k)
ans+=k
arr[ind]-=1
temp-=1
arr2.sort()
ans2=0
i=0
while i<m and ans2<n:
ans2+=sum(range(arr2[i]+1))
i+=1
print(ans,ans2) | Title: Airport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.
Output Specification:
Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.
Demo Input:
['4 3\n2 1 1\n', '4 3\n2 2 2\n']
Demo Output:
['5 5\n', '7 6\n']
Note:
In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane. | ```python
n,m=map(int,input().split())
arr=list(map(int,input().split()))
if sum(arr)==n:
ans=0
for i in arr:
ans+=sum(range(i+1))
print(ans)
else:
arr2=arr.copy()
temp=n
ans=0
while temp>0:
k=max(arr)
ind=arr.index(k)
ans+=k
arr[ind]-=1
temp-=1
arr2.sort()
ans2=0
i=0
while i<m and ans2<n:
ans2+=sum(range(arr2[i]+1))
i+=1
print(ans,ans2)
``` | 0 | |
262 | A | Roma and Lucky Numbers | PROGRAMMING | 800 | [
"implementation"
] | null | null | Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem. | The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has.
The numbers in the lines are separated by single spaces. | In a single line print a single integer — the answer to the problem. | [
"3 4\n1 2 4\n",
"3 2\n447 44 77\n"
] | [
"3\n",
"2\n"
] | In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2. | 500 | [
{
"input": "3 4\n1 2 4",
"output": "3"
},
{
"input": "3 2\n447 44 77",
"output": "2"
},
{
"input": "2 2\n507978501 180480073",
"output": "2"
},
{
"input": "9 6\n655243746 167613748 1470546 57644035 176077477 56984809 44677 215706823 369042089",
"output": "9"
},
{
... | 1,639,427,566 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | n,k = map(int,input().split())
summa = 0
a = input().split()
for i in a:
if i.count("4")+i.count("7")<=k:
summa+=1
print(summa) | Title: Roma and Lucky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem.
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has.
The numbers in the lines are separated by single spaces.
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['3 4\n1 2 4\n', '3 2\n447 44 77\n']
Demo Output:
['3\n', '2\n']
Note:
In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2. | ```python
n,k = map(int,input().split())
summa = 0
a = input().split()
for i in a:
if i.count("4")+i.count("7")<=k:
summa+=1
print(summa)
``` | 3 | |
302 | A | Eugeny and Array | PROGRAMMING | 800 | [
"implementation"
] | null | null | Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries. | The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). | Print *m* integers — the responses to Eugene's queries in the order they occur in the input. | [
"2 3\n1 -1\n1 1\n1 2\n2 2\n",
"5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n"
] | [
"0\n1\n0\n",
"0\n1\n0\n1\n0\n"
] | none | 500 | [
{
"input": "2 3\n1 -1\n1 1\n1 2\n2 2",
"output": "0\n1\n0"
},
{
"input": "5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5",
"output": "0\n1\n0\n1\n0"
},
{
"input": "3 3\n1 1 1\n2 2\n1 1\n1 1",
"output": "0\n0\n0"
},
{
"input": "4 4\n-1 -1 -1 -1\n1 3\n1 2\n1 2\n1 1",
"output": "... | 1,574,190,323 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 11,673,600 | n, m = map(int, input().split())
l = list(map(int, input().split()))
ans = min(l.count(1), l.count(-1))
for _ in range(m):
l, r = map(int, input().split())
temp = r-l+1
if temp % 2 == 0 and temp//2 <= ans:
print(1)
else:
print(0) | Title: Eugeny and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
Input Specification:
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
Demo Input:
['2 3\n1 -1\n1 1\n1 2\n2 2\n', '5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n']
Demo Output:
['0\n1\n0\n', '0\n1\n0\n1\n0\n']
Note:
none | ```python
n, m = map(int, input().split())
l = list(map(int, input().split()))
ans = min(l.count(1), l.count(-1))
for _ in range(m):
l, r = map(int, input().split())
temp = r-l+1
if temp % 2 == 0 and temp//2 <= ans:
print(1)
else:
print(0)
``` | 0 | |
552 | B | Vanya and Books | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | null | null | Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library. | Print the number of digits needed to number all the books. | [
"13\n",
"4\n"
] | [
"17\n",
"4\n"
] | Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | 1,000 | [
{
"input": "13",
"output": "17"
},
{
"input": "4",
"output": "4"
},
{
"input": "100",
"output": "192"
},
{
"input": "99",
"output": "189"
},
{
"input": "1000000000",
"output": "8888888899"
},
{
"input": "1000000",
"output": "5888896"
},
{
"... | 1,591,693,712 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 108 | 307,200 | from sys import stdin, stdout
n=int(stdin.readline())
h=str(n)
a=[]
a.append(9)
c=1
for i in range(1,19):
c*=10
a.append((a[i-1]+(9*c)))
if(n%10==0):
k=n-1
l=len(h)-1
else:
k=n
l=len(h)
b=[]
b.append(0)
c=10
for i in range(1,19):
b.append(c)
c*=10
if(l==1 or l==0):
print(n)
else:
s=a[l-2]
s+=((((k-b[l-1])+1)*(l)))
if(n%10==0):
print(s+len(h))
else:
print(s)
| Title: Vanya and Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library.
Output Specification:
Print the number of digits needed to number all the books.
Demo Input:
['13\n', '4\n']
Demo Output:
['17\n', '4\n']
Note:
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | ```python
from sys import stdin, stdout
n=int(stdin.readline())
h=str(n)
a=[]
a.append(9)
c=1
for i in range(1,19):
c*=10
a.append((a[i-1]+(9*c)))
if(n%10==0):
k=n-1
l=len(h)-1
else:
k=n
l=len(h)
b=[]
b.append(0)
c=10
for i in range(1,19):
b.append(c)
c*=10
if(l==1 or l==0):
print(n)
else:
s=a[l-2]
s+=((((k-b[l-1])+1)*(l)))
if(n%10==0):
print(s+len(h))
else:
print(s)
``` | 0 | |
379 | A | New Year Candles | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. | The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). | Print a single integer — the number of hours Vasily can light up the room for. | [
"4 2\n",
"6 3\n"
] | [
"7\n",
"8\n"
] | Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | 500 | [
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": ... | 1,678,929,801 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | import sys
import math
inputs = []
for line in sys.stdin:
inputs.append(line)
a = int(inputs[0].strip().split()[0])
b = int(inputs[0].strip().split()[1])
totalCandlesLeft = a
aCandlesTaken = 0
candlesToReuse = 0
hours = 0
while totalCandlesLeft > 0:
totalCandlesLeft-=1
aCandlesTaken+=1
hours+=1
if aCandlesTaken == b:
aCandlesTaken-=b
totalCandlesLeft+=1
print(hours)
| Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | ```python
import sys
import math
inputs = []
for line in sys.stdin:
inputs.append(line)
a = int(inputs[0].strip().split()[0])
b = int(inputs[0].strip().split()[1])
totalCandlesLeft = a
aCandlesTaken = 0
candlesToReuse = 0
hours = 0
while totalCandlesLeft > 0:
totalCandlesLeft-=1
aCandlesTaken+=1
hours+=1
if aCandlesTaken == b:
aCandlesTaken-=b
totalCandlesLeft+=1
print(hours)
``` | 3 | |
29 | A | Spit Problem | PROGRAMMING | 1,000 | [
"brute force"
] | A. Spit Problem | 2 | 256 | In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position. | If there are two camels, which spitted at each other, output YES. Otherwise, output NO. | [
"2\n0 1\n1 -1\n",
"3\n0 1\n1 1\n2 -2\n",
"5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "2\n0 1\n1 -1",
"output": "YES"
},
{
"input": "3\n0 1\n1 1\n2 -2",
"output": "NO"
},
{
"input": "5\n2 -10\n3 10\n0 5\n5 -5\n10 1",
"output": "YES"
},
{
"input": "10\n-9897 -1144\n-4230 -6350\n2116 -3551\n-3635 4993\n3907 -9071\n-2362 4120\n-6542 984\n5807 3745\n759... | 1,698,991,335 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 30 | 124 | 1,536,000 | def sol():
n = int(input())
l = []
for i in range(n):
l.append(list(map(int,input().split())))
for i in range(n-1):
x, d = l[i]
for j in range(i+1,n):
xj, dj = l[j]
if x+d == xj and x == xj+dj:
print("YES")
return
print("NO")
sol() | Title: Spit Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
Output Specification:
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
Demo Input:
['2\n0 1\n1 -1\n', '3\n0 1\n1 1\n2 -2\n', '5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
def sol():
n = int(input())
l = []
for i in range(n):
l.append(list(map(int,input().split())))
for i in range(n-1):
x, d = l[i]
for j in range(i+1,n):
xj, dj = l[j]
if x+d == xj and x == xj+dj:
print("YES")
return
print("NO")
sol()
``` | 3.966139 |
844 | A | Diversity | PROGRAMMING | 1,000 | [
"greedy",
"implementation",
"strings"
] | null | null | Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too. | First line of input contains string *s*, consisting only of lowercase Latin letters (1<=≤<=|*s*|<=≤<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=≤<=*k*<=≤<=26). | Print single line with a minimum number of necessary changes, or the word «impossible» (without quotes) if it is impossible. | [
"yandex\n6\n",
"yahoo\n5\n",
"google\n7\n"
] | [
"0\n",
"1\n",
"impossible\n"
] | In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*', '*h*', '*o*', '*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6. | 500 | [
{
"input": "yandex\n6",
"output": "0"
},
{
"input": "yahoo\n5",
"output": "1"
},
{
"input": "google\n7",
"output": "impossible"
},
{
"input": "a\n1",
"output": "0"
},
{
"input": "z\n2",
"output": "impossible"
},
{
"input": "fwgfrwgkuwghfiruhewgirueguhe... | 1,627,988,260 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 62 | 6,963,200 | s=input()
n=int(input())
set1=set(s)
if len(s)==n:
a=len(s)-len(set1)
print(a)
elif len(s)>n:
z=n-len(set1)
if z<0:
print("0")
else:
print(n-len(set1))
elif len(s)<n:
print("impossible") | Title: Diversity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.
Input Specification:
First line of input contains string *s*, consisting only of lowercase Latin letters (1<=≤<=|*s*|<=≤<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=≤<=*k*<=≤<=26).
Output Specification:
Print single line with a minimum number of necessary changes, or the word «impossible» (without quotes) if it is impossible.
Demo Input:
['yandex\n6\n', 'yahoo\n5\n', 'google\n7\n']
Demo Output:
['0\n', '1\n', 'impossible\n']
Note:
In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*', '*h*', '*o*', '*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6. | ```python
s=input()
n=int(input())
set1=set(s)
if len(s)==n:
a=len(s)-len(set1)
print(a)
elif len(s)>n:
z=n-len(set1)
if z<0:
print("0")
else:
print(n-len(set1))
elif len(s)<n:
print("impossible")
``` | 3 | |
43 | A | Football | PROGRAMMING | 1,000 | [
"strings"
] | A. Football | 2 | 256 | One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. | Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. | [
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] | [
"ABC\n",
"A\n"
] | none | 500 | [
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
... | 1,689,249,901 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | t=int(input())
scores={}
for _ in range(t):
T=input()
if T in scores:
scores[T]+=1
else:
scores[T]=1
X=max(scores,key=scores.get)
print(X) | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none | ```python
t=int(input())
scores={}
for _ in range(t):
T=input()
if T in scores:
scores[T]+=1
else:
scores[T]=1
X=max(scores,key=scores.get)
print(X)
``` | 3.977 |
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",
... | 1,681,935,225 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 77 | 0 | n, k = [int(x) for x in input().split()]
time_left = 240 - k
time_spent = 0
n_tasks = n
for num in range(1, n + 1):
time_spent += num * 5
if time_spent > time_left:
n_tasks = num - 1
break
print(n_tasks) | 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 = [int(x) for x in input().split()]
time_left = 240 - k
time_spent = 0
n_tasks = n
for num in range(1, n + 1):
time_spent += num * 5
if time_spent > time_left:
n_tasks = num - 1
break
print(n_tasks)
``` | 3 | |
356 | A | Knight Tournament | PROGRAMMING | 1,500 | [
"data structures",
"dsu"
] | null | null | Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.
As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:
- There are *n* knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to *n*. - The tournament consisted of *m* fights, in the *i*-th fight the knights that were still in the game with numbers at least *l**i* and at most *r**i* have fought for the right to continue taking part in the tournament. - After the *i*-th fight among all participants of the fight only one knight won — the knight number *x**i*, he continued participating in the tournament. Other knights left the tournament. - The winner of the last (the *m*-th) fight (the knight number *x**m*) became the winner of the tournament.
You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number *b* was conquered by the knight number *a*, if there was a fight with both of these knights present and the winner was the knight number *a*.
Write the code that calculates for each knight, the name of the knight that beat him. | The first line contains two integers *n*, *m* (2<=≤<=*n*<=≤<=3·105; 1<=≤<=*m*<=≤<=3·105) — the number of knights and the number of fights. Each of the following *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*; *l**i*<=≤<=*x**i*<=≤<=*r**i*) — the description of the *i*-th fight.
It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle. | Print *n* integers. If the *i*-th knight lost, then the *i*-th number should equal the number of the knight that beat the knight number *i*. If the *i*-th knight is the winner, then the *i*-th number must equal 0. | [
"4 3\n1 2 1\n1 3 3\n1 4 4\n",
"8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1\n"
] | [
"3 1 4 0 ",
"0 8 4 6 4 8 6 1 "
] | Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won. | 500 | [
{
"input": "4 3\n1 2 1\n1 3 3\n1 4 4",
"output": "3 1 4 0 "
},
{
"input": "8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1",
"output": "0 8 4 6 4 8 6 1 "
},
{
"input": "2 1\n1 2 1",
"output": "0 1 "
},
{
"input": "2 1\n1 2 2",
"output": "2 0 "
},
{
"input": "3 1\n1 3 1",
"out... | 1,601,247,774 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | def knight_tournament(oponents,results):
knights=[]
derrotados=dict()
survivors=[]
for k in range(1,oponents+1):
knights.append(k)
for s in range(1,oponents+1):
survivors.append(str(s))
for battle in results:
for defeat in survivors:
ini=0
mid = len(survivors)/2
fin=len(survivors)
while(survivors[ini]<=ini):
ini=(ini+mid)/2
if(survivors[ini-1]>=ini):
ini=ini/2
while(survivors[fin]>=fin):
fin=fin/2
if(survivors[fin+1]<=fin):
fin=(fin+mid)/2
for defeat in survivors[ini:fin+1]:
if(defeat!=int(battle[2]):
derrotados[defeat]=battle[2]
for k in range(1,oponents+1):
print(derrotados.get(k,0))
nm=input().split()
battles=[]
for b in range(0,int(nm[1])):
info=input().split()
round=[]
round.append(info[0])
round.append(info[1])
round.append(info[2])
battles.append(round)
knight_tournament(int(nm[0]),battles)
| Title: Knight Tournament
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.
As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:
- There are *n* knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to *n*. - The tournament consisted of *m* fights, in the *i*-th fight the knights that were still in the game with numbers at least *l**i* and at most *r**i* have fought for the right to continue taking part in the tournament. - After the *i*-th fight among all participants of the fight only one knight won — the knight number *x**i*, he continued participating in the tournament. Other knights left the tournament. - The winner of the last (the *m*-th) fight (the knight number *x**m*) became the winner of the tournament.
You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number *b* was conquered by the knight number *a*, if there was a fight with both of these knights present and the winner was the knight number *a*.
Write the code that calculates for each knight, the name of the knight that beat him.
Input Specification:
The first line contains two integers *n*, *m* (2<=≤<=*n*<=≤<=3·105; 1<=≤<=*m*<=≤<=3·105) — the number of knights and the number of fights. Each of the following *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*; *l**i*<=≤<=*x**i*<=≤<=*r**i*) — the description of the *i*-th fight.
It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle.
Output Specification:
Print *n* integers. If the *i*-th knight lost, then the *i*-th number should equal the number of the knight that beat the knight number *i*. If the *i*-th knight is the winner, then the *i*-th number must equal 0.
Demo Input:
['4 3\n1 2 1\n1 3 3\n1 4 4\n', '8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1\n']
Demo Output:
['3 1 4 0 ', '0 8 4 6 4 8 6 1 ']
Note:
Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won. | ```python
def knight_tournament(oponents,results):
knights=[]
derrotados=dict()
survivors=[]
for k in range(1,oponents+1):
knights.append(k)
for s in range(1,oponents+1):
survivors.append(str(s))
for battle in results:
for defeat in survivors:
ini=0
mid = len(survivors)/2
fin=len(survivors)
while(survivors[ini]<=ini):
ini=(ini+mid)/2
if(survivors[ini-1]>=ini):
ini=ini/2
while(survivors[fin]>=fin):
fin=fin/2
if(survivors[fin+1]<=fin):
fin=(fin+mid)/2
for defeat in survivors[ini:fin+1]:
if(defeat!=int(battle[2]):
derrotados[defeat]=battle[2]
for k in range(1,oponents+1):
print(derrotados.get(k,0))
nm=input().split()
battles=[]
for b in range(0,int(nm[1])):
info=input().split()
round=[]
round.append(info[0])
round.append(info[1])
round.append(info[2])
battles.append(round)
knight_tournament(int(nm[0]),battles)
``` | -1 | |
370 | A | Rook, Bishop and King | PROGRAMMING | 1,100 | [
"graphs",
"math",
"shortest paths"
] | null | null | Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8<=×<=8 table. A field is represented by a pair of integers (*r*,<=*c*) — the number of the row and the number of the column (in a classical game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:
- A rook moves any number of fields horizontally or vertically. - A bishop moves any number of fields diagonally. - A king moves one field in any direction — horizontally, vertically or diagonally.
Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2)? At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem. | The input contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*c*1,<=*r*2,<=*c*2<=≤<=8) — the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.
You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8. | Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2). If a piece cannot make such a move, print a 0 instead of the corresponding number. | [
"4 3 1 6\n",
"5 5 5 6\n"
] | [
"2 1 3\n",
"1 0 1\n"
] | none | 500 | [
{
"input": "4 3 1 6",
"output": "2 1 3"
},
{
"input": "5 5 5 6",
"output": "1 0 1"
},
{
"input": "1 1 8 8",
"output": "2 1 7"
},
{
"input": "1 1 8 1",
"output": "1 0 7"
},
{
"input": "1 1 1 8",
"output": "1 0 7"
},
{
"input": "8 1 1 1",
"output": "... | 1,621,653,462 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 61 | 0 | import math
r1, c1, r2, c2 = [int(item) for item in input().split()]
# rei pode se movimentar para todos os lados
# r1 - r2 é quanto ele precisa se movimentar na linha
# c1 - c2 é quanto ele precisa se movimentar na coluna
ReiMovimentoLinha = abs(r1-r2)
ReiMovimentoColuna = abs(c1-c2)
reiResposta = math.ceil((ReiMovimentoColuna + ReiMovimentoLinha)/2)
# o bismo pode se mover 1 field pode se movimentar apenas diagonalmente
# logo, ele só se movimenta nos quadrados coloridos igual ele
# pegando o mod da soma de r+c, podemos saber se pode ou não
# depois disso é só ver quantas vezes é necessário
# para isso, é só ver se r1-c1 é igual a r2-c2
somaBispoInicio = r1+c1
somaBispoFinal = r2+c2
respostaBispo = 0
if(somaBispoInicio%2 == somaBispoFinal%2):
respostaBispo = abs(r1-r2) == abs(c1-c2)
if(respostaBispo):
respostaBispo = 1
else:
respostaBispo = 0
# para a torre
# se estiver na mesma coluna ou linha é 1
# se não é 2
respostaTorre = 0
if(r1 == r2 or c1 == c2):
respostaTorre = 1
else:
respostaTorre = 2
print(respostaTorre, respostaBispo, reiResposta)
| Title: Rook, Bishop and King
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8<=×<=8 table. A field is represented by a pair of integers (*r*,<=*c*) — the number of the row and the number of the column (in a classical game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:
- A rook moves any number of fields horizontally or vertically. - A bishop moves any number of fields diagonally. - A king moves one field in any direction — horizontally, vertically or diagonally.
Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2)? At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem.
Input Specification:
The input contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*c*1,<=*r*2,<=*c*2<=≤<=8) — the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.
You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8.
Output Specification:
Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2). If a piece cannot make such a move, print a 0 instead of the corresponding number.
Demo Input:
['4 3 1 6\n', '5 5 5 6\n']
Demo Output:
['2 1 3\n', '1 0 1\n']
Note:
none | ```python
import math
r1, c1, r2, c2 = [int(item) for item in input().split()]
# rei pode se movimentar para todos os lados
# r1 - r2 é quanto ele precisa se movimentar na linha
# c1 - c2 é quanto ele precisa se movimentar na coluna
ReiMovimentoLinha = abs(r1-r2)
ReiMovimentoColuna = abs(c1-c2)
reiResposta = math.ceil((ReiMovimentoColuna + ReiMovimentoLinha)/2)
# o bismo pode se mover 1 field pode se movimentar apenas diagonalmente
# logo, ele só se movimenta nos quadrados coloridos igual ele
# pegando o mod da soma de r+c, podemos saber se pode ou não
# depois disso é só ver quantas vezes é necessário
# para isso, é só ver se r1-c1 é igual a r2-c2
somaBispoInicio = r1+c1
somaBispoFinal = r2+c2
respostaBispo = 0
if(somaBispoInicio%2 == somaBispoFinal%2):
respostaBispo = abs(r1-r2) == abs(c1-c2)
if(respostaBispo):
respostaBispo = 1
else:
respostaBispo = 0
# para a torre
# se estiver na mesma coluna ou linha é 1
# se não é 2
respostaTorre = 0
if(r1 == r2 or c1 == c2):
respostaTorre = 1
else:
respostaTorre = 2
print(respostaTorre, respostaBispo, reiResposta)
``` | 0 | |
38 | C | Blinds | PROGRAMMING | 1,400 | [
"brute force"
] | C. Blinds | 2 | 256 | The blinds are known to consist of opaque horizontal stripes that can be rotated thus regulating the amount of light flowing in the room. There are *n* blind stripes with the width of 1 in the factory warehouse for blind production. The problem is that all of them are spare details from different orders, that is, they may not have the same length (it is even possible for them to have different lengths)
Every stripe can be cut into two or more parts. The cuttings are made perpendicularly to the side along which the length is measured. Thus the cuttings do not change the width of a stripe but each of the resulting pieces has a lesser length (the sum of which is equal to the length of the initial stripe)
After all the cuttings the blinds are constructed through consecutive joining of several parts, similar in length, along sides, along which length is measured. Also, apart from the resulting pieces an initial stripe can be used as a blind if it hasn't been cut. It is forbidden to construct blinds in any other way.
Thus, if the blinds consist of *k* pieces each *d* in length, then they are of form of a rectangle of *k*<=×<=*d* bourlemeters.
Your task is to find for what window possessing the largest possible area the blinds can be made from the given stripes if on technical grounds it is forbidden to use pieces shorter than *l* bourlemeter. The window is of form of a rectangle with side lengths as positive integers. | The first output line contains two space-separated integers *n* and *l* (1<=≤<=*n*,<=*l*<=≤<=100). They are the number of stripes in the warehouse and the minimal acceptable length of a blind stripe in bourlemeters. The second line contains space-separated *n* integers *a**i*. They are the lengths of initial stripes in bourlemeters (1<=≤<=*a**i*<=≤<=100). | Print the single number — the maximal area of the window in square bourlemeters that can be completely covered. If no window with a positive area that can be covered completely without breaking any of the given rules exist, then print the single number 0. | [
"4 2\n1 2 3 4\n",
"5 3\n5 5 7 3 1\n",
"2 3\n1 2\n"
] | [
"8\n",
"15\n",
"0\n"
] | In the first sample test the required window is 2 × 4 in size and the blinds for it consist of 4 parts, each 2 bourlemeters long. One of the parts is the initial stripe with the length of 2, the other one is a part of a cut stripe with the length of 3 and the two remaining stripes are parts of a stripe with the length of 4 cut in halves. | 0 | [
{
"input": "4 2\n1 2 3 4",
"output": "8"
},
{
"input": "5 3\n5 5 7 3 1",
"output": "15"
},
{
"input": "2 3\n1 2",
"output": "0"
},
{
"input": "2 2\n3 3",
"output": "6"
},
{
"input": "5 2\n2 4 1 1 3",
"output": "8"
},
{
"input": "7 4\n3 2 1 1 1 3 2",
... | 1,675,506,421 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 92 | 4,300,800 | # LUOGU_RID: 101448633
n, m, *a = map(int, open(0).read().split())
ans = []
for i in range(m, 101):
ans += i * sum(x // i for x in a),
print(max(ans)) | Title: Blinds
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The blinds are known to consist of opaque horizontal stripes that can be rotated thus regulating the amount of light flowing in the room. There are *n* blind stripes with the width of 1 in the factory warehouse for blind production. The problem is that all of them are spare details from different orders, that is, they may not have the same length (it is even possible for them to have different lengths)
Every stripe can be cut into two or more parts. The cuttings are made perpendicularly to the side along which the length is measured. Thus the cuttings do not change the width of a stripe but each of the resulting pieces has a lesser length (the sum of which is equal to the length of the initial stripe)
After all the cuttings the blinds are constructed through consecutive joining of several parts, similar in length, along sides, along which length is measured. Also, apart from the resulting pieces an initial stripe can be used as a blind if it hasn't been cut. It is forbidden to construct blinds in any other way.
Thus, if the blinds consist of *k* pieces each *d* in length, then they are of form of a rectangle of *k*<=×<=*d* bourlemeters.
Your task is to find for what window possessing the largest possible area the blinds can be made from the given stripes if on technical grounds it is forbidden to use pieces shorter than *l* bourlemeter. The window is of form of a rectangle with side lengths as positive integers.
Input Specification:
The first output line contains two space-separated integers *n* and *l* (1<=≤<=*n*,<=*l*<=≤<=100). They are the number of stripes in the warehouse and the minimal acceptable length of a blind stripe in bourlemeters. The second line contains space-separated *n* integers *a**i*. They are the lengths of initial stripes in bourlemeters (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the single number — the maximal area of the window in square bourlemeters that can be completely covered. If no window with a positive area that can be covered completely without breaking any of the given rules exist, then print the single number 0.
Demo Input:
['4 2\n1 2 3 4\n', '5 3\n5 5 7 3 1\n', '2 3\n1 2\n']
Demo Output:
['8\n', '15\n', '0\n']
Note:
In the first sample test the required window is 2 × 4 in size and the blinds for it consist of 4 parts, each 2 bourlemeters long. One of the parts is the initial stripe with the length of 2, the other one is a part of a cut stripe with the length of 3 and the two remaining stripes are parts of a stripe with the length of 4 cut in halves. | ```python
# LUOGU_RID: 101448633
n, m, *a = map(int, open(0).read().split())
ans = []
for i in range(m, 101):
ans += i * sum(x // i for x in a),
print(max(ans))
``` | 3.968989 |
254 | A | Cards with Numbers | PROGRAMMING | 1,200 | [
"constructive algorithms",
"sortings"
] | null | null | Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that. | The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces. | If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs.
Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them. | [
"3\n20 30 10 30 20 10\n",
"1\n1 2\n"
] | [
"4 2\n1 5\n6 3\n",
"-1"
] | none | 500 | [
{
"input": "3\n20 30 10 30 20 10",
"output": "4 2\n1 5\n6 3"
},
{
"input": "1\n1 2",
"output": "-1"
},
{
"input": "5\n2 2 2 2 2 1 2 2 1 2",
"output": "2 1\n3 4\n7 5\n6 9\n10 8"
},
{
"input": "5\n2 1 2 2 1 1 1 1 1 2",
"output": "3 1\n2 5\n7 6\n8 9\n10 4"
},
{
"inpu... | 1,576,255,124 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | with open('input.txt', 'r') as f:
line = f.readlines()
line1, line2 = line
n = int(line1.strip())
arr = [(int(card), i) for i, card in enumerate(line2.strip().split())]
#arr = [int(card) for i, card in input().split()]
#arr.sort()
for i, card in enumerate(arr):
if card in d:
d[card].append(i)
else:
d[card] = [i]
for card, indices in d.items():
if len(indices)%2:
answer = '-1'
break
else:
answer += str(indices[i])+' '+str(indices[i+1])+'\n' for i in range(0, len(indices), 2)
'''
for i in range(0, 2*n, 2):
if arr[i][0] == arr[i+1][0]:
answer += str(arr[i][1]+1) + ' ' + str(arr[i+1][1]+1) + '\n'
#print(i)
else:
answer = '-1'
break
'''
with open('output.txt', 'w') as f:
f.write(answer) | Title: Cards with Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.
Output Specification:
If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs.
Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.
Demo Input:
['3\n20 30 10 30 20 10\n', '1\n1 2\n']
Demo Output:
['4 2\n1 5\n6 3\n', '-1']
Note:
none | ```python
with open('input.txt', 'r') as f:
line = f.readlines()
line1, line2 = line
n = int(line1.strip())
arr = [(int(card), i) for i, card in enumerate(line2.strip().split())]
#arr = [int(card) for i, card in input().split()]
#arr.sort()
for i, card in enumerate(arr):
if card in d:
d[card].append(i)
else:
d[card] = [i]
for card, indices in d.items():
if len(indices)%2:
answer = '-1'
break
else:
answer += str(indices[i])+' '+str(indices[i+1])+'\n' for i in range(0, len(indices), 2)
'''
for i in range(0, 2*n, 2):
if arr[i][0] == arr[i+1][0]:
answer += str(arr[i][1]+1) + ' ' + str(arr[i+1][1]+1) + '\n'
#print(i)
else:
answer = '-1'
break
'''
with open('output.txt', 'w') as f:
f.write(answer)
``` | -1 | |
803 | B | Distances to Zero | PROGRAMMING | 1,200 | [
"constructive algorithms"
] | null | null | You are given the array of integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array. | The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — length of the array *a*. The second line contains integer elements of the array separated by single spaces (<=-<=109<=≤<=*a**i*<=≤<=109). | Print the sequence *d*0,<=*d*1,<=...,<=*d**n*<=-<=1, where *d**i* is the difference of indices between *i* and nearest *j* such that *a**j*<==<=0. It is possible that *i*<==<=*j*. | [
"9\n2 1 0 3 0 0 3 2 4\n",
"5\n0 1 2 3 4\n",
"7\n5 6 0 1 -2 3 4\n"
] | [
"2 1 0 1 0 0 1 2 3 ",
"0 1 2 3 4 ",
"2 1 0 1 2 3 4 "
] | none | 0 | [
{
"input": "9\n2 1 0 3 0 0 3 2 4",
"output": "2 1 0 1 0 0 1 2 3 "
},
{
"input": "5\n0 1 2 3 4",
"output": "0 1 2 3 4 "
},
{
"input": "7\n5 6 0 1 -2 3 4",
"output": "2 1 0 1 2 3 4 "
},
{
"input": "1\n0",
"output": "0 "
},
{
"input": "2\n0 0",
"output": "0 0 "
... | 1,628,924,093 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 14 | 467 | 24,576,000 | i = int(input())
l = list(map(int,input().split()))
d= 1000
n = []
for x in l:
if x==0:d = 0;n.append(d)
else:d +=1;n.append(d)
ans = []
d = 1000
for x in list(reversed(n)):
if x == 0: d=0
else: d+=1
ans.append(min(x,d))
print(*ans[::-1]) | Title: Distances to Zero
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given the array of integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — length of the array *a*. The second line contains integer elements of the array separated by single spaces (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print the sequence *d*0,<=*d*1,<=...,<=*d**n*<=-<=1, where *d**i* is the difference of indices between *i* and nearest *j* such that *a**j*<==<=0. It is possible that *i*<==<=*j*.
Demo Input:
['9\n2 1 0 3 0 0 3 2 4\n', '5\n0 1 2 3 4\n', '7\n5 6 0 1 -2 3 4\n']
Demo Output:
['2 1 0 1 0 0 1 2 3 ', '0 1 2 3 4 ', '2 1 0 1 2 3 4 ']
Note:
none | ```python
i = int(input())
l = list(map(int,input().split()))
d= 1000
n = []
for x in l:
if x==0:d = 0;n.append(d)
else:d +=1;n.append(d)
ans = []
d = 1000
for x in list(reversed(n)):
if x == 0: d=0
else: d+=1
ans.append(min(x,d))
print(*ans[::-1])
``` | 0 | |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,560,849,680 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 109 | 307,200 | n=int(input())
a=list(map(int,input().split()))
dp1={}
dp2={}
c1=0
c2=0
x=a.count(0)
dp1[0]=a[0]
if(a[0]==3):
dp1[0]=1
for i in range(1,n):
if(a[i]==3):
dp1[i]=({1,2}-{dp1[i-1]}).pop()
else:
dp1[i]=a[i]
if(dp1[i]==dp1[i-1])and((dp1[i]==1)or(dp1[i]==2)):
c1+=1
c1+=x
if(a[0]!=3):
print(c1)
exit()
dp2[0]=a[0]
if(a[0]==3):
dp2[0]=2
for i in range(1,n):
if(a[i]==3):
dp2[i]=({1,2}-{dp2[i-1]}).pop()
else:
dp2[i]=a[i]
if(dp2[i]==dp2[i-1])and((dp2[i]==1)or(dp2[i]==2)):
c2+=1
c2+=x
if(a[0]!=3):
print(c2)
exit()
print(min(c1,c2))
| Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
n=int(input())
a=list(map(int,input().split()))
dp1={}
dp2={}
c1=0
c2=0
x=a.count(0)
dp1[0]=a[0]
if(a[0]==3):
dp1[0]=1
for i in range(1,n):
if(a[i]==3):
dp1[i]=({1,2}-{dp1[i-1]}).pop()
else:
dp1[i]=a[i]
if(dp1[i]==dp1[i-1])and((dp1[i]==1)or(dp1[i]==2)):
c1+=1
c1+=x
if(a[0]!=3):
print(c1)
exit()
dp2[0]=a[0]
if(a[0]==3):
dp2[0]=2
for i in range(1,n):
if(a[i]==3):
dp2[i]=({1,2}-{dp2[i-1]}).pop()
else:
dp2[i]=a[i]
if(dp2[i]==dp2[i-1])and((dp2[i]==1)or(dp2[i]==2)):
c2+=1
c2+=x
if(a[0]!=3):
print(c2)
exit()
print(min(c1,c2))
``` | 0 | |
336 | A | Vasily the Bear and Triangle | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique. | The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0). | Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points. | [
"10 5\n",
"-10 5\n"
] | [
"0 15 15 0\n",
"-15 0 0 15\n"
] | <img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample | 500 | [
{
"input": "10 5",
"output": "0 15 15 0"
},
{
"input": "-10 5",
"output": "-15 0 0 15"
},
{
"input": "20 -10",
"output": "0 -30 30 0"
},
{
"input": "-10 -1000000000",
"output": "-1000000010 0 0 -1000000010"
},
{
"input": "-1000000000 -1000000000",
"output": "-... | 1,649,438,828 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | x,y=map(int,input().split())
a=abs(x)+abs(y)
if x>0 and y>0:
ans=[0,a,a,0]
elif x>0 and y<0:
ans=[-a,0,0,a]
elif x<0 and y>0:
ans=[-a,0,0,a]
else:
ans=[-a,0,0,-a]
for i in ans:
print(i, end=" ")
| Title: Vasily the Bear and Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique.
Input Specification:
The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0).
Output Specification:
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points.
Demo Input:
['10 5\n', '-10 5\n']
Demo Output:
['0 15 15 0\n', '-15 0 0 15\n']
Note:
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample | ```python
x,y=map(int,input().split())
a=abs(x)+abs(y)
if x>0 and y>0:
ans=[0,a,a,0]
elif x>0 and y<0:
ans=[-a,0,0,a]
elif x<0 and y>0:
ans=[-a,0,0,a]
else:
ans=[-a,0,0,-a]
for i in ans:
print(i, end=" ")
``` | 0 | |
34 | A | Reconnaissance 2 | PROGRAMMING | 800 | [
"implementation"
] | A. Reconnaissance 2 | 2 | 256 | *n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit. | The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction. | Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle. | [
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] | [
"5 1\n",
"1 2\n"
] | none | 500 | [
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output"... | 1,689,453,413 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 92 | 0 | n = int(input())
# how do you read the list?
a = [int(x) for x in input().split()]
# (a0, a1), (a1, a2), ... (an-1, a0)
diff = 10000
for i in range(n): # 0-based
# i from 0 to n-1
# j = i + 1, if i == n-1, j = 0
j = i + 1
if j == n:
j = 0
if abs(a[i]-a[j]) < diff:
diff = abs(a[i]-a[j])
x1, x2 = i, j
print(x1+1, x2+1) # convert to 1-based | Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none | ```python
n = int(input())
# how do you read the list?
a = [int(x) for x in input().split()]
# (a0, a1), (a1, a2), ... (an-1, a0)
diff = 10000
for i in range(n): # 0-based
# i from 0 to n-1
# j = i + 1, if i == n-1, j = 0
j = i + 1
if j == n:
j = 0
if abs(a[i]-a[j]) < diff:
diff = abs(a[i]-a[j])
x1, x2 = i, j
print(x1+1, x2+1) # convert to 1-based
``` | 3.977 |
718 | A | Efim and Strange Grade | PROGRAMMING | 1,700 | [
"dp",
"implementation",
"math"
] | null | null | Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappointed, as he expected a way more pleasant result. Then, he developed a tricky plan. Each second, he can ask his teacher to round the grade at any place after the decimal point (also, he can ask to round to the nearest integer).
There are *t* seconds left till the end of the break, so Efim has to act fast. Help him find what is the maximum grade he can get in no more than *t* seconds. Note, that he can choose to not use all *t* seconds. Moreover, he can even choose to not round the grade at all.
In this problem, classic rounding rules are used: while rounding number to the *n*-th digit one has to take a look at the digit *n*<=+<=1. If it is less than 5 than the *n*-th digit remain unchanged while all subsequent digits are replaced with 0. Otherwise, if the *n*<=+<=1 digit is greater or equal to 5, the digit at the position *n* is increased by 1 (this might also change some other digits, if this one was equal to 9) and all subsequent digits are replaced with 0. At the end, all trailing zeroes are thrown away.
For example, if the number 1.14 is rounded to the first decimal place, the result is 1.1, while if we round 1.5 to the nearest integer, the result is 2. Rounding number 1.299996121 in the fifth decimal place will result in number 1.3. | The first line of the input contains two integers *n* and *t* (1<=≤<=*n*<=≤<=200<=000, 1<=≤<=*t*<=≤<=109) — the length of Efim's grade and the number of seconds till the end of the break respectively.
The second line contains the grade itself. It's guaranteed that the grade is a positive number, containing at least one digit after the decimal points, and it's representation doesn't finish with 0. | Print the maximum grade that Efim can get in *t* seconds. Do not print trailing zeroes. | [
"6 1\n10.245\n",
"6 2\n10.245\n",
"3 100\n9.2\n"
] | [
"10.25\n",
"10.3\n",
"9.2\n"
] | In the first two samples Efim initially has grade 10.245.
During the first second Efim can obtain grade 10.25, and then 10.3 during the next second. Note, that the answer 10.30 will be considered incorrect.
In the third sample the optimal strategy is to not perform any rounding at all. | 500 | [
{
"input": "6 1\n10.245",
"output": "10.25"
},
{
"input": "6 2\n10.245",
"output": "10.3"
},
{
"input": "3 100\n9.2",
"output": "9.2"
},
{
"input": "12 5\n872.04488525",
"output": "872.1"
},
{
"input": "35 8\n984227318.2031144444444444494637612",
"output": "98... | 1,475,155,873 | 373 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | a = input()
a = a.split()
n = int(a[0])
t = int(a[1])
grade = float(input())
for j in range(t):
l = []
for i in range(100):
l += [round(grade, i)]
grade = max(l)
print(grade) | Title: Efim and Strange Grade
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappointed, as he expected a way more pleasant result. Then, he developed a tricky plan. Each second, he can ask his teacher to round the grade at any place after the decimal point (also, he can ask to round to the nearest integer).
There are *t* seconds left till the end of the break, so Efim has to act fast. Help him find what is the maximum grade he can get in no more than *t* seconds. Note, that he can choose to not use all *t* seconds. Moreover, he can even choose to not round the grade at all.
In this problem, classic rounding rules are used: while rounding number to the *n*-th digit one has to take a look at the digit *n*<=+<=1. If it is less than 5 than the *n*-th digit remain unchanged while all subsequent digits are replaced with 0. Otherwise, if the *n*<=+<=1 digit is greater or equal to 5, the digit at the position *n* is increased by 1 (this might also change some other digits, if this one was equal to 9) and all subsequent digits are replaced with 0. At the end, all trailing zeroes are thrown away.
For example, if the number 1.14 is rounded to the first decimal place, the result is 1.1, while if we round 1.5 to the nearest integer, the result is 2. Rounding number 1.299996121 in the fifth decimal place will result in number 1.3.
Input Specification:
The first line of the input contains two integers *n* and *t* (1<=≤<=*n*<=≤<=200<=000, 1<=≤<=*t*<=≤<=109) — the length of Efim's grade and the number of seconds till the end of the break respectively.
The second line contains the grade itself. It's guaranteed that the grade is a positive number, containing at least one digit after the decimal points, and it's representation doesn't finish with 0.
Output Specification:
Print the maximum grade that Efim can get in *t* seconds. Do not print trailing zeroes.
Demo Input:
['6 1\n10.245\n', '6 2\n10.245\n', '3 100\n9.2\n']
Demo Output:
['10.25\n', '10.3\n', '9.2\n']
Note:
In the first two samples Efim initially has grade 10.245.
During the first second Efim can obtain grade 10.25, and then 10.3 during the next second. Note, that the answer 10.30 will be considered incorrect.
In the third sample the optimal strategy is to not perform any rounding at all. | ```python
a = input()
a = a.split()
n = int(a[0])
t = int(a[1])
grade = float(input())
for j in range(t):
l = []
for i in range(100):
l += [round(grade, i)]
grade = max(l)
print(grade)
``` | 0 | |
616 | A | Comparing Two Long Integers | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input(). | The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits. | Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=". | [
"9\n10\n",
"11\n10\n",
"00012345\n12345\n",
"0123\n9\n",
"0123\n111\n"
] | [
"<\n",
">\n",
"=\n",
">\n",
">\n"
] | none | 0 | [
{
"input": "9\n10",
"output": "<"
},
{
"input": "11\n10",
"output": ">"
},
{
"input": "00012345\n12345",
"output": "="
},
{
"input": "0123\n9",
"output": ">"
},
{
"input": "0123\n111",
"output": ">"
},
{
"input": "9\n9",
"output": "="
},
{
... | 1,686,032,021 | 1,121 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 13 | 2,000 | 183,398,400 | import sys
a = list(input())
b = list(input())
idx1 = 0
idx2 = 0
for i in range(len(a)-1):
if a[i] == '0':
idx1+=1
else:
break
for i in range(len(b)-1):
if b[i] == '0':
idx2 += 1
else:
break
if len(a) - idx1 != len(b)- idx2:
if len(a)- idx1 > len(b) - idx2:
print(">")
else:
print("<")
sys.exit(0)
numa = int("".join(a[idx1:]))
numb = int("".join(b[idx2:]))
if numa == numb:
print("=")
elif numa > numb:
print(">")
else:
print("<")
| Title: Comparing Two Long Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input().
Input Specification:
The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits.
Output Specification:
Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=".
Demo Input:
['9\n10\n', '11\n10\n', '00012345\n12345\n', '0123\n9\n', '0123\n111\n']
Demo Output:
['<\n', '>\n', '=\n', '>\n', '>\n']
Note:
none | ```python
import sys
a = list(input())
b = list(input())
idx1 = 0
idx2 = 0
for i in range(len(a)-1):
if a[i] == '0':
idx1+=1
else:
break
for i in range(len(b)-1):
if b[i] == '0':
idx2 += 1
else:
break
if len(a) - idx1 != len(b)- idx2:
if len(a)- idx1 > len(b) - idx2:
print(">")
else:
print("<")
sys.exit(0)
numa = int("".join(a[idx1:]))
numb = int("".join(b[idx2:]))
if numa == numb:
print("=")
elif numa > numb:
print(">")
else:
print("<")
``` | 0 | |
224 | A | Parallelepiped | PROGRAMMING | 1,100 | [
"brute force",
"geometry",
"math"
] | null | null | You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped. | The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement. | Print a single number — the sum of all edges of the parallelepiped. | [
"1 1 1\n",
"4 6 6\n"
] | [
"12\n",
"28\n"
] | In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3. | 500 | [
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{... | 1,580,045,885 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 218 | 512,000 | x,y,z = map(int, input().split())
fxy = []
fz = []
mini = min(x,y)
for i in range(1,mini+1):
if(x%i==0 and y%i==0):
fxy.append(i)
for i in range(1,z+1):
if(z%i==0):
fz.append(i)
for i in fxy:
if(x/i in fz and y/i in fz):
print(int(4*(x/i) + 4*(y/i) + 4*(i)))
break
| Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number — the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3. | ```python
x,y,z = map(int, input().split())
fxy = []
fz = []
mini = min(x,y)
for i in range(1,mini+1):
if(x%i==0 and y%i==0):
fxy.append(i)
for i in range(1,z+1):
if(z%i==0):
fz.append(i)
for i in fxy:
if(x/i in fz and y/i in fz):
print(int(4*(x/i) + 4*(y/i) + 4*(i)))
break
``` | 0 | |
729 | A | Interview with Oleg | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking! | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters. | Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences. | [
"7\naogogob\n",
"13\nogogmgogogogo\n",
"9\nogoogoogo\n"
] | [
"a***b\n",
"***gmg***\n",
"*********\n"
] | The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | 500 | [
{
"input": "7\naogogob",
"output": "a***b"
},
{
"input": "13\nogogmgogogogo",
"output": "***gmg***"
},
{
"input": "9\nogoogoogo",
"output": "*********"
},
{
"input": "32\nabcdefogoghijklmnogoopqrstuvwxyz",
"output": "abcdef***ghijklmn***opqrstuvwxyz"
},
{
"input":... | 1,479,633,942 | 1,242 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 30 | 0 | n=int(input())
s=input()
cut='ogo'+'go'*49
while True:
if cut!='ogo':
s=s.replace(cut,'***')
cut=str(cut[0:-2])
print(cut)
else:
s=s.replace(cut,'***')
break
print(s)
| Title: Interview with Oleg
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
Output Specification:
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Demo Input:
['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n']
Demo Output:
['a***b\n', '***gmg***\n', '*********\n']
Note:
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | ```python
n=int(input())
s=input()
cut='ogo'+'go'*49
while True:
if cut!='ogo':
s=s.replace(cut,'***')
cut=str(cut[0:-2])
print(cut)
else:
s=s.replace(cut,'***')
break
print(s)
``` | 0 | |
129 | A | Cookies | PROGRAMMING | 900 | [
"implementation"
] | null | null | Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. | Print in the only line the only number — the sought number of ways. If there are no such ways print 0. | [
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] | [
"1\n",
"8\n",
"1\n"
] | In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | 500 | [
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"o... | 1,617,709,760 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 124 | 0 | cases = int(input())
while cases:
cases -= 1
arr = list(map(int, input().split()))
ans = 0
total = sum(arr)
for n in arr:
if (total - n) % 2 == 0:
ans += 1
print(ans)
| Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | ```python
cases = int(input())
while cases:
cases -= 1
arr = list(map(int, input().split()))
ans = 0
total = sum(arr)
for n in arr:
if (total - n) % 2 == 0:
ans += 1
print(ans)
``` | -1 | |
676 | C | Vasya and String | PROGRAMMING | 1,500 | [
"binary search",
"dp",
"strings",
"two pointers"
] | null | null | High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve? | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only. | Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters. | [
"4 2\nabba\n",
"8 1\naabaabaa\n"
] | [
"4\n",
"5\n"
] | In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | 1,500 | [
{
"input": "4 2\nabba",
"output": "4"
},
{
"input": "8 1\naabaabaa",
"output": "5"
},
{
"input": "1 0\na",
"output": "1"
},
{
"input": "1 1\nb",
"output": "1"
},
{
"input": "1 0\nb",
"output": "1"
},
{
"input": "1 1\na",
"output": "1"
},
{
... | 1,612,643,671 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 1 | 1,000 | 307,200 | # collaborated with Rudransh
def retMaxConsecutive(string, n, k, choice):
if string.count(choice) <= k:
return n
maxi = -1
index = 0
count = 0
for i in range(n):
if string[i] == choice:
count += 1
while count > k:
if string[index] == choice:
index += 1
count -= 1
maxi = max(maxi, i - index + 1)
return (maxi)
strLen, maxChange = input().split(" ")
string = input()
strLen = int(strLen)
maxChange = int(maxChange)
print(max(retMaxConsecutive(string, strLen, maxChange, "a"), retMaxConsecutive(string, strLen, maxChange, "b")))
| Title: Vasya and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
Output Specification:
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
Demo Input:
['4 2\nabba\n', '8 1\naabaabaa\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | ```python
# collaborated with Rudransh
def retMaxConsecutive(string, n, k, choice):
if string.count(choice) <= k:
return n
maxi = -1
index = 0
count = 0
for i in range(n):
if string[i] == choice:
count += 1
while count > k:
if string[index] == choice:
index += 1
count -= 1
maxi = max(maxi, i - index + 1)
return (maxi)
strLen, maxChange = input().split(" ")
string = input()
strLen = int(strLen)
maxChange = int(maxChange)
print(max(retMaxConsecutive(string, strLen, maxChange, "a"), retMaxConsecutive(string, strLen, maxChange, "b")))
``` | 0 | |
689 | A | Mike and Cellphone | PROGRAMMING | 1,400 | [
"brute force",
"constructive algorithms",
"implementation"
] | null | null | While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only ten digital equal-sized keys, located in the following way:
Together with his old phone, he lost all his contacts and now he can only remember the way his fingers moved when he put some number in. One can formally consider finger movements as a sequence of vectors connecting centers of keys pressed consecutively to put in a number. For example, the finger movements for number "586" are the same as finger movements for number "253":
Mike has already put in a number by his "finger memory" and started calling it, so he is now worrying, can he be sure that he is calling the correct number? In other words, is there any other number, that has the same finger movements? | The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=9) — the number of digits in the phone number that Mike put in.
The second line contains the string consisting of *n* digits (characters from '0' to '9') representing the number that Mike put in. | If there is no other phone number with the same finger movements and Mike can be sure he is calling the correct number, print "YES" (without quotes) in the only line.
Otherwise print "NO" (without quotes) in the first line. | [
"3\n586\n",
"2\n09\n",
"9\n123456789\n",
"3\n911\n"
] | [
"NO\n",
"NO\n",
"YES\n",
"YES\n"
] | You can find the picture clarifying the first sample case in the statement above. | 500 | [
{
"input": "3\n586",
"output": "NO"
},
{
"input": "2\n09",
"output": "NO"
},
{
"input": "9\n123456789",
"output": "YES"
},
{
"input": "3\n911",
"output": "YES"
},
{
"input": "3\n089",
"output": "NO"
},
{
"input": "3\n159",
"output": "YES"
},
{
... | 1,467,823,745 | 845 | Python 3 | CHALLENGED | CHALLENGES | 10 | 62 | 0 | import sys
def main():
n = int(sys.stdin.readline())
s = sys.stdin.readline().rstrip()
m = {}
for i in range(0,10):
m[i] = False
for i in range(n):
m[int(s[i])] = True
if m[0]:
if m[1] or m[2] or m[3]:
print("YES")
return
print("NO")
return
if (m[1] or m[2] or m[3]) and (m[1] or m[4] or m[7]) and (m[3] or m[6] or m[9]) and (m[7] or m[8] or m[9]):
print("YES")
return
print("NO")
main() | Title: Mike and Cellphone
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only ten digital equal-sized keys, located in the following way:
Together with his old phone, he lost all his contacts and now he can only remember the way his fingers moved when he put some number in. One can formally consider finger movements as a sequence of vectors connecting centers of keys pressed consecutively to put in a number. For example, the finger movements for number "586" are the same as finger movements for number "253":
Mike has already put in a number by his "finger memory" and started calling it, so he is now worrying, can he be sure that he is calling the correct number? In other words, is there any other number, that has the same finger movements?
Input Specification:
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=9) — the number of digits in the phone number that Mike put in.
The second line contains the string consisting of *n* digits (characters from '0' to '9') representing the number that Mike put in.
Output Specification:
If there is no other phone number with the same finger movements and Mike can be sure he is calling the correct number, print "YES" (without quotes) in the only line.
Otherwise print "NO" (without quotes) in the first line.
Demo Input:
['3\n586\n', '2\n09\n', '9\n123456789\n', '3\n911\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n', 'YES\n']
Note:
You can find the picture clarifying the first sample case in the statement above. | ```python
import sys
def main():
n = int(sys.stdin.readline())
s = sys.stdin.readline().rstrip()
m = {}
for i in range(0,10):
m[i] = False
for i in range(n):
m[int(s[i])] = True
if m[0]:
if m[1] or m[2] or m[3]:
print("YES")
return
print("NO")
return
if (m[1] or m[2] or m[3]) and (m[1] or m[4] or m[7]) and (m[3] or m[6] or m[9]) and (m[7] or m[8] or m[9]):
print("YES")
return
print("NO")
main()
``` | -1 | |
556 | A | Case of the Zeros and Ones | PROGRAMMING | 900 | [
"greedy"
] | null | null | Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones. | Output the minimum length of the string that may remain after applying the described operations several times. | [
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] | [
"0\n",
"1\n",
"6\n"
] | In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 250 | [
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input"... | 1,663,035,565 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 49 | 62 | 1,945,600 | # Case of the Zeros and Ones
n = int(input())
s = input()
counter_0 = 0
counter_1 = 0
for i in s:
if i == "1":
counter_1 += 1
elif i == "0":
counter_0 += 1
min_1_0 = min(counter_0,counter_1)
print(n-(min_1_0*2)) | Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
# Case of the Zeros and Ones
n = int(input())
s = input()
counter_0 = 0
counter_1 = 0
for i in s:
if i == "1":
counter_1 += 1
elif i == "0":
counter_0 += 1
min_1_0 = min(counter_0,counter_1)
print(n-(min_1_0*2))
``` | 3 | |
429 | B | Working out | PROGRAMMING | 1,600 | [
"dp"
] | null | null | Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix *a* with *n* lines and *m* columns. Let number *a*[*i*][*j*] represents the calories burned by performing workout at the cell of gym in the *i*-th line and the *j*-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout *a*[*n*][*m*]. After finishing workout *a*[*i*][*j*], he can go to workout *a*[*i*<=+<=1][*j*] or *a*[*i*][*j*<=+<=1]. Similarly, Iahubina starts with workout *a*[*n*][1] and she needs to finish with workout *a*[1][*m*]. After finishing workout from cell *a*[*i*][*j*], she goes to either *a*[*i*][*j*<=+<=1] or *a*[*i*<=-<=1][*j*].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs. | The first line of the input contains two integers *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=1000). Each of the next *n* lines contains *m* integers: *j*-th number from *i*-th line denotes element *a*[*i*][*j*] (0<=≤<=*a*[*i*][*j*]<=≤<=105). | The output contains a single number — the maximum total gain possible. | [
"3 3\n100 100 100\n100 1 100\n100 100 100\n"
] | [
"800"
] | Iahub will choose exercises *a*[1][1] → *a*[1][2] → *a*[2][2] → *a*[3][2] → *a*[3][3]. Iahubina will choose exercises *a*[3][1] → *a*[2][1] → *a*[2][2] → *a*[2][3] → *a*[1][3]. | 1,000 | [
{
"input": "3 3\n100 100 100\n100 1 100\n100 100 100",
"output": "800"
},
{
"input": "4 5\n87882 40786 3691 85313 46694\n28884 16067 3242 97367 78518\n4250 35501 9780 14435 19004\n64673 65438 56977 64495 27280",
"output": "747898"
},
{
"input": "3 3\n3 1 2\n3 2 0\n2 3 2",
"output": "... | 1,623,581,675 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 93 | 0 | import sys
input=sys.stdin.readline
n,m=map(int,input().split())
a=[list(map(int,input().split())) for i in range(n)]
p1=[[0]*m for i in range(n)]
p1[0][0]=a[0][0]
for j in range(1,m):
p1[0][j]=p1[0][j-1]+a[0][j]
for i in range(1,n):
p1[i][0]=p1[i-1][0]+a[i][0]
for i in range(1,n):
for j in range(1,m):
p1[i][j]=max(p1[i-1][j],p1[i][j-1])+a[i][j]
p2=[[0]*m for i in range(n)]
p2[n-1][m-1]=a[n-1][m-1]
for i in range(n-1)[::-1]:
p2[i][m-1]=p2[i+1][m-1]+a[i][m-1]
for j in range(m-1)[::-1]:
p2[n-1][j]=p2[n-1][j+1]+a[n-1][j]
for i in range(n-1)[::-1]:
for j in range(m-1)[::-1]:
p2[i][j]=max(p2[i+1][j],p2[i][j+1])+a[i][j]
q1=[[0]*m for i in range(n)]
q1[n-1][0]=a[n-1][0]
for j in range(1,m):
q1[n-1][j]=q1[n-1][j-1]+a[n-1][j]
for i in range(n-1)[::-1]:
q1[i][0]=q1[i+1][0]+a[i][0]
for i in range(n-1)[::-1]:
for j in range(1,m):
q1[i][j]=max(q1[i+1][j],q1[i][j-1])+a[i][j]
print(q1)
q2=[[0]*m for i in range(n)]
q2[0][m-1]=a[0][m-1]
for j in range(m-1)[::-1]:
q2[0][j]=q2[0][j+1]+a[0][j]
for i in range(n-1)[::-1]:
q2[i][m-1]=q2[i+1][m-1]+a[i][m-1]
for i in range(1,n):
for j in range(m-1)[::-1]:
q2[i][j]=max(q2[i-1][j],q2[i][j+1])
ans=-1
for i in range(1,n-1):
for j in range(1,m-1):
ans=max(ans,p1[i-1][j]+p2[i+1][j]+q1[i][j-1]+q2[i][j+1])
ans=max(ans,p1[i][j-1]+p2[i][j+1]+q1[i+1][j]+q2[i-1][j])
print(ans) | Title: Working out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix *a* with *n* lines and *m* columns. Let number *a*[*i*][*j*] represents the calories burned by performing workout at the cell of gym in the *i*-th line and the *j*-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout *a*[*n*][*m*]. After finishing workout *a*[*i*][*j*], he can go to workout *a*[*i*<=+<=1][*j*] or *a*[*i*][*j*<=+<=1]. Similarly, Iahubina starts with workout *a*[*n*][1] and she needs to finish with workout *a*[1][*m*]. After finishing workout from cell *a*[*i*][*j*], she goes to either *a*[*i*][*j*<=+<=1] or *a*[*i*<=-<=1][*j*].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input Specification:
The first line of the input contains two integers *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=1000). Each of the next *n* lines contains *m* integers: *j*-th number from *i*-th line denotes element *a*[*i*][*j*] (0<=≤<=*a*[*i*][*j*]<=≤<=105).
Output Specification:
The output contains a single number — the maximum total gain possible.
Demo Input:
['3 3\n100 100 100\n100 1 100\n100 100 100\n']
Demo Output:
['800']
Note:
Iahub will choose exercises *a*[1][1] → *a*[1][2] → *a*[2][2] → *a*[3][2] → *a*[3][3]. Iahubina will choose exercises *a*[3][1] → *a*[2][1] → *a*[2][2] → *a*[2][3] → *a*[1][3]. | ```python
import sys
input=sys.stdin.readline
n,m=map(int,input().split())
a=[list(map(int,input().split())) for i in range(n)]
p1=[[0]*m for i in range(n)]
p1[0][0]=a[0][0]
for j in range(1,m):
p1[0][j]=p1[0][j-1]+a[0][j]
for i in range(1,n):
p1[i][0]=p1[i-1][0]+a[i][0]
for i in range(1,n):
for j in range(1,m):
p1[i][j]=max(p1[i-1][j],p1[i][j-1])+a[i][j]
p2=[[0]*m for i in range(n)]
p2[n-1][m-1]=a[n-1][m-1]
for i in range(n-1)[::-1]:
p2[i][m-1]=p2[i+1][m-1]+a[i][m-1]
for j in range(m-1)[::-1]:
p2[n-1][j]=p2[n-1][j+1]+a[n-1][j]
for i in range(n-1)[::-1]:
for j in range(m-1)[::-1]:
p2[i][j]=max(p2[i+1][j],p2[i][j+1])+a[i][j]
q1=[[0]*m for i in range(n)]
q1[n-1][0]=a[n-1][0]
for j in range(1,m):
q1[n-1][j]=q1[n-1][j-1]+a[n-1][j]
for i in range(n-1)[::-1]:
q1[i][0]=q1[i+1][0]+a[i][0]
for i in range(n-1)[::-1]:
for j in range(1,m):
q1[i][j]=max(q1[i+1][j],q1[i][j-1])+a[i][j]
print(q1)
q2=[[0]*m for i in range(n)]
q2[0][m-1]=a[0][m-1]
for j in range(m-1)[::-1]:
q2[0][j]=q2[0][j+1]+a[0][j]
for i in range(n-1)[::-1]:
q2[i][m-1]=q2[i+1][m-1]+a[i][m-1]
for i in range(1,n):
for j in range(m-1)[::-1]:
q2[i][j]=max(q2[i-1][j],q2[i][j+1])
ans=-1
for i in range(1,n-1):
for j in range(1,m-1):
ans=max(ans,p1[i-1][j]+p2[i+1][j]+q1[i][j-1]+q2[i][j+1])
ans=max(ans,p1[i][j-1]+p2[i][j+1]+q1[i+1][j]+q2[i-1][j])
print(ans)
``` | 0 | |
246 | B | Increase and Decrease | PROGRAMMING | 1,300 | [
"greedy",
"math"
] | null | null | Polycarpus has an array, consisting of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
- he chooses two elements of the array *a**i*, *a**j* (*i*<=≠<=*j*); - he simultaneously increases number *a**i* by 1 and decreases number *a**j* by 1, that is, executes *a**i*<==<=*a**i*<=+<=1 and *a**j*<==<=*a**j*<=-<=1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the array size. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=104) — the original array. | Print a single integer — the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation. | [
"2\n2 1\n",
"3\n1 4 1\n"
] | [
"1\n",
"3\n"
] | none | 1,000 | [
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 4 1",
"output": "3"
},
{
"input": "4\n2 -7 -2 -6",
"output": "3"
},
{
"input": "4\n2 0 -2 -1",
"output": "3"
},
{
"input": "6\n-1 1 0 0 -1 -1",
"output": "5"
},
{
"input": "5\n0 0 0 0 0",
"outp... | 1,669,405,783 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 30 | 154 | 7,577,600 | n=int(input())
nums=list(map(int,input().split()))
ans=0
S=abs(sum(nums))
if S%n==0:
print(n)
else:
print(n-1)
| Title: Increase and Decrease
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has an array, consisting of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
- he chooses two elements of the array *a**i*, *a**j* (*i*<=≠<=*j*); - he simultaneously increases number *a**i* by 1 and decreases number *a**j* by 1, that is, executes *a**i*<==<=*a**i*<=+<=1 and *a**j*<==<=*a**j*<=-<=1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the array size. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=104) — the original array.
Output Specification:
Print a single integer — the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Demo Input:
['2\n2 1\n', '3\n1 4 1\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
n=int(input())
nums=list(map(int,input().split()))
ans=0
S=abs(sum(nums))
if S%n==0:
print(n)
else:
print(n-1)
``` | 3 | |
960 | A | Check the string | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes). | The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'. | Print "YES" or "NO", according to the condition. | [
"aaabccc\n",
"bbacc\n",
"aabc\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | 500 | [
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"in... | 1,622,907,411 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 109 | 0 | s = input()
l = len(s)
count = 0
count2 = 0
count3 = 0
count4 = 0
odd = ["ba", "ca", "ac", "cb"]
lenght = len(odd)
if(s.islower()):
for j in range(lenght):
if s.find(odd[j])>0:
count4+=1
if s[0]=="a" and s[l-1]=="c" and count4==0:
count+=1
for i in range(1,l):
if s[i]=="a":
count+=1
if s[i]=="b":
count2+=1
if s[i]=="c":
count3+=1
if count >= 1 and count2 >=1 and (count3==count or count3==count2 or (count3==count and count3==count3)):
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
| Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | ```python
s = input()
l = len(s)
count = 0
count2 = 0
count3 = 0
count4 = 0
odd = ["ba", "ca", "ac", "cb"]
lenght = len(odd)
if(s.islower()):
for j in range(lenght):
if s.find(odd[j])>0:
count4+=1
if s[0]=="a" and s[l-1]=="c" and count4==0:
count+=1
for i in range(1,l):
if s[i]=="a":
count+=1
if s[i]=="b":
count2+=1
if s[i]=="c":
count3+=1
if count >= 1 and count2 >=1 and (count3==count or count3==count2 or (count3==count and count3==count3)):
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
``` | 3 | |
911 | C | Three Garlands | PROGRAMMING | 1,400 | [
"brute force",
"constructive algorithms"
] | null | null | Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if *i*-th garland is switched on during *x*-th second, then it is lit only during seconds *x*, *x*<=+<=*k**i*, *x*<=+<=2*k**i*, *x*<=+<=3*k**i* and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers *x*1, *x*2 and *x*3 (not necessarily distinct) so that he will switch on the first garland during *x*1-th second, the second one — during *x*2-th second, and the third one — during *x*3-th second, respectively, and during each second starting from *max*(*x*1,<=*x*2,<=*x*3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this! | The first line contains three integers *k*1, *k*2 and *k*3 (1<=≤<=*k**i*<=≤<=1500) — time intervals of the garlands. | If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO. | [
"2 2 3\n",
"4 2 3\n"
] | [
"YES\n",
"NO\n"
] | In the first example Mishka can choose *x*<sub class="lower-index">1</sub> = 1, *x*<sub class="lower-index">2</sub> = 2, *x*<sub class="lower-index">3</sub> = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what *x*<sub class="lower-index">3</sub> is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit. | 0 | [
{
"input": "2 2 3",
"output": "YES"
},
{
"input": "4 2 3",
"output": "NO"
},
{
"input": "1499 1498 1500",
"output": "NO"
},
{
"input": "1500 1500 1500",
"output": "NO"
},
{
"input": "100 4 1",
"output": "YES"
},
{
"input": "4 2 4",
"output": "YES"
... | 1,658,234,053 | 2,147,483,647 | Python 3 | OK | TESTS | 67 | 46 | 0 | # https://algoprog.ru/material/pc911pC
k1, k2, k3 = map(int, input().split())
k1, k2, k3 = min(k1, k2, k3), k1 + k2 + k3 - min(k1, k2, k3) - max(k1, k2, k3), max(k1, k2, k3)
if k1 * k2 + k1 * k3 + k2 * k3 < k1 * k2 * k3:
print('NO')
elif k1 == 2 and k2 == 2 or k1 == 1:
print('YES')
else:
st1 = 0
st2 = 1
for st3 in range(2, 10):
for i in range(1, 2000):
if (i+st1) % k1 != 0:
if (i+st2) % k2 != 0:
if (i + st3) % k3 != 0:
break
else:
print('YES')
break
else:
print('NO') | Title: Three Garlands
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if *i*-th garland is switched on during *x*-th second, then it is lit only during seconds *x*, *x*<=+<=*k**i*, *x*<=+<=2*k**i*, *x*<=+<=3*k**i* and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers *x*1, *x*2 and *x*3 (not necessarily distinct) so that he will switch on the first garland during *x*1-th second, the second one — during *x*2-th second, and the third one — during *x*3-th second, respectively, and during each second starting from *max*(*x*1,<=*x*2,<=*x*3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this!
Input Specification:
The first line contains three integers *k*1, *k*2 and *k*3 (1<=≤<=*k**i*<=≤<=1500) — time intervals of the garlands.
Output Specification:
If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO.
Demo Input:
['2 2 3\n', '4 2 3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Mishka can choose *x*<sub class="lower-index">1</sub> = 1, *x*<sub class="lower-index">2</sub> = 2, *x*<sub class="lower-index">3</sub> = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what *x*<sub class="lower-index">3</sub> is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit. | ```python
# https://algoprog.ru/material/pc911pC
k1, k2, k3 = map(int, input().split())
k1, k2, k3 = min(k1, k2, k3), k1 + k2 + k3 - min(k1, k2, k3) - max(k1, k2, k3), max(k1, k2, k3)
if k1 * k2 + k1 * k3 + k2 * k3 < k1 * k2 * k3:
print('NO')
elif k1 == 2 and k2 == 2 or k1 == 1:
print('YES')
else:
st1 = 0
st2 = 1
for st3 in range(2, 10):
for i in range(1, 2000):
if (i+st1) % k1 != 0:
if (i+st2) % k2 != 0:
if (i + st3) % k3 != 0:
break
else:
print('YES')
break
else:
print('NO')
``` | 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... | 1,699,695,989 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 62 | 0 | n = int(input())
p = list(map(int, input().split()))
i = 0
sum = 0
while i < n:
sum += p[i]
i += 1
print(sum/n)
| 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
n = int(input())
p = list(map(int, input().split()))
i = 0
sum = 0
while i < n:
sum += p[i]
i += 1
print(sum/n)
``` | 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"
},
{
"... | 1,689,741,314 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 46 | 0 | t=int(input())
count=0
for i in range(t):
ls=list(map(int,input().split()))
k=ls[0]
j=ls[-1]
if abs(k-j)>=2:
count+=1
print(count) | 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
t=int(input())
count=0
for i in range(t):
ls=list(map(int,input().split()))
k=ls[0]
j=ls[-1]
if abs(k-j)>=2:
count+=1
print(count)
``` | 3 |
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