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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
869 | A | The Artful Expedient | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game. | The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*. | Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization. | [
"3\n1 2 3\n4 5 6\n",
"5\n2 4 6 8 10\n9 7 5 3 1\n"
] | [
"Karen\n",
"Karen\n"
] | In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again. | 500 | [
{
"input": "3\n1 2 3\n4 5 6",
"output": "Karen"
},
{
"input": "5\n2 4 6 8 10\n9 7 5 3 1",
"output": "Karen"
},
{
"input": "1\n1\n2000000",
"output": "Karen"
},
{
"input": "2\n97153 2000000\n1999998 254",
"output": "Karen"
},
{
"input": "15\n31 30 29 28 27 26 25 24... | 1,507,668,679 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | KOYOMI = 'Koyomi'
KAREN = 'Karen'
def main():
input()
first_str = input().split()
first = list(map(int, first_str))
second_str = input().split()
second = list(map(int, second_str))
print(game(first, second))
def game(first, second):
numbers_dict = {}
for lst in [first, second]:
for i in lst:
numbers_dict[i] = 0
for i in first:
for j in second:
result = i ^ j
if numbers_dict.get(result) is not None:
numbers_dict[result] += 1
print(numbers_dict)
return KOYOMI if sum(numbers_dict.values()) % 2 else KAREN
main()
| Title: The Artful Expedient
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.
Input Specification:
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*.
Output Specification:
Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization.
Demo Input:
['3\n1 2 3\n4 5 6\n', '5\n2 4 6 8 10\n9 7 5 3 1\n']
Demo Output:
['Karen\n', 'Karen\n']
Note:
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again. | ```python
KOYOMI = 'Koyomi'
KAREN = 'Karen'
def main():
input()
first_str = input().split()
first = list(map(int, first_str))
second_str = input().split()
second = list(map(int, second_str))
print(game(first, second))
def game(first, second):
numbers_dict = {}
for lst in [first, second]:
for i in lst:
numbers_dict[i] = 0
for i in first:
for j in second:
result = i ^ j
if numbers_dict.get(result) is not None:
numbers_dict[result] += 1
print(numbers_dict)
return KOYOMI if sum(numbers_dict.values()) % 2 else KAREN
main()
``` | 0 | |
977 | A | Wrong Subtraction | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number. | The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly. | Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number. | [
"512 4\n",
"1000000000 9\n"
] | [
"50\n",
"1\n"
] | The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | 0 | [
{
"input": "512 4",
"output": "50"
},
{
"input": "1000000000 9",
"output": "1"
},
{
"input": "131203 11",
"output": "12"
},
{
"input": "999999999 50",
"output": "9999"
},
{
"input": "999999999 49",
"output": "99990"
},
{
"input": "131203 9",
"outpu... | 1,696,270,530 | 2,147,483,647 | Python 3 | OK | TESTS | 11 | 46 | 0 | n,k=map(int,input().split())
for i in range (k) :
if str(n)[-1]=='0' :
n//=10
else :
n-=1
print(n) | Title: Wrong Subtraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
Input Specification:
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly.
Output Specification:
Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
Demo Input:
['512 4\n', '1000000000 9\n']
Demo Output:
['50\n', '1\n']
Note:
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | ```python
n,k=map(int,input().split())
for i in range (k) :
if str(n)[-1]=='0' :
n//=10
else :
n-=1
print(n)
``` | 3 | |
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "... | 1,678,032,927 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 0 | def wub(a):
return x.replace('WUB', ' ')
b = input()
ans = wub(a)
print(ans) | Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
def wub(a):
return x.replace('WUB', ' ')
b = input()
ans = wub(a)
print(ans)
``` | -1 | |
621 | A | Wet Shark and Odd and Even | PROGRAMMING | 900 | [
"implementation"
] | null | null | Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. | The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. | Print the maximum possible even sum that can be obtained if we use some of the given integers. | [
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] | [
"6",
"3999999996"
] | In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | 500 | [
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\... | 1,670,689,444 | 2,147,483,647 | Python 3 | OK | TESTS | 87 | 140 | 7,782,400 | n = int(input())
arr = [int(i) for i in input().split()]
mn = 10000000000
for i in range(n):
if arr[i] % 2 != 0:
mn = min(arr[i], mn)
sum_arr = sum(arr)
if sum_arr % 2 != 0:
sum_arr -= mn
print(sum_arr)
| Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | ```python
n = int(input())
arr = [int(i) for i in input().split()]
mn = 10000000000
for i in range(n):
if arr[i] % 2 != 0:
mn = min(arr[i], mn)
sum_arr = sum(arr)
if sum_arr % 2 != 0:
sum_arr -= mn
print(sum_arr)
``` | 3 | |
383 | D | Antimatter | PROGRAMMING | 2,300 | [
"dp"
] | null | null | Iahub accidentally discovered a secret lab. He found there *n* devices ordered in a line, numbered from 1 to *n* from left to right. Each device *i* (1<=≤<=*i*<=≤<=*n*) can create either *a**i* units of matter or *a**i* units of antimatter.
Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).
You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000).
The sum *a*1<=+<=*a*2<=+<=...<=+<=*a**n* will be less than or equal to 10000. | Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (109<=+<=7). | [
"4\n1 1 1 1\n"
] | [
"12\n"
] | The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "*i*+" means that the *i*-th element produces matter, and "*i*-" means that the *i*-th element produces antimatter. | 2,000 | [
{
"input": "4\n1 1 1 1",
"output": "12"
},
{
"input": "10\n16 9 9 11 10 12 9 6 10 8",
"output": "86"
},
{
"input": "50\n2 1 5 2 1 3 1 2 3 2 1 1 5 2 2 2 3 2 1 2 2 2 3 3 1 3 1 1 2 2 2 2 1 2 3 1 2 4 1 1 1 3 2 1 1 1 3 2 1 3",
"output": "115119382"
},
{
"input": "100\n8 3 3 7 3 6 ... | 1,657,242,790 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 1,000 | 13,619,200 | from sys import stdin, stdout
MOD = 10**9 + 7
def solve(A, n):
assert n == len(A)
# calculate prefix sum
ps = prefix_sum(A)
# calculate suffix sum
ss = suffix_sum(A)
# total sum
ST = ps[-1]
# count ways of bipartite A with equal sum
count = 0
# iterate over each subarray
for sub, i, j in subarrays(A):
# take sum of subarray
M = ST - ps[i] - ss[j + 1]
# if sum of subarray is even then calculate dp
if not M % 2:
count += calculate_photos(sub, M)
return count % MOD
def prefix_sum(A):
# this method calculate the prefix sum of the array, ps[i] is the sum of the elements in [0, i)
ps = [0] * (len(A) + 1)
for i in range(1, len(A) + 1):
ps[i] = ps[i - 1] + A[i-1]
return ps
def suffix_sum(A):
# this method calculate the suffix sum of the array, ss[i] is the sum of the elements in (i, n-1]
ss = [0] * (len(A) + 1)
for i in range(n-1,-1,-1):
ss[i] = ss[i + 1] + A[i]
return ss
def subarrays(A):
# this method calculate all the subarray of lenght >=2
for i in range(n):
for j in range(i + 1, n):
yield A[i:j + 1], i, j
def calculate_photos(A, S):
# this method calculate the number of ways of bipartite A with equal sum S
dp = [[0] * (S + 1) for _ in range(len(A) + 1)]
# how many photo can take end in elemnt i with sum 0, just 1 with empty subset
for i in range(len(A)):
dp[i][0] = 1
for i in range(1, len(A) + 1):
for j in range(1, S + 1):
# dont take element i in subset
dp[i][j] = dp[i - 1][j]
# take element i in subset if it is possible
if A[i - 1] <= j:
dp[i][j] += dp[i-1][j - A[i-1]]
# then return dp[n][S/2]
return dp[len(A)][S//2]
if __name__ == "__main__":
n = int(stdin.readline().split()[0])
A = [*map(int, stdin.readline().split())]
stdout.write(str(solve(A, n))+"\n") | Title: Antimatter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub accidentally discovered a secret lab. He found there *n* devices ordered in a line, numbered from 1 to *n* from left to right. Each device *i* (1<=≤<=*i*<=≤<=*n*) can create either *a**i* units of matter or *a**i* units of antimatter.
Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).
You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000).
The sum *a*1<=+<=*a*2<=+<=...<=+<=*a**n* will be less than or equal to 10000.
Output Specification:
Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (109<=+<=7).
Demo Input:
['4\n1 1 1 1\n']
Demo Output:
['12\n']
Note:
The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "*i*+" means that the *i*-th element produces matter, and "*i*-" means that the *i*-th element produces antimatter. | ```python
from sys import stdin, stdout
MOD = 10**9 + 7
def solve(A, n):
assert n == len(A)
# calculate prefix sum
ps = prefix_sum(A)
# calculate suffix sum
ss = suffix_sum(A)
# total sum
ST = ps[-1]
# count ways of bipartite A with equal sum
count = 0
# iterate over each subarray
for sub, i, j in subarrays(A):
# take sum of subarray
M = ST - ps[i] - ss[j + 1]
# if sum of subarray is even then calculate dp
if not M % 2:
count += calculate_photos(sub, M)
return count % MOD
def prefix_sum(A):
# this method calculate the prefix sum of the array, ps[i] is the sum of the elements in [0, i)
ps = [0] * (len(A) + 1)
for i in range(1, len(A) + 1):
ps[i] = ps[i - 1] + A[i-1]
return ps
def suffix_sum(A):
# this method calculate the suffix sum of the array, ss[i] is the sum of the elements in (i, n-1]
ss = [0] * (len(A) + 1)
for i in range(n-1,-1,-1):
ss[i] = ss[i + 1] + A[i]
return ss
def subarrays(A):
# this method calculate all the subarray of lenght >=2
for i in range(n):
for j in range(i + 1, n):
yield A[i:j + 1], i, j
def calculate_photos(A, S):
# this method calculate the number of ways of bipartite A with equal sum S
dp = [[0] * (S + 1) for _ in range(len(A) + 1)]
# how many photo can take end in elemnt i with sum 0, just 1 with empty subset
for i in range(len(A)):
dp[i][0] = 1
for i in range(1, len(A) + 1):
for j in range(1, S + 1):
# dont take element i in subset
dp[i][j] = dp[i - 1][j]
# take element i in subset if it is possible
if A[i - 1] <= j:
dp[i][j] += dp[i-1][j - A[i-1]]
# then return dp[n][S/2]
return dp[len(A)][S//2]
if __name__ == "__main__":
n = int(stdin.readline().split()[0])
A = [*map(int, stdin.readline().split())]
stdout.write(str(solve(A, n))+"\n")
``` | 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... | 1,660,543,460 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 30 | 1,000 | 33,177,600 | from heapq import heappush, heappop
n, m = map(int, input().split())
g = {i: [] for i in range(1, n + 1)}
while m:
a, b, w = map(int, input().split())
g[a].append((w, b))
g[b].append((w, a))
m = m - 1
start = 1
end = n
heap = [(0, start)]
dist = [float('inf') for _ in range(n + 1)]
v = set()
present_c = float('inf')
ans = []
parent = [-1 for _ in range(n + 1)]
while heap:
curr_c, curr = heappop(heap)
if curr not in v:
v.add(curr)
for i_c, i in g[curr]:
if i not in v:
if curr_c + i_c < dist[i]:
parent[i] = curr
dist[i] = curr_c + i_c
heappush(heap, (curr_c + i_c, i))
if parent[end] == -1:
print(-1)
exit()
curr = end
ans = []
while curr != start:
ans.append(curr)
curr = parent[curr]
ans.append(start)
ans.reverse()
print(*ans) | Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none | ```python
from heapq import heappush, heappop
n, m = map(int, input().split())
g = {i: [] for i in range(1, n + 1)}
while m:
a, b, w = map(int, input().split())
g[a].append((w, b))
g[b].append((w, a))
m = m - 1
start = 1
end = n
heap = [(0, start)]
dist = [float('inf') for _ in range(n + 1)]
v = set()
present_c = float('inf')
ans = []
parent = [-1 for _ in range(n + 1)]
while heap:
curr_c, curr = heappop(heap)
if curr not in v:
v.add(curr)
for i_c, i in g[curr]:
if i not in v:
if curr_c + i_c < dist[i]:
parent[i] = curr
dist[i] = curr_c + i_c
heappush(heap, (curr_c + i_c, i))
if parent[end] == -1:
print(-1)
exit()
curr = end
ans = []
while curr != start:
ans.append(curr)
curr = parent[curr]
ans.append(start)
ans.reverse()
print(*ans)
``` | 0 |
702 | B | Powers of Two | PROGRAMMING | 1,500 | [
"brute force",
"data structures",
"implementation",
"math"
] | null | null | You are given *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Find the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2 (i. e. some integer *x* exists so that *a**i*<=+<=*a**j*<==<=2*x*). | The first line contains the single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | Print the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2. | [
"4\n7 3 2 1\n",
"3\n1 1 1\n"
] | [
"2\n",
"3\n"
] | In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).
In the second example all pairs of indexes (*i*, *j*) (where *i* < *j*) include in answer. | 0 | [
{
"input": "4\n7 3 2 1",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "10\n2827343 1373647 96204862 723505 796619138 71550121 799843967 5561265 402690754 446173607",
"output": "2"
},
{
"input": "1... | 1,688,721,628 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 43 | 826 | 98,713,600 | from collections import defaultdict
n = int(input())
ls = list(map(int,input().split()))
ls.sort()
dc = defaultdict(int)
for i in range(n) :
dc[ls[i]] += 1
def check(val) :
while val%2 == 0 :
val //= 2
if val == 1 :
return True
return False
arr = []
for i,j in dc.items() :
arr.append([i,j])
ans = 0
for i in arr :
if check(i[0]) :
ans += (i[1]*(i[1]-1))//2
else :
k = 1
while 2**k < i[0] :
k += 1
while k < 31 :
ans += i[1]*dc[2**k-i[0]]
k += 1
dc[i[0]] = 0
print(ans) | Title: Powers of Two
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Find the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2 (i. e. some integer *x* exists so that *a**i*<=+<=*a**j*<==<=2*x*).
Input Specification:
The first line contains the single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print the number of pairs of indexes *i*,<=*j* (*i*<=<<=*j*) that *a**i*<=+<=*a**j* is a power of 2.
Demo Input:
['4\n7 3 2 1\n', '3\n1 1 1\n']
Demo Output:
['2\n', '3\n']
Note:
In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).
In the second example all pairs of indexes (*i*, *j*) (where *i* < *j*) include in answer. | ```python
from collections import defaultdict
n = int(input())
ls = list(map(int,input().split()))
ls.sort()
dc = defaultdict(int)
for i in range(n) :
dc[ls[i]] += 1
def check(val) :
while val%2 == 0 :
val //= 2
if val == 1 :
return True
return False
arr = []
for i,j in dc.items() :
arr.append([i,j])
ans = 0
for i in arr :
if check(i[0]) :
ans += (i[1]*(i[1]-1))//2
else :
k = 1
while 2**k < i[0] :
k += 1
while k < 31 :
ans += i[1]*dc[2**k-i[0]]
k += 1
dc[i[0]] = 0
print(ans)
``` | 3 | |
702 | A | Maximum Increase | PROGRAMMING | 800 | [
"dp",
"greedy",
"implementation"
] | null | null | You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous. | The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | Print the maximum length of an increasing subarray of the given array. | [
"5\n1 7 2 11 15\n",
"6\n100 100 100 100 100 100\n",
"3\n1 2 3\n"
] | [
"3\n",
"1\n",
"3\n"
] | none | 0 | [
{
"input": "5\n1 7 2 11 15",
"output": "3"
},
{
"input": "6\n100 100 100 100 100 100",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "10\n802030518 598196518 640274071 983359971 71550121 96204862 7... | 1,683,017,630 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 31 | 0 | #Maximum Increase
n = int(input())
arrayList = list(map(int,input().split()))
count = 0
for num in range(len(arrayList)):
if(num == len(arrayList) - 1):
if(arrayList[num-1] >= arrayList[num - 2]):
count+=1
elif(arrayList[num] <= arrayList[num+1]):
count+=1
else:
count = 0
if (count == n) and (arrayList[0] == arrayList[n-1]):
print(1)
else:
print(count)
| Title: Maximum Increase
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
Input Specification:
The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print the maximum length of an increasing subarray of the given array.
Demo Input:
['5\n1 7 2 11 15\n', '6\n100 100 100 100 100 100\n', '3\n1 2 3\n']
Demo Output:
['3\n', '1\n', '3\n']
Note:
none | ```python
#Maximum Increase
n = int(input())
arrayList = list(map(int,input().split()))
count = 0
for num in range(len(arrayList)):
if(num == len(arrayList) - 1):
if(arrayList[num-1] >= arrayList[num - 2]):
count+=1
elif(arrayList[num] <= arrayList[num+1]):
count+=1
else:
count = 0
if (count == n) and (arrayList[0] == arrayList[n-1]):
print(1)
else:
print(count)
``` | -1 | |
99 | A | Help Far Away Kingdom | PROGRAMMING | 800 | [
"strings"
] | A. Help Far Away Kingdom | 2 | 256 | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order? | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data. | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | [
"0.0\n",
"1.49\n",
"1.50\n",
"2.71828182845904523536\n",
"3.14159265358979323846\n",
"12345678901234567890.1\n",
"123456789123456789.999\n"
] | [
"0",
"1",
"2",
"3",
"3",
"12345678901234567890",
"GOTO Vasilisa."
] | none | 500 | [
{
"input": "0.0",
"output": "0"
},
{
"input": "1.49",
"output": "1"
},
{
"input": "1.50",
"output": "2"
},
{
"input": "2.71828182845904523536",
"output": "3"
},
{
"input": "3.14159265358979323846",
"output": "3"
},
{
"input": "12345678901234567890.1",
... | 1,597,079,113 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 134 | 218 | 6,656,000 | a,b=input().split('.')
if a[-1]=='9':
print('GOTO Vasilisa.')
elif float("0."+b)<0.5:
print(a)
else:
print(str(int(a)+1))
| Title: Help Far Away Kingdom
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order?
Input Specification:
The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data.
Output Specification:
If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes).
Demo Input:
['0.0\n', '1.49\n', '1.50\n', '2.71828182845904523536\n', '3.14159265358979323846\n', '12345678901234567890.1\n', '123456789123456789.999\n']
Demo Output:
['0', '1', '2', '3', '3', '12345678901234567890', 'GOTO Vasilisa.']
Note:
none | ```python
a,b=input().split('.')
if a[-1]=='9':
print('GOTO Vasilisa.')
elif float("0."+b)<0.5:
print(a)
else:
print(str(int(a)+1))
``` | 0 |
716 | B | Complete the Word | PROGRAMMING | 1,300 | [
"greedy",
"two pointers"
] | null | null | ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him? | The first and only line of the input contains a single string *s* (1<=≤<=|*s*|<=≤<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember. | If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them. | [
"ABC??FGHIJK???OPQR?TUVWXY?\n",
"WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n",
"??????????????????????????\n",
"AABCDEFGHIJKLMNOPQRSTUVW??M\n"
] | [
"ABCDEFGHIJKLMNOPQRZTUVWXYS",
"-1",
"MNBVCXZLKJHGFDSAQPWOEIRUYT",
"-1"
] | In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is - 1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer. | 1,000 | [
{
"input": "ABC??FGHIJK???OPQR?TUVWXY?",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO",
"output": "-1"
},
{
"input": "??????????????????????????",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "AABCDEFGHIJKLMNO... | 1,635,233,392 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 4,505,600 | x = list(input())
v = ""
y = set(x)
z = list("ABCDEFGHIJKLMNOPQRZTUVWXYS")
if x.count("?") ==0 and len(y) != 26 :
print(-1)
else :
for i in range(len(x)) :
if x[i] == "?" :
for j in z :
if j not in x :
x[i] = j
if len(x) != 26 :
print(-1)
else :
for i in x :
v+= i
print(v)
| Title: Complete the Word
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?
Input Specification:
The first and only line of the input contains a single string *s* (1<=≤<=|*s*|<=≤<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.
Output Specification:
If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them.
Demo Input:
['ABC??FGHIJK???OPQR?TUVWXY?\n', 'WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n', '??????????????????????????\n', 'AABCDEFGHIJKLMNOPQRSTUVW??M\n']
Demo Output:
['ABCDEFGHIJKLMNOPQRZTUVWXYS', '-1', 'MNBVCXZLKJHGFDSAQPWOEIRUYT', '-1']
Note:
In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is - 1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer. | ```python
x = list(input())
v = ""
y = set(x)
z = list("ABCDEFGHIJKLMNOPQRZTUVWXYS")
if x.count("?") ==0 and len(y) != 26 :
print(-1)
else :
for i in range(len(x)) :
if x[i] == "?" :
for j in z :
if j not in x :
x[i] = j
if len(x) != 26 :
print(-1)
else :
for i in x :
v+= i
print(v)
``` | 0 | |
599 | A | Patrick and Shopping | PROGRAMMING | 800 | [
"implementation"
] | null | null | Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled. | The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops. | Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house. | [
"10 20 30\n",
"1 1 5\n"
] | [
"60\n",
"4\n"
] | The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | 500 | [
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
... | 1,617,899,737 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | a,b,c=map(int,input().split())
print(2*min((a+b),(a+c),(b+c))) | Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | ```python
a,b,c=map(int,input().split())
print(2*min((a+b),(a+c),(b+c)))
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter. | The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase. | Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"4 3\n1 2 3 4\n",
"4 2\n-3 -2 -1 0\n",
"5 19\n1 10 20 30 50\n"
] | [
"4\n",
"2\n",
"1\n"
] | In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | 0 | [
{
"input": "4 3\n1 2 3 4",
"output": "4"
},
{
"input": "4 2\n-3 -2 -1 0",
"output": "2"
},
{
"input": "5 19\n1 10 20 30 50",
"output": "1"
},
{
"input": "10 5\n31 36 43 47 48 50 56 69 71 86",
"output": "2"
},
{
"input": "10 50\n1 4 20 27 65 79 82 83 99 100",
"... | 1,673,934,026 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 1,528 | 8,089,600 | n,d = map(int,input().split())
arr = list(map(int,input().split()))
def binary_search(arr,low,high,target):
while low<=high:
mid = (low+high)>>1
# print(low,high,mid)
if arr[mid]>target:
high = mid-1
else:
if mid==high:
return mid
if arr[mid+1]>target:
return mid
low = mid+1
# print(binary_search(arr,0,n-1,0))
res = 0
i = 0
while i<n:
# print(d+arr[i])
j = binary_search(arr,i,n-1,d+arr[i])
# print(j)
if (j-i)>=2:
cnt = 1+j-i
res+= ((cnt-2)*(cnt-1))//2
i+=1
print(res) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input Specification:
The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output Specification:
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['4 3\n1 2 3 4\n', '4 2\n-3 -2 -1 0\n', '5 19\n1 10 20 30 50\n']
Demo Output:
['4\n', '2\n', '1\n']
Note:
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | ```python
n,d = map(int,input().split())
arr = list(map(int,input().split()))
def binary_search(arr,low,high,target):
while low<=high:
mid = (low+high)>>1
# print(low,high,mid)
if arr[mid]>target:
high = mid-1
else:
if mid==high:
return mid
if arr[mid+1]>target:
return mid
low = mid+1
# print(binary_search(arr,0,n-1,0))
res = 0
i = 0
while i<n:
# print(d+arr[i])
j = binary_search(arr,i,n-1,d+arr[i])
# print(j)
if (j-i)>=2:
cnt = 1+j-i
res+= ((cnt-2)*(cnt-1))//2
i+=1
print(res)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Programmers working on a large project have just received a task to write exactly *m* lines of code. There are *n* programmers working on a project, the *i*-th of them makes exactly *a**i* bugs in every line of code that he writes.
Let's call a sequence of non-negative integers *v*1,<=*v*2,<=...,<=*v**n* a plan, if *v*1<=+<=*v*2<=+<=...<=+<=*v**n*<==<=*m*. The programmers follow the plan like that: in the beginning the first programmer writes the first *v*1 lines of the given task, then the second programmer writes *v*2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most *b* bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer *mod*. | The first line contains four integers *n*, *m*, *b*, *mod* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*b*<=≤<=500; 1<=≤<=*mod*<=≤<=109<=+<=7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=500) — the number of bugs per line for each programmer. | Print a single integer — the answer to the problem modulo *mod*. | [
"3 3 3 100\n1 1 1\n",
"3 6 5 1000000007\n1 2 3\n",
"3 5 6 11\n1 2 1\n"
] | [
"10\n",
"0\n",
"0\n"
] | none | 0 | [
{
"input": "3 3 3 100\n1 1 1",
"output": "10"
},
{
"input": "3 6 5 1000000007\n1 2 3",
"output": "0"
},
{
"input": "3 5 6 11\n1 2 1",
"output": "0"
},
{
"input": "2 3 3 1000\n1 2",
"output": "1"
},
{
"input": "3 10 10 150691913\n8 7 10",
"output": "0"
},
{... | 1,431,022,615 | 6,415 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 5 | 3,000 | 0 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
def test(i, k, sum, a, b, n, m):
if ( i == n-1 ):
if sum+(m-k)*a[i] > b:
return 0
else:
return 1
acc = 0
for j in range(k, m+1):
if ( sum > b ):
break
acc += test(i+1, j, sum, a, b, n, m)
sum += a[i]
return acc
(n, m, b, mod) = [ int(i) for i in input().split() ]
a = [ int(i) for i in input().split() ]
if m < min(a)*n:
print(0)
else:
res = test(0, 0, 0, a, b, n, m)
print(divmod(res, mod)[1])
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmers working on a large project have just received a task to write exactly *m* lines of code. There are *n* programmers working on a project, the *i*-th of them makes exactly *a**i* bugs in every line of code that he writes.
Let's call a sequence of non-negative integers *v*1,<=*v*2,<=...,<=*v**n* a plan, if *v*1<=+<=*v*2<=+<=...<=+<=*v**n*<==<=*m*. The programmers follow the plan like that: in the beginning the first programmer writes the first *v*1 lines of the given task, then the second programmer writes *v*2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most *b* bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer *mod*.
Input Specification:
The first line contains four integers *n*, *m*, *b*, *mod* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*b*<=≤<=500; 1<=≤<=*mod*<=≤<=109<=+<=7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=500) — the number of bugs per line for each programmer.
Output Specification:
Print a single integer — the answer to the problem modulo *mod*.
Demo Input:
['3 3 3 100\n1 1 1\n', '3 6 5 1000000007\n1 2 3\n', '3 5 6 11\n1 2 1\n']
Demo Output:
['10\n', '0\n', '0\n']
Note:
none | ```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def test(i, k, sum, a, b, n, m):
if ( i == n-1 ):
if sum+(m-k)*a[i] > b:
return 0
else:
return 1
acc = 0
for j in range(k, m+1):
if ( sum > b ):
break
acc += test(i+1, j, sum, a, b, n, m)
sum += a[i]
return acc
(n, m, b, mod) = [ int(i) for i in input().split() ]
a = [ int(i) for i in input().split() ]
if m < min(a)*n:
print(0)
else:
res = test(0, 0, 0, a, b, n, m)
print(divmod(res, mod)[1])
``` | 0 | |
122 | A | Lucky Division | PROGRAMMING | 1,000 | [
"brute force",
"number theory"
] | null | null | Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky. | The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked. | In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes). | [
"47\n",
"16\n",
"78\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4. | 500 | [
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"ou... | 1,688,461,779 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 154 | 0 | n = int(input())
l = [7,4,47,74,444,777,447,474,744,774,747,477]
count =0
for i in l :
if (n % i == 0):
count += 1
if (count > 0):
print("YES")
else:
print("NO") | Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4. | ```python
n = int(input())
l = [7,4,47,74,444,777,447,474,744,774,747,477]
count =0
for i in l :
if (n % i == 0):
count += 1
if (count > 0):
print("YES")
else:
print("NO")
``` | 3 | |
758 | A | Holiday Of Equality | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. | In the only line print the integer *S* — the minimum number of burles which are had to spend. | [
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] | [
"10",
"1",
"4",
"0"
] | In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | 500 | [
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853... | 1,652,532,193 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | n = int(input())
b = list(map(int, input().split()))
t = 0
ma = max(b)
for x in range(n):
a = b[n-1] - b[x]
t += a
print(t) | Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | ```python
n = int(input())
b = list(map(int, input().split()))
t = 0
ma = max(b)
for x in range(n):
a = b[n-1] - b[x]
t += a
print(t)
``` | 0 | |
124 | A | The number of positions | PROGRAMMING | 1,000 | [
"math"
] | null | null | Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy. | The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100). | Print the single number — the number of the sought positions. | [
"3 1 1\n",
"5 2 3\n"
] | [
"2\n",
"3\n"
] | The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | 500 | [
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 ... | 1,663,810,235 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n,a,b=map(int,input().split())
if n-b-a>0:
print(n-a+1)
else:
print(n-b+1) | Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | ```python
n,a,b=map(int,input().split())
if n-b-a>0:
print(n-a+1)
else:
print(n-b+1)
``` | 0 | |
915 | A | Garden | PROGRAMMING | 900 | [
"implementation"
] | null | null | Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding. | The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket. | Print one integer number — the minimum number of hours required to water the garden. | [
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] | [
"2\n",
"7\n"
] | In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1. | 0 | [
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 5... | 1,650,681,695 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 46 | 0 | import sys
input = sys.stdin.readline
n, k = map(int, input().split())
w = list(map(int, input().split()))
for i in range(n-1, -1, -1):
if k % w[i] == 0:
print(k//w[i])
break | Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number — the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1. | ```python
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
w = list(map(int, input().split()))
for i in range(n-1, -1, -1):
if k % w[i] == 0:
print(k//w[i])
break
``` | 0 | |
427 | B | Prison Transfer | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | The prison of your city has *n* prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer *c* of the prisoners to a prison located in another city.
For this reason, he made the *n* prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.
Then, the mayor told you to choose the *c* prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,
- The chosen *c* prisoners has to form a contiguous segment of prisoners. - Any of the chosen prisoner's crime level should not be greater then *t*. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.
Find the number of ways you can choose the *c* prisoners. | The first line of input will contain three space separated integers *n* (1<=≤<=*n*<=≤<=2·105), *t* (0<=≤<=*t*<=≤<=109) and *c* (1<=≤<=*c*<=≤<=*n*). The next line will contain *n* space separated integers, the *i**th* integer is the severity *i**th* prisoner's crime. The value of crime severities will be non-negative and will not exceed 109. | Print a single integer — the number of ways you can choose the *c* prisoners. | [
"4 3 3\n2 3 1 1\n",
"1 1 1\n2\n",
"11 4 2\n2 2 0 7 3 2 2 4 9 1 4\n"
] | [
"2\n",
"0\n",
"6\n"
] | none | 1,000 | [
{
"input": "4 3 3\n2 3 1 1",
"output": "2"
},
{
"input": "1 1 1\n2",
"output": "0"
},
{
"input": "11 4 2\n2 2 0 7 3 2 2 4 9 1 4",
"output": "6"
},
{
"input": "57 2 10\n7 5 2 7 4 1 0 5 2 9 2 9 8 6 6 5 9 6 8 1 0 1 0 3 2 6 5 2 8 8 8 8 0 9 4 3 6 6 2 4 5 1 2 0 1 7 1 1 5 4 5 0 7 5 ... | 1,695,835,862 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | do=input().split()
mi=input().split()
n=int(do[0])
t=int(do[1])
c=int(do[2])
ni=[int(j) for j in mi]
k=0
s=0
for j in ni:
if (j>t and k>=c) or k==n-1:
s=s+k-c+2
print(s)
print(k)
k=0
else:
k+=1
print(s) | Title: Prison Transfer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The prison of your city has *n* prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer *c* of the prisoners to a prison located in another city.
For this reason, he made the *n* prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.
Then, the mayor told you to choose the *c* prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,
- The chosen *c* prisoners has to form a contiguous segment of prisoners. - Any of the chosen prisoner's crime level should not be greater then *t*. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.
Find the number of ways you can choose the *c* prisoners.
Input Specification:
The first line of input will contain three space separated integers *n* (1<=≤<=*n*<=≤<=2·105), *t* (0<=≤<=*t*<=≤<=109) and *c* (1<=≤<=*c*<=≤<=*n*). The next line will contain *n* space separated integers, the *i**th* integer is the severity *i**th* prisoner's crime. The value of crime severities will be non-negative and will not exceed 109.
Output Specification:
Print a single integer — the number of ways you can choose the *c* prisoners.
Demo Input:
['4 3 3\n2 3 1 1\n', '1 1 1\n2\n', '11 4 2\n2 2 0 7 3 2 2 4 9 1 4\n']
Demo Output:
['2\n', '0\n', '6\n']
Note:
none | ```python
do=input().split()
mi=input().split()
n=int(do[0])
t=int(do[1])
c=int(do[2])
ni=[int(j) for j in mi]
k=0
s=0
for j in ni:
if (j>t and k>=c) or k==n-1:
s=s+k-c+2
print(s)
print(k)
k=0
else:
k+=1
print(s)
``` | 0 | |
976 | A | Minimum Binary Number | PROGRAMMING | 800 | [
"implementation"
] | null | null | String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all). | The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct. | Print one string — the minimum correct string that you can obtain from the given one. | [
"4\n1001\n",
"1\n1\n"
] | [
"100\n",
"1\n"
] | In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | 0 | [
{
"input": "4\n1001",
"output": "100"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "100\n1110111100001111011111111010110011111111011110000111101101011100110110001011000000101010110101011100",
"output": "1000000000000000000000000000000000000000"
},
{
"input": "100\n100000... | 1,548,781,782 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 109 | 0 | input()
a = input()
print('1' + a.count('0') * '0') | Title: Minimum Binary Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
Output Specification:
Print one string — the minimum correct string that you can obtain from the given one.
Demo Input:
['4\n1001\n', '1\n1\n']
Demo Output:
['100\n', '1\n']
Note:
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | ```python
input()
a = input()
print('1' + a.count('0') * '0')
``` | 0 | |
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"
},
{
"... | 1,690,975,545 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 46 | 0 | n = int(input())
x = 0
for i in range(n):
c, m = input().split(" ")
if int(m) - int(c) >= 2:
x += 1
print(x)
| 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
n = int(input())
x = 0
for i in range(n):
c, m = input().split(" ")
if int(m) - int(c) >= 2:
x += 1
print(x)
``` | 3 | |
527 | A | Playing with Paper | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | null | null | One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson? | The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper. | Print a single integer — the number of ships that Vasya will make. | [
"2 1\n",
"10 7\n",
"1000000000000 1\n"
] | [
"2\n",
"6\n",
"1000000000000\n"
] | Pictures to the first and second sample test. | 500 | [
{
"input": "2 1",
"output": "2"
},
{
"input": "10 7",
"output": "6"
},
{
"input": "1000000000000 1",
"output": "1000000000000"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "3 2",
"output": "3"
},
{
"in... | 1,479,450,542 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 2,000 | 0 | a, b = map(int, input().split())
count = 1
while a != b:
if a < b:
a, b = b, a
a -= b
count += 1
print(count) | Title: Playing with Paper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
Input Specification:
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
Output Specification:
Print a single integer — the number of ships that Vasya will make.
Demo Input:
['2 1\n', '10 7\n', '1000000000000 1\n']
Demo Output:
['2\n', '6\n', '1000000000000\n']
Note:
Pictures to the first and second sample test. | ```python
a, b = map(int, input().split())
count = 1
while a != b:
if a < b:
a, b = b, a
a -= b
count += 1
print(count)
``` | 0 | |
601 | A | The Two Routes | PROGRAMMING | 1,600 | [
"graphs",
"shortest paths"
] | null | null | In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so. | The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns. | Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1. | [
"4 2\n1 3\n3 4\n",
"4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n",
"5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n"
] | [
"2\n",
"-1\n",
"3\n"
] | In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4. | 500 | [
{
"input": "4 2\n1 3\n3 4",
"output": "2"
},
{
"input": "4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4",
"output": "-1"
},
{
"input": "5 5\n4 2\n3 5\n4 5\n5 1\n1 2",
"output": "3"
},
{
"input": "5 4\n1 2\n3 2\n3 4\n5 4",
"output": "4"
},
{
"input": "3 1\n1 2",
"output": "... | 1,678,717,163 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 59 | 155 | 15,872,000 | import sys
input=sys.stdin.readline
from collections import defaultdict,deque
n,m=map(int,input().split())
g=defaultdict(list)
s1=set()
for _ in range(m):
x,y=map(int,input().split())
g[x].append(y)
g[y].append(x)
s1.add((x,y))
g2=defaultdict(list)
for i in range(1,n+1):
for j in range(i+1,n+1):
if (i,j) not in s1 and (j,i) not in s1:
g2[i].append(j)
g2[j].append(i)
time={}
def bfs(src):
visited=set()
q=deque()
parent=[-1]*(n+1)
q.append([src])
visited.add(src)
c=0
time[src]=0
while len(q)>0:
f=q.popleft()
l=[]
for node in f:
for nbr in g[node]:
if nbr not in visited:
visited.add(nbr)
l.append(nbr)
parent[nbr]=node
time[nbr]=c+1
if len(l)==0:break
c+=1
q.append(l)
if n in visited:break
if n not in visited:return -1
s=set()
cur=n
while parent[cur]!=1:
s.add(cur)
cur=parent[cur]
s.add(1)
for x in time:
if x not in s:time[x]=-1
return c
rail=bfs(1)
def bfs2(src):
visited=set()
q=deque()
q.append([src])
visited.add(src)
c=0
while len(q)>0:
f=q.popleft()
l=[]
for node in f:
for nbr in g2[node]:
if nbr not in visited:
if nbr in time:
if time[nbr]==c+1:
if nbr!=n:pass
else:
visited.add(nbr)
l.append(nbr)
else:
visited.add(nbr)
l.append(nbr)
else:
visited.add(nbr)
l.append(nbr)
if len(l)==0:break
c+=1
q.append(l)
if n in visited:return c
if n not in visited:return -1
road=bfs2(1)
if road==-1 or rail==-1:print(-1)
else:print(max(rail,road)) | Title: The Two Routes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so.
Input Specification:
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns.
Output Specification:
Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1.
Demo Input:
['4 2\n1 3\n3 4\n', '4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n', '5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n']
Demo Output:
['2\n', '-1\n', '3\n']
Note:
In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4. | ```python
import sys
input=sys.stdin.readline
from collections import defaultdict,deque
n,m=map(int,input().split())
g=defaultdict(list)
s1=set()
for _ in range(m):
x,y=map(int,input().split())
g[x].append(y)
g[y].append(x)
s1.add((x,y))
g2=defaultdict(list)
for i in range(1,n+1):
for j in range(i+1,n+1):
if (i,j) not in s1 and (j,i) not in s1:
g2[i].append(j)
g2[j].append(i)
time={}
def bfs(src):
visited=set()
q=deque()
parent=[-1]*(n+1)
q.append([src])
visited.add(src)
c=0
time[src]=0
while len(q)>0:
f=q.popleft()
l=[]
for node in f:
for nbr in g[node]:
if nbr not in visited:
visited.add(nbr)
l.append(nbr)
parent[nbr]=node
time[nbr]=c+1
if len(l)==0:break
c+=1
q.append(l)
if n in visited:break
if n not in visited:return -1
s=set()
cur=n
while parent[cur]!=1:
s.add(cur)
cur=parent[cur]
s.add(1)
for x in time:
if x not in s:time[x]=-1
return c
rail=bfs(1)
def bfs2(src):
visited=set()
q=deque()
q.append([src])
visited.add(src)
c=0
while len(q)>0:
f=q.popleft()
l=[]
for node in f:
for nbr in g2[node]:
if nbr not in visited:
if nbr in time:
if time[nbr]==c+1:
if nbr!=n:pass
else:
visited.add(nbr)
l.append(nbr)
else:
visited.add(nbr)
l.append(nbr)
else:
visited.add(nbr)
l.append(nbr)
if len(l)==0:break
c+=1
q.append(l)
if n in visited:return c
if n not in visited:return -1
road=bfs2(1)
if road==-1 or rail==-1:print(-1)
else:print(max(rail,road))
``` | 3 | |
168 | A | Wizards and Demonstration | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | Some country is populated by wizards. They want to organize a demonstration.
There are *n* people living in the city, *x* of them are the wizards who will surely go to the demonstration. Other city people (*n*<=-<=*x* people) do not support the wizards and aren't going to go to the demonstration. We know that the city administration will react only to the demonstration involving at least *y* percent of the city people. Having considered the matter, the wizards decided to create clone puppets which can substitute the city people on the demonstration.
So all in all, the demonstration will involve only the wizards and their puppets. The city administration cannot tell the difference between a puppet and a person, so, as they calculate the percentage, the administration will consider the city to be consisting of only *n* people and not containing any clone puppets.
Help the wizards and find the minimum number of clones to create to that the demonstration had no less than *y* percent of the city people. | The first line contains three space-separated integers, *n*, *x*, *y* (1<=≤<=*n*,<=*x*,<=*y*<=≤<=104,<=*x*<=≤<=*n*) — the number of citizens in the city, the number of wizards and the percentage the administration needs, correspondingly.
Please note that *y* can exceed 100 percent, that is, the administration wants to see on a demonstration more people that actually live in the city (<=><=*n*). | Print a single integer — the answer to the problem, the minimum number of clones to create, so that the demonstration involved no less than *y* percent of *n* (the real total city population). | [
"10 1 14\n",
"20 10 50\n",
"1000 352 146\n"
] | [
"1\n",
"0\n",
"1108\n"
] | In the first sample it is necessary that at least 14% of 10 people came to the demonstration. As the number of people should be integer, then at least two people should come. There is only one wizard living in the city and he is going to come. That isn't enough, so he needs to create one clone.
In the second sample 10 people should come to the demonstration. The city has 10 wizards. They will all come to the demonstration, so nobody has to create any clones. | 500 | [
{
"input": "10 1 14",
"output": "1"
},
{
"input": "20 10 50",
"output": "0"
},
{
"input": "1000 352 146",
"output": "1108"
},
{
"input": "68 65 20",
"output": "0"
},
{
"input": "78 28 27",
"output": "0"
},
{
"input": "78 73 58",
"output": "0"
},
... | 1,578,848,153 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 218 | 0 | n, x, y = map(int, input().split())
print(max(0, -x--n*y//100)) | Title: Wizards and Demonstration
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some country is populated by wizards. They want to organize a demonstration.
There are *n* people living in the city, *x* of them are the wizards who will surely go to the demonstration. Other city people (*n*<=-<=*x* people) do not support the wizards and aren't going to go to the demonstration. We know that the city administration will react only to the demonstration involving at least *y* percent of the city people. Having considered the matter, the wizards decided to create clone puppets which can substitute the city people on the demonstration.
So all in all, the demonstration will involve only the wizards and their puppets. The city administration cannot tell the difference between a puppet and a person, so, as they calculate the percentage, the administration will consider the city to be consisting of only *n* people and not containing any clone puppets.
Help the wizards and find the minimum number of clones to create to that the demonstration had no less than *y* percent of the city people.
Input Specification:
The first line contains three space-separated integers, *n*, *x*, *y* (1<=≤<=*n*,<=*x*,<=*y*<=≤<=104,<=*x*<=≤<=*n*) — the number of citizens in the city, the number of wizards and the percentage the administration needs, correspondingly.
Please note that *y* can exceed 100 percent, that is, the administration wants to see on a demonstration more people that actually live in the city (<=><=*n*).
Output Specification:
Print a single integer — the answer to the problem, the minimum number of clones to create, so that the demonstration involved no less than *y* percent of *n* (the real total city population).
Demo Input:
['10 1 14\n', '20 10 50\n', '1000 352 146\n']
Demo Output:
['1\n', '0\n', '1108\n']
Note:
In the first sample it is necessary that at least 14% of 10 people came to the demonstration. As the number of people should be integer, then at least two people should come. There is only one wizard living in the city and he is going to come. That isn't enough, so he needs to create one clone.
In the second sample 10 people should come to the demonstration. The city has 10 wizards. They will all come to the demonstration, so nobody has to create any clones. | ```python
n, x, y = map(int, input().split())
print(max(0, -x--n*y//100))
``` | 3 | |
779 | B | Weird Rounding | PROGRAMMING | 1,100 | [
"brute force",
"greedy"
] | null | null | Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10*k*.
In the given number of *n* Polycarp wants to remove the least number of digits to get a number that is divisible by 10*k*. For example, if *k*<==<=3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103<==<=1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number *n*, so that the result is divisible by 10*k*. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists. | The only line of the input contains two integer numbers *n* and *k* (0<=≤<=*n*<=≤<=2<=000<=000<=000, 1<=≤<=*k*<=≤<=9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros. | Print *w* — the required minimal number of digits to erase. After removing the appropriate *w* digits from the number *n*, the result should have a value that is divisible by 10*k*. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0). | [
"30020 3\n",
"100 9\n",
"10203049 2\n"
] | [
"1\n",
"2\n",
"3\n"
] | In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number. | 1,000 | [
{
"input": "30020 3",
"output": "1"
},
{
"input": "100 9",
"output": "2"
},
{
"input": "10203049 2",
"output": "3"
},
{
"input": "0 1",
"output": "0"
},
{
"input": "0 9",
"output": "0"
},
{
"input": "100 2",
"output": "0"
},
{
"input": "102... | 1,612,218,296 | 2,147,483,647 | Python 3 | OK | TESTS | 161 | 77 | 307,200 | a,b=map(int,input().split())
z=list(str(a))
zeroc=z.count('0')
if zeroc<b:
print(len(str(a))-1)
exit()
c=0
x=len(z)-1
while int(''.join(z)) %10**b>0:
if z[x]!='0':
del z[x]
c+=1
x-=1
print(c) | Title: Weird Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10*k*.
In the given number of *n* Polycarp wants to remove the least number of digits to get a number that is divisible by 10*k*. For example, if *k*<==<=3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103<==<=1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number *n*, so that the result is divisible by 10*k*. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input Specification:
The only line of the input contains two integer numbers *n* and *k* (0<=≤<=*n*<=≤<=2<=000<=000<=000, 1<=≤<=*k*<=≤<=9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output Specification:
Print *w* — the required minimal number of digits to erase. After removing the appropriate *w* digits from the number *n*, the result should have a value that is divisible by 10*k*. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Demo Input:
['30020 3\n', '100 9\n', '10203049 2\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number. | ```python
a,b=map(int,input().split())
z=list(str(a))
zeroc=z.count('0')
if zeroc<b:
print(len(str(a))-1)
exit()
c=0
x=len(z)-1
while int(''.join(z)) %10**b>0:
if z[x]!='0':
del z[x]
c+=1
x-=1
print(c)
``` | 3 | |
758 | A | Holiday Of Equality | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. | In the only line print the integer *S* — the minimum number of burles which are had to spend. | [
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] | [
"10",
"1",
"4",
"0"
] | In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | 500 | [
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853... | 1,691,673,613 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 46 | 0 | nik=int(input())
*n,=map(int,input().split())
print(nik*max(n)-sum(n)) | Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | ```python
nik=int(input())
*n,=map(int,input().split())
print(nik*max(n)-sum(n))
``` | 3 | |
275 | A | Lights Out | PROGRAMMING | 900 | [
"implementation"
] | null | null | Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light. | The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed. | Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0". | [
"1 0 0\n0 0 0\n0 0 1\n",
"1 0 1\n8 8 8\n2 0 3\n"
] | [
"001\n010\n100\n",
"010\n011\n100\n"
] | none | 500 | [
{
"input": "1 0 0\n0 0 0\n0 0 1",
"output": "001\n010\n100"
},
{
"input": "1 0 1\n8 8 8\n2 0 3",
"output": "010\n011\n100"
},
{
"input": "13 85 77\n25 50 45\n65 79 9",
"output": "000\n010\n000"
},
{
"input": "96 95 5\n8 84 74\n67 31 61",
"output": "011\n011\n101"
},
{... | 1,665,132,257 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 46 | 0 |
light_a = [[1 for i in range(3)] for j in range(3)]
press_a = []
for i in range(3):
l1 = list(map(int,input().split()))
l2 = list(map(lambda x: x%2,l1))
press_a.append(l2)
for i in range(3):
for j in range(3):
if press_a[i][j]:
try:
if i == 0:
pass
elif light_a[i-1][j] == 1:
light_a[i-1][j] = 0
else:
light_a[i-1][j] = 1
except Exception:
pass
try:
if light_a[i][j] == 1:
light_a[i][j] = 0
else:
light_a[i][j] = 1
except Exception:
pass
try:
if light_a[i + 1][j] == 1:
light_a[i + 1][j] = 0
else:
light_a[i + 1][j] = 1
except Exception:
pass
try:
if light_a[i][j+1] == 1:
light_a[i][j+1] = 0
else:
light_a[i][j+1] = 1
except Exception:
pass
try:
if j == 0:
pass
elif light_a[i][j - 1] == 1:
light_a[i][j - 1] = 0
else:
light_a[i][j - 1] = 1
except Exception:
pass
for i in range(3):
for j in range(3):
print(light_a[i][j],end="")
print() | Title: Lights Out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
Input Specification:
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Output Specification:
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
Demo Input:
['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n']
Demo Output:
['001\n010\n100\n', '010\n011\n100\n']
Note:
none | ```python
light_a = [[1 for i in range(3)] for j in range(3)]
press_a = []
for i in range(3):
l1 = list(map(int,input().split()))
l2 = list(map(lambda x: x%2,l1))
press_a.append(l2)
for i in range(3):
for j in range(3):
if press_a[i][j]:
try:
if i == 0:
pass
elif light_a[i-1][j] == 1:
light_a[i-1][j] = 0
else:
light_a[i-1][j] = 1
except Exception:
pass
try:
if light_a[i][j] == 1:
light_a[i][j] = 0
else:
light_a[i][j] = 1
except Exception:
pass
try:
if light_a[i + 1][j] == 1:
light_a[i + 1][j] = 0
else:
light_a[i + 1][j] = 1
except Exception:
pass
try:
if light_a[i][j+1] == 1:
light_a[i][j+1] = 0
else:
light_a[i][j+1] = 1
except Exception:
pass
try:
if j == 0:
pass
elif light_a[i][j - 1] == 1:
light_a[i][j - 1] = 0
else:
light_a[i][j - 1] = 1
except Exception:
pass
for i in range(3):
for j in range(3):
print(light_a[i][j],end="")
print()
``` | 3 | |
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": ... | 1,672,304,243 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 61 | 0 | a, b = map(int, input().split())
i = a
while a != 0:
a //= b
i += a
print(i) | 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
a, b = map(int, input().split())
i = a
while a != 0:
a //= b
i += a
print(i)
``` | 0 | |
520 | B | Two Buttons | PROGRAMMING | 1,400 | [
"dfs and similar",
"graphs",
"greedy",
"implementation",
"math",
"shortest paths"
] | null | null | Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number *n*.
Bob wants to get number *m* on the display. What minimum number of clicks he has to make in order to achieve this result? | The first and the only line of the input contains two distinct integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=104), separated by a space . | Print a single number — the minimum number of times one needs to push the button required to get the number *m* out of number *n*. | [
"4 6\n",
"10 1\n"
] | [
"2\n",
"9\n"
] | In the first example you need to push the blue button once, and then push the red button once.
In the second example, doubling the number is unnecessary, so we need to push the blue button nine times. | 1,000 | [
{
"input": "4 6",
"output": "2"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "2 10",
"outpu... | 1,694,264,877 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 4 | 31 | 0 | # -*- coding: utf-8 -*-
"""
Created on Sat Sep 9 17:11:06 2023
@author: mac
"""
import math
n, m = map(int, input().split())
depth = 0
go = True
while go:
depth += 1
for i in range(depth + 1):
left = n * 2**(depth - i)
for j in range(i):
left -= 2**min(int(math.log(left,2)), depth - i)
if left == m:
print(depth)
go = False
break | Title: Two Buttons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number *n*.
Bob wants to get number *m* on the display. What minimum number of clicks he has to make in order to achieve this result?
Input Specification:
The first and the only line of the input contains two distinct integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=104), separated by a space .
Output Specification:
Print a single number — the minimum number of times one needs to push the button required to get the number *m* out of number *n*.
Demo Input:
['4 6\n', '10 1\n']
Demo Output:
['2\n', '9\n']
Note:
In the first example you need to push the blue button once, and then push the red button once.
In the second example, doubling the number is unnecessary, so we need to push the blue button nine times. | ```python
# -*- coding: utf-8 -*-
"""
Created on Sat Sep 9 17:11:06 2023
@author: mac
"""
import math
n, m = map(int, input().split())
depth = 0
go = True
while go:
depth += 1
for i in range(depth + 1):
left = n * 2**(depth - i)
for j in range(i):
left -= 2**min(int(math.log(left,2)), depth - i)
if left == m:
print(depth)
go = False
break
``` | -1 | |
749 | A | Bachgold Problem | PROGRAMMING | 800 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*. | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000). | The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them. | [
"5\n",
"6\n"
] | [
"2\n2 3\n",
"3\n2 2 2\n"
] | none | 500 | [
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ... | 1,688,335,471 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 41 | 77 | 6,041,600 | n = int(input())
if (n%2 == 0):
print(n//2)
print(*[2 for i in range(n//2)])
else:
k = (n - 3)//2
print(k+1)
print(*[2 for i in range(k)], end = " ")
print(3)# 1688335471.126259 | Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none | ```python
n = int(input())
if (n%2 == 0):
print(n//2)
print(*[2 for i in range(n//2)])
else:
k = (n - 3)//2
print(k+1)
print(*[2 for i in range(k)], end = " ")
print(3)# 1688335471.126259
``` | 3 | |
95 | B | Lucky Numbers | PROGRAMMING | 1,800 | [
"dp",
"greedy"
] | B. Lucky Numbers | 2 | 256 | Petya loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.
One day Petya came across a positive integer *n*. Help him to find the least super lucky number which is not less than *n*. | The only line contains a positive integer *n* (1<=≤<=*n*<=≤<=10100000). This number doesn't have leading zeroes. | Output the least super lucky number that is more than or equal to *n*. | [
"4500\n",
"47\n"
] | [
"4747\n",
"47\n"
] | none | 1,000 | [
{
"input": "4500",
"output": "4747"
},
{
"input": "47",
"output": "47"
},
{
"input": "1",
"output": "47"
},
{
"input": "12",
"output": "47"
},
{
"input": "4587",
"output": "4747"
},
{
"input": "100",
"output": "4477"
},
{
"input": "1007",
... | 1,627,393,711 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 34,406,400 | n=int(input())
q=__import__('collections').deque([0])
while True:
x=q.popleft()
s=str(x)
if s.count('4')==s.count('7') and x>=n: exit(print(x))
q.append(x*10+4)
q.append(x*10+7) | Title: Lucky Numbers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.
One day Petya came across a positive integer *n*. Help him to find the least super lucky number which is not less than *n*.
Input Specification:
The only line contains a positive integer *n* (1<=≤<=*n*<=≤<=10100000). This number doesn't have leading zeroes.
Output Specification:
Output the least super lucky number that is more than or equal to *n*.
Demo Input:
['4500\n', '47\n']
Demo Output:
['4747\n', '47\n']
Note:
none | ```python
n=int(input())
q=__import__('collections').deque([0])
while True:
x=q.popleft()
s=str(x)
if s.count('4')==s.count('7') and x>=n: exit(print(x))
q.append(x*10+4)
q.append(x*10+7)
``` | 0 |
934 | A | A Compatible Pair | PROGRAMMING | 1,400 | [
"brute force",
"games"
] | null | null | Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109. | Print a single integer — the brightness of the chosen pair. | [
"2 2\n20 18\n2 14\n",
"5 3\n-1 0 1 2 3\n-1 0 1\n"
] | [
"252\n",
"2\n"
] | In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself. | 500 | [
{
"input": "2 2\n20 18\n2 14",
"output": "252"
},
{
"input": "5 3\n-1 0 1 2 3\n-1 0 1",
"output": "2"
},
{
"input": "10 2\n1 6 2 10 2 3 2 10 6 4\n5 7",
"output": "70"
},
{
"input": "50 50\n1 6 2 10 2 3 2 10 6 4 5 0 3 1 7 3 2 4 4 2 1 5 0 6 10 1 8 0 10 9 0 4 10 5 5 7 4 9 9 5 5 ... | 1,699,446,493 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 11 | 93 | 1,740,800 | n=int(input().split()[0])
a=list(map(int,input().split()))
b=list(map(int,input().split()))
mn=1<<50
def f():
global a,b
mx=-(1<<50)
for i in a:
for j in b:
mx=max(mx,i*j)
return mx
for i in range(n):
t=a[0]
del a[0]
mn=min(mn,f())
a.append(t)
print(mn) | Title: A Compatible Pair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109.
Output Specification:
Print a single integer — the brightness of the chosen pair.
Demo Input:
['2 2\n20 18\n2 14\n', '5 3\n-1 0 1 2 3\n-1 0 1\n']
Demo Output:
['252\n', '2\n']
Note:
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself. | ```python
n=int(input().split()[0])
a=list(map(int,input().split()))
b=list(map(int,input().split()))
mn=1<<50
def f():
global a,b
mx=-(1<<50)
for i in a:
for j in b:
mx=max(mx,i*j)
return mx
for i in range(n):
t=a[0]
del a[0]
mn=min(mn,f())
a.append(t)
print(mn)
``` | 0 | |
433 | B | Kuriyama Mirai's Stones | PROGRAMMING | 1,200 | [
"dp",
"implementation",
"sortings"
] | null | null | Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one. | Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input. | [
"6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n",
"4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n"
] | [
"24\n9\n28\n",
"10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n"
] | Please note that the answers to the questions may overflow 32-bit integer type. | 1,500 | [
{
"input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6",
"output": "24\n9\n28"
},
{
"input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2",
"output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5"
},
{
"input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1... | 1,668,202,388 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 1,060 | 10,137,600 | n = int(input())
v = list(map(int, input().split()))
m = int(input())
vs = list(v)
vs.sort()
dpv = []
dpvs = []
dpv.append(v[0])
dpvs.append(vs[0])
for i in range(1, len(v)):
dpv.append(dpv[i - 1] + v[i])
dpvs.append(dpvs[i - 1] + vs[i])
for j in range(m):
t, l, r = map(int, input().split())
l -= 2
r -= 1
if t == 1:
print(dpv[r] - (0 if l == -1 else dpv[l]))
else:
print(dpvs[r] - (0 if l == -1 else dpvs[l]))
| Title: Kuriyama Mirai's Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output Specification:
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Demo Input:
['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n']
Demo Output:
['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n']
Note:
Please note that the answers to the questions may overflow 32-bit integer type. | ```python
n = int(input())
v = list(map(int, input().split()))
m = int(input())
vs = list(v)
vs.sort()
dpv = []
dpvs = []
dpv.append(v[0])
dpvs.append(vs[0])
for i in range(1, len(v)):
dpv.append(dpv[i - 1] + v[i])
dpvs.append(dpvs[i - 1] + vs[i])
for j in range(m):
t, l, r = map(int, input().split())
l -= 2
r -= 1
if t == 1:
print(dpv[r] - (0 if l == -1 else dpv[l]))
else:
print(dpvs[r] - (0 if l == -1 else dpvs[l]))
``` | 3 | |
727 | A | Transformation: from A to B | PROGRAMMING | 1,000 | [
"brute force",
"dfs and similar",
"math"
] | null | null | Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*. | The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have. | If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them. | [
"2 162\n",
"4 42\n",
"100 40021\n"
] | [
"YES\n5\n2 4 8 81 162 \n",
"NO\n",
"YES\n5\n100 200 2001 4002 40021 \n"
] | none | 1,000 | [
{
"input": "2 162",
"output": "YES\n5\n2 4 8 81 162 "
},
{
"input": "4 42",
"output": "NO"
},
{
"input": "100 40021",
"output": "YES\n5\n100 200 2001 4002 40021 "
},
{
"input": "1 111111111",
"output": "YES\n9\n1 11 111 1111 11111 111111 1111111 11111111 111111111 "
},
... | 1,617,627,636 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 62 | 0 | def from_a_2_b(a, b):
count = 1
nums = [b]
while(a < b):
if b % 2 == 0:
b = b / 2
count += 1
nums.insert(0, int(b))
elif b % 10 == 1:
b = b // 10
count += 1
nums.insert(0, int(b))
else:
b = 0
if(a != b):
print("NO")
else:
print("YES")
print(count)
print(" ".join(map(str, nums)))
a, b = map(int, input().split())
from_a_2_b(a, b)
| Title: Transformation: from A to B
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
Input Specification:
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
Output Specification:
If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them.
Demo Input:
['2 162\n', '4 42\n', '100 40021\n']
Demo Output:
['YES\n5\n2 4 8 81 162 \n', 'NO\n', 'YES\n5\n100 200 2001 4002 40021 \n']
Note:
none | ```python
def from_a_2_b(a, b):
count = 1
nums = [b]
while(a < b):
if b % 2 == 0:
b = b / 2
count += 1
nums.insert(0, int(b))
elif b % 10 == 1:
b = b // 10
count += 1
nums.insert(0, int(b))
else:
b = 0
if(a != b):
print("NO")
else:
print("YES")
print(count)
print(" ".join(map(str, nums)))
a, b = map(int, input().split())
from_a_2_b(a, b)
``` | 3 | |
635 | A | Orchestra | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Paul is at the orchestra. The string section is arranged in an *r*<=×<=*c* rectangular grid and is filled with violinists with the exception of *n* violists. Paul really likes violas, so he would like to take a picture including at least *k* of them. Paul can take a picture of any axis-parallel rectangle in the orchestra. Count the number of possible pictures that Paul can take.
Two pictures are considered to be different if the coordinates of corresponding rectangles are different. | The first line of input contains four space-separated integers *r*, *c*, *n*, *k* (1<=≤<=*r*,<=*c*,<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=*n*) — the number of rows and columns of the string section, the total number of violas, and the minimum number of violas Paul would like in his photograph, respectively.
The next *n* lines each contain two integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=*r*, 1<=≤<=*y**i*<=≤<=*c*): the position of the *i*-th viola. It is guaranteed that no location appears more than once in the input. | Print a single integer — the number of photographs Paul can take which include at least *k* violas. | [
"2 2 1 1\n1 2\n",
"3 2 3 3\n1 1\n3 1\n2 2\n",
"3 2 3 2\n1 1\n3 1\n2 2\n"
] | [
"4\n",
"1\n",
"4\n"
] | We will use '*' to denote violinists and '#' to denote violists.
In the first sample, the orchestra looks as follows
In the second sample, the orchestra looks as follows
In the third sample, the orchestra looks the same as in the second sample. | 500 | [
{
"input": "2 2 1 1\n1 2",
"output": "4"
},
{
"input": "3 2 3 3\n1 1\n3 1\n2 2",
"output": "1"
},
{
"input": "3 2 3 2\n1 1\n3 1\n2 2",
"output": "4"
},
{
"input": "1 1 1 1\n1 1",
"output": "1"
},
{
"input": "10 10 10 10\n6 1\n3 8\n10 6\n10 3\n10 4\n8 9\n2 3\n5 7\n... | 1,456,683,651 | 651 | Python 3 | OK | TESTS | 22 | 77 | 0 | r, c, n, k = map(int, input().split())
g = set([tuple(map(int, input().split())) for _ in range(n)])
ret = 0
for i in range(r):
for j in range(c):
for l in range(1, r-i+1):
for w in range(1, c-j+1):
count = 0
for a in range(i, i+l):
for b in range(j, j+w):
if (a+1, b+1) in g:
count += 1
if count >= k:
ret += 1
print(ret)
| Title: Orchestra
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Paul is at the orchestra. The string section is arranged in an *r*<=×<=*c* rectangular grid and is filled with violinists with the exception of *n* violists. Paul really likes violas, so he would like to take a picture including at least *k* of them. Paul can take a picture of any axis-parallel rectangle in the orchestra. Count the number of possible pictures that Paul can take.
Two pictures are considered to be different if the coordinates of corresponding rectangles are different.
Input Specification:
The first line of input contains four space-separated integers *r*, *c*, *n*, *k* (1<=≤<=*r*,<=*c*,<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=*n*) — the number of rows and columns of the string section, the total number of violas, and the minimum number of violas Paul would like in his photograph, respectively.
The next *n* lines each contain two integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=*r*, 1<=≤<=*y**i*<=≤<=*c*): the position of the *i*-th viola. It is guaranteed that no location appears more than once in the input.
Output Specification:
Print a single integer — the number of photographs Paul can take which include at least *k* violas.
Demo Input:
['2 2 1 1\n1 2\n', '3 2 3 3\n1 1\n3 1\n2 2\n', '3 2 3 2\n1 1\n3 1\n2 2\n']
Demo Output:
['4\n', '1\n', '4\n']
Note:
We will use '*' to denote violinists and '#' to denote violists.
In the first sample, the orchestra looks as follows
In the second sample, the orchestra looks as follows
In the third sample, the orchestra looks the same as in the second sample. | ```python
r, c, n, k = map(int, input().split())
g = set([tuple(map(int, input().split())) for _ in range(n)])
ret = 0
for i in range(r):
for j in range(c):
for l in range(1, r-i+1):
for w in range(1, c-j+1):
count = 0
for a in range(i, i+l):
for b in range(j, j+w):
if (a+1, b+1) in g:
count += 1
if count >= k:
ret += 1
print(ret)
``` | 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",
... | 1,631,007,396 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 124 | 6,758,400 | n = int(input())
team = [input() for i in range(n)]
win=team[0]
for i in team :
if team.count(i) > team.count(win):
win=i
print(win) | 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
n = int(input())
team = [input() for i in range(n)]
win=team[0]
for i in team :
if team.count(i) > team.count(win):
win=i
print(win)
``` | 3.956411 |
454 | A | Little Pony and Crystal Mine | PROGRAMMING | 800 | [
"implementation"
] | null | null | Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it.
You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. | The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd). | Output a crystal of size *n*. | [
"3\n",
"5\n",
"7\n"
] | [
"*D*\nDDD\n*D*\n",
"**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n",
"***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n"
] | none | 500 | [
{
"input": "3",
"output": "*D*\nDDD\n*D*"
},
{
"input": "5",
"output": "**D**\n*DDD*\nDDDDD\n*DDD*\n**D**"
},
{
"input": "7",
"output": "***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***"
},
{
"input": "11",
"output": "*****D*****\n****DDD****\n***DDDDD***\n**... | 1,690,991,693 | 2,147,483,647 | Python 3 | OK | TESTS | 12 | 46 | 0 | n = int(input())
inc = d = 1
for _ in range(n):
star = (n - d) // 2
print(('*' * star) + ('D' * d) + ('*' * star))
if d == n:
inc = -1
d += (2 * inc)
| Title: Little Pony and Crystal Mine
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it.
You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.
Input Specification:
The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd).
Output Specification:
Output a crystal of size *n*.
Demo Input:
['3\n', '5\n', '7\n']
Demo Output:
['*D*\nDDD\n*D*\n', '**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n', '***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n']
Note:
none | ```python
n = int(input())
inc = d = 1
for _ in range(n):
star = (n - d) // 2
print(('*' * star) + ('D' * d) + ('*' * star))
if d == n:
inc = -1
d += (2 * inc)
``` | 3 | |
898 | A | Rounding | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it. | The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has. | Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer. | [
"5\n",
"113\n",
"1000000000\n",
"5432359\n"
] | [
"0\n",
"110\n",
"1000000000\n",
"5432360\n"
] | In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | 500 | [
{
"input": "5",
"output": "0"
},
{
"input": "113",
"output": "110"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "5432359",
"output": "5432360"
},
{
"input": "999999994",
"output": "999999990"
},
{
"input": "10",
"output": "10"
... | 1,685,457,839 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 6 | 62 | 2,764,800 | n = str(input())
if n[len(n)-1] <= '5':
print(n[0:len(n)-1]+'0')
else:
print((int(n[0:len(n)-1])+1)*10)
| Title: Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
Input Specification:
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
Output Specification:
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
Demo Input:
['5\n', '113\n', '1000000000\n', '5432359\n']
Demo Output:
['0\n', '110\n', '1000000000\n', '5432360\n']
Note:
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | ```python
n = str(input())
if n[len(n)-1] <= '5':
print(n[0:len(n)-1]+'0')
else:
print((int(n[0:len(n)-1])+1)*10)
``` | -1 | |
158 | A | Next Round | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | "Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round. | The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1). | Output the number of participants who advance to the next round. | [
"8 5\n10 9 8 7 7 7 5 5\n",
"4 2\n0 0 0 0\n"
] | [
"6\n",
"0\n"
] | In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | 500 | [
{
"input": "8 5\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "4 2\n0 0 0 0",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "1 1\n10",
"output": "1"
},
{
"input": "17 14\n16 15... | 1,696,697,972 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 92 | 0 | n,k=map(int,input().split())
list2=list(map(int,input().split()))
count=0
k=list2[k-1]
for i in range(0,len(list2)):
if list2[i]>=k and list2[i]>0:
count=count+1
print(count)
| Title: Next Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output Specification:
Output the number of participants who advance to the next round.
Demo Input:
['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n']
Demo Output:
['6\n', '0\n']
Note:
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | ```python
n,k=map(int,input().split())
list2=list(map(int,input().split()))
count=0
k=list2[k-1]
for i in range(0,len(list2)):
if list2[i]>=k and list2[i]>0:
count=count+1
print(count)
``` | 3 | |
680 | B | Bear and Finding Criminals | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation"
] | null | null | There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD. | The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city. | Print the number of criminals Limak will catch. | [
"6 3\n1 1 1 0 1 0\n",
"5 2\n0 0 0 1 0\n"
] | [
"3\n",
"1\n"
] | In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | 1,000 | [
{
"input": "6 3\n1 1 1 0 1 0",
"output": "3"
},
{
"input": "5 2\n0 0 0 1 0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "9 3\n1 1 1 1 1 1 1 1 0",
"output": "8"
},
{
"input": "9 5\n1 0 1 0 1 0... | 1,662,983,244 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | n, a = map(int, input().split())
l = list(map(int, input().split()))
c = 0
for i in range((n - a) + 1):
if l[a - 1] and i == 0: c += 1
elif l[(a - 1) + i] and l[(a - 1) - i]: c += 2
elif ((a - 1) + i) < n and ((a - 1) - i) < 0 and l[(a - 1) + i]:c += 1
elif ((a - 1) + i) >= n and ((a - 1) - i) >= 0 and l[(a - 1) - i]: c += 1
print(c) | Title: Bear and Finding Criminals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
Input Specification:
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Output Specification:
Print the number of criminals Limak will catch.
Demo Input:
['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | ```python
n, a = map(int, input().split())
l = list(map(int, input().split()))
c = 0
for i in range((n - a) + 1):
if l[a - 1] and i == 0: c += 1
elif l[(a - 1) + i] and l[(a - 1) - i]: c += 2
elif ((a - 1) + i) < n and ((a - 1) - i) < 0 and l[(a - 1) + i]:c += 1
elif ((a - 1) + i) >= n and ((a - 1) - i) >= 0 and l[(a - 1) - i]: c += 1
print(c)
``` | 0 | |
5 | A | Chat Servers Outgoing Traffic | PROGRAMMING | 1,000 | [
"implementation"
] | A. Chat Server's Outgoing Traffic | 1 | 64 | Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop and implemented a chat server that can process three types of commands:
- Include a person to the chat ('Add' command). - Remove a person from the chat ('Remove' command). - Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send' command).
Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands.
Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sends *l* bytes to each participant of the chat, where *l* is the length of the message.
As Polycarp has no time, he is asking for your help in solving this problem. | Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following:
- +<name> for 'Add' command. - -<name> for 'Remove' command. - <sender_name>:<message_text> for 'Send' command.
<name> and <sender_name> is a non-empty sequence of Latin letters and digits. <message_text> can contain letters, digits and spaces, but can't start or end with a space. <message_text> can be an empty line.
It is guaranteed, that input data are correct, i.e. there will be no 'Add' command if person with such a name is already in the chat, there will be no 'Remove' command if there is no person with such a name in the chat etc.
All names are case-sensitive. | Print a single number — answer to the problem. | [
"+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate\n",
"+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate\n"
] | [
"9\n",
"14\n"
] | none | 0 | [
{
"input": "+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate",
"output": "9"
},
{
"input": "+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate",
"output": "14"
},
{
"input": "+Dmitry\n+Mike\nDmitry:All letters will be used\nDmitry:qwertyuiopasdfghjklzxcvbnm QWERTYUIO... | 1,684,591,886 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 30 | 154 | 0 | current_users = 0
res = 0
while True:
try:
s = input()
if s[0] == '+': current_users += 1
elif s[0] == '-': current_users -= 1
else: res += len(s.split(':')[1]) * current_users
except EOFError:
print(res)
break | Title: Chat Servers Outgoing Traffic
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop and implemented a chat server that can process three types of commands:
- Include a person to the chat ('Add' command). - Remove a person from the chat ('Remove' command). - Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send' command).
Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands.
Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sends *l* bytes to each participant of the chat, where *l* is the length of the message.
As Polycarp has no time, he is asking for your help in solving this problem.
Input Specification:
Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following:
- +<name> for 'Add' command. - -<name> for 'Remove' command. - <sender_name>:<message_text> for 'Send' command.
<name> and <sender_name> is a non-empty sequence of Latin letters and digits. <message_text> can contain letters, digits and spaces, but can't start or end with a space. <message_text> can be an empty line.
It is guaranteed, that input data are correct, i.e. there will be no 'Add' command if person with such a name is already in the chat, there will be no 'Remove' command if there is no person with such a name in the chat etc.
All names are case-sensitive.
Output Specification:
Print a single number — answer to the problem.
Demo Input:
['+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate\n', '+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate\n']
Demo Output:
['9\n', '14\n']
Note:
none | ```python
current_users = 0
res = 0
while True:
try:
s = input()
if s[0] == '+': current_users += 1
elif s[0] == '-': current_users -= 1
else: res += len(s.split(':')[1]) * current_users
except EOFError:
print(res)
break
``` | 3.923 |
109 | B | Lucky Probability | PROGRAMMING | 1,900 | [
"brute force",
"probabilities"
] | B. Lucky Probability | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer *p* from the interval [*p**l*,<=*p**r*] and Vasya chooses an integer *v* from the interval [*v**l*,<=*v**r*] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [*min*(*v*,<=*p*),<=*max*(*v*,<=*p*)] contains exactly *k* lucky numbers. | The single line contains five integers *p**l*, *p**r*, *v**l*, *v**r* and *k* (1<=≤<=*p**l*<=≤<=*p**r*<=≤<=109,<=1<=≤<=*v**l*<=≤<=*v**r*<=≤<=109,<=1<=≤<=*k*<=≤<=1000). | On the single line print the result with an absolute error of no more than 10<=-<=9. | [
"1 10 1 10 2\n",
"5 6 8 10 1\n"
] | [
"0.320000000000\n",
"1.000000000000\n"
] | Consider that [*a*, *b*] denotes an interval of integers; this interval includes the boundaries. That is, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18b4a6012d95ad18891561410f0314497a578d63.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | 1,000 | [
{
"input": "1 10 1 10 2",
"output": "0.320000000000"
},
{
"input": "5 6 8 10 1",
"output": "1.000000000000"
},
{
"input": "1 20 100 120 5",
"output": "0.150000000000"
},
{
"input": "1 10 1 10 3",
"output": "0.000000000000"
},
{
"input": "1 100 1 100 2",
"outpu... | 1,668,386,570 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 92 | 0 | def is_lucky(x):
while x != 0:
n = x % 10
if n != 4 and n != 7:
return 0
x //= 10
return 1
def intersect(a1, b1, a2, b2):
if max(a1, a2) > min(b1, b2):
return 0
return min(b1, b2) - max(a1, a2) + 1
p1, p2, v1, v2, k = map(int, input().split())
l, r, t = min(p1, v1), max(p2, v2), 0
ind = [l - 1]
d1, d2 = p2 - p1 + 1, v2 - v1 + 1
prob = 0
q = 0
prob_left_p = prob_left_v = 0
for i in range(l, r):
if is_lucky(i):
ind.append(i)
if t != k:
t += 1
# if t == k:
# first, last = ind[q + 1], ind[-1]
# print(first, last)
# if p1 <= first and last <= v2:
# prob1 = intersect(p1, p2, ind[q], ind[q + 1]) / d1
# prob2 = intersect(v1, v2, ind[-2], ind[-1]) / d2
# prob += prob1 * prob2
# if v1 <= first and last <= p2:
# prob1 = intersect(v1, v2, ind[q], ind[q + 1]) / d2
# prob2 = intersect(p1, p2, ind[-2], ind[-1]) / d1
# prob += prob1 * prob2
# q += 1
if t == k:
if q != 0:
prob_right_v = intersect(v1, v2, ind[-2], ind[-1]) / d2
prob_right_p = intersect(p1, p2, ind[-2], ind[-1]) / d1
prob += prob_left_p * prob_right_v
prob += prob_left_v * prob_right_p
prob_left_p = intersect(p1, p2, ind[q] + 1, ind[q + 1]) / d1
prob_left_v = intersect(v1, v2, ind[q] + 1, ind[q + 1]) / d2
q += 1
ind.append(r)
prob_right_v = intersect(v1, v2, ind[-2], ind[-1]) / d2
prob_right_p = intersect(p1, p2, ind[-2], ind[-1]) / d1
prob += prob_left_p * prob_right_v
prob += prob_left_v * prob_right_p
print(prob)
| Title: Lucky Probability
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer *p* from the interval [*p**l*,<=*p**r*] and Vasya chooses an integer *v* from the interval [*v**l*,<=*v**r*] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [*min*(*v*,<=*p*),<=*max*(*v*,<=*p*)] contains exactly *k* lucky numbers.
Input Specification:
The single line contains five integers *p**l*, *p**r*, *v**l*, *v**r* and *k* (1<=≤<=*p**l*<=≤<=*p**r*<=≤<=109,<=1<=≤<=*v**l*<=≤<=*v**r*<=≤<=109,<=1<=≤<=*k*<=≤<=1000).
Output Specification:
On the single line print the result with an absolute error of no more than 10<=-<=9.
Demo Input:
['1 10 1 10 2\n', '5 6 8 10 1\n']
Demo Output:
['0.320000000000\n', '1.000000000000\n']
Note:
Consider that [*a*, *b*] denotes an interval of integers; this interval includes the boundaries. That is, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18b4a6012d95ad18891561410f0314497a578d63.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.
In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number. | ```python
def is_lucky(x):
while x != 0:
n = x % 10
if n != 4 and n != 7:
return 0
x //= 10
return 1
def intersect(a1, b1, a2, b2):
if max(a1, a2) > min(b1, b2):
return 0
return min(b1, b2) - max(a1, a2) + 1
p1, p2, v1, v2, k = map(int, input().split())
l, r, t = min(p1, v1), max(p2, v2), 0
ind = [l - 1]
d1, d2 = p2 - p1 + 1, v2 - v1 + 1
prob = 0
q = 0
prob_left_p = prob_left_v = 0
for i in range(l, r):
if is_lucky(i):
ind.append(i)
if t != k:
t += 1
# if t == k:
# first, last = ind[q + 1], ind[-1]
# print(first, last)
# if p1 <= first and last <= v2:
# prob1 = intersect(p1, p2, ind[q], ind[q + 1]) / d1
# prob2 = intersect(v1, v2, ind[-2], ind[-1]) / d2
# prob += prob1 * prob2
# if v1 <= first and last <= p2:
# prob1 = intersect(v1, v2, ind[q], ind[q + 1]) / d2
# prob2 = intersect(p1, p2, ind[-2], ind[-1]) / d1
# prob += prob1 * prob2
# q += 1
if t == k:
if q != 0:
prob_right_v = intersect(v1, v2, ind[-2], ind[-1]) / d2
prob_right_p = intersect(p1, p2, ind[-2], ind[-1]) / d1
prob += prob_left_p * prob_right_v
prob += prob_left_v * prob_right_p
prob_left_p = intersect(p1, p2, ind[q] + 1, ind[q + 1]) / d1
prob_left_v = intersect(v1, v2, ind[q] + 1, ind[q + 1]) / d2
q += 1
ind.append(r)
prob_right_v = intersect(v1, v2, ind[-2], ind[-1]) / d2
prob_right_p = intersect(p1, p2, ind[-2], ind[-1]) / d1
prob += prob_left_p * prob_right_v
prob += prob_left_v * prob_right_p
print(prob)
``` | 0 |
608 | B | Hamming Distance Sum | PROGRAMMING | 1,500 | [
"combinatorics",
"strings"
] | null | null | Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. | The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only. | Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. | [
"01\n00111\n",
"0011\n0110\n"
] | [
"3\n",
"2\n"
] | For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement. | 1,000 | [
{
"input": "01\n00111",
"output": "3"
},
{
"input": "0011\n0110",
"output": "2"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "0\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1... | 1,455,555,448 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 921,600 | a=str(input())
b=str(input())
count=0
al=len(a)
bl=len(b)
for i in range(al):
s=b[i:bl-al+i+1]
if a[i]=='1':
count+=s.count('0')
else:
count+=s.count('1')
print(count)
| Title: Hamming Distance Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Input Specification:
The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only.
Output Specification:
Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Demo Input:
['01\n00111\n', '0011\n0110\n']
Demo Output:
['3\n', '2\n']
Note:
For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement. | ```python
a=str(input())
b=str(input())
count=0
al=len(a)
bl=len(b)
for i in range(al):
s=b[i:bl-al+i+1]
if a[i]=='1':
count+=s.count('0')
else:
count+=s.count('1')
print(count)
``` | 0 | |
496 | A | Minimum Difficulty | PROGRAMMING | 900 | [
"brute force",
"implementation",
"math"
] | null | null | Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold. | The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one). | Print a single number — the minimum difficulty of the track after removing a single hold. | [
"3\n1 4 6\n",
"5\n1 2 3 4 5\n",
"5\n1 2 3 7 8\n"
] | [
"5\n",
"2\n",
"4\n"
] | In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4. | 500 | [
{
"input": "3\n1 4 6",
"output": "5"
},
{
"input": "5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5\n1 2 3 7 8",
"output": "4"
},
{
"input": "3\n1 500 1000",
"output": "999"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "2"
},
{
"input": "10\n1 4 9... | 1,644,935,172 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 46 | 0 |
def solve(n,arr):
mn=1000
for i in arr[1:-1] :
l2=list(arr)
l2.remove(i)
mx=0
for j in range(n-2) :
mx= max(mx , l2[j+1]-l2[j])
mn=min(mn,mx)
return mn
from sys import stdin
input = stdin.readline
n= int(input())
l=[int(x) for x in input().split()]
print(solve(n,l))
| Title: Minimum Difficulty
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
Input Specification:
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
Output Specification:
Print a single number — the minimum difficulty of the track after removing a single hold.
Demo Input:
['3\n1 4 6\n', '5\n1 2 3 4 5\n', '5\n1 2 3 7 8\n']
Demo Output:
['5\n', '2\n', '4\n']
Note:
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4. | ```python
def solve(n,arr):
mn=1000
for i in arr[1:-1] :
l2=list(arr)
l2.remove(i)
mx=0
for j in range(n-2) :
mx= max(mx , l2[j+1]-l2[j])
mn=min(mn,mx)
return mn
from sys import stdin
input = stdin.readline
n= int(input())
l=[int(x) for x in input().split()]
print(solve(n,l))
``` | 3 | |
120 | B | Quiz League | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | A team quiz game called "What? Where? When?" is very popular in Berland. The game is centered on two teams competing. They are the team of six Experts versus the team of the Audience. A person from the audience asks a question and the experts are allowed a minute on brainstorming and finding the right answer to the question. All it takes to answer a typical question is general knowledge and common logic. The question sent be the audience are in envelops lain out in a circle on a round table. Each envelop is marked by the name of the asker's town. Each question is positioned in a separate sector. In the centre of the table is a spinning arrow. Thus, the table rather resembles a roulette table with no ball but with a spinning arrow instead. The host sets off the spinning arrow to choose a question for the experts: when the arrow stops spinning, the question it is pointing at is chosen. If the arrow points at the question that has already been asked, the host chooses the next unanswered question in the clockwise direction. Your task is to determine which will be the number of the next asked question if the arrow points at sector number *k*. | The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=1000 and 1<=≤<=*k*<=≤<=*n*) — the numbers of sectors on the table and the number of the sector where the arrow is pointing. The second line contains *n* numbers: *a**i*<==<=0 if the question from sector *i* has already been asked and *a**i*<==<=1 if the question from sector *i* hasn't been asked yet (1<=≤<=*i*<=≤<=*n*). The sectors are given in the clockwise order, the first sector follows after the *n*-th one. | Print the single number — the number of the sector containing the question the experts will be asked. It is guaranteed that the answer exists, that is that not all the questions have already been asked. | [
"5 5\n0 1 0 1 0\n",
"2 1\n1 1\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "5 5\n0 1 0 1 0",
"output": "2"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "3 2\n1 0 0",
"output": "1"
},
{
"input": "3 3\n0 1 0",
"output": "2"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "6 3\n0 0 1 1 0 1",
"output":... | 1,586,351,740 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 78 | 0 | n,k = map(int,input().split())
n = list(map(int,input().split()))
k = k - 1
while(1):
if(n[k] == 0):
k = k + 1
k = k % len(n)
if(n[k] == 1):
break
print(k+1) | Title: Quiz League
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A team quiz game called "What? Where? When?" is very popular in Berland. The game is centered on two teams competing. They are the team of six Experts versus the team of the Audience. A person from the audience asks a question and the experts are allowed a minute on brainstorming and finding the right answer to the question. All it takes to answer a typical question is general knowledge and common logic. The question sent be the audience are in envelops lain out in a circle on a round table. Each envelop is marked by the name of the asker's town. Each question is positioned in a separate sector. In the centre of the table is a spinning arrow. Thus, the table rather resembles a roulette table with no ball but with a spinning arrow instead. The host sets off the spinning arrow to choose a question for the experts: when the arrow stops spinning, the question it is pointing at is chosen. If the arrow points at the question that has already been asked, the host chooses the next unanswered question in the clockwise direction. Your task is to determine which will be the number of the next asked question if the arrow points at sector number *k*.
Input Specification:
The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=1000 and 1<=≤<=*k*<=≤<=*n*) — the numbers of sectors on the table and the number of the sector where the arrow is pointing. The second line contains *n* numbers: *a**i*<==<=0 if the question from sector *i* has already been asked and *a**i*<==<=1 if the question from sector *i* hasn't been asked yet (1<=≤<=*i*<=≤<=*n*). The sectors are given in the clockwise order, the first sector follows after the *n*-th one.
Output Specification:
Print the single number — the number of the sector containing the question the experts will be asked. It is guaranteed that the answer exists, that is that not all the questions have already been asked.
Demo Input:
['5 5\n0 1 0 1 0\n', '2 1\n1 1\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
n,k = map(int,input().split())
n = list(map(int,input().split()))
k = k - 1
while(1):
if(n[k] == 0):
k = k + 1
k = k % len(n)
if(n[k] == 1):
break
print(k+1)
``` | -1 | |
919 | D | Substring | PROGRAMMING | 1,700 | [
"dfs and similar",
"dp",
"graphs"
] | null | null | You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest. | The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges.
The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node.
Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected. | Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead. | [
"5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n",
"6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n",
"10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n"
] | [
"3\n",
"-1\n",
"4\n"
] | In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times. | 1,500 | [
{
"input": "5 4\nabaca\n1 2\n1 3\n3 4\n4 5",
"output": "3"
},
{
"input": "6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4",
"output": "-1"
},
{
"input": "10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7",
"output": "4"
},
{
"input": "1 1\nf... | 1,688,239,091 | 2,147,483,647 | PyPy 3-64 | MEMORY_LIMIT_EXCEEDED | TESTS | 5 | 982 | 268,390,400 | """
Дан граф из n вершин и m ориентированных ребер. В каждой вершине записана некоторая строчная
латинская буква. Определим величину пути как наибольшее количество раз, которое какая-то буква
встречалась на этом пути. Например, если буквы на пути образуют строку «abaca», то величина
этого пути равна 3. Ваша задача — найти путь с наибольшей величиной.
"""
from collections import deque
import copy
class Node:
def __init__(self):
self.adjacent_out = list()
self.indegree = 0
self.lengths = dict()
for i in 'qwertyuiopasdfghjklzxcvbnm':
self.lengths[i] = 0
class Graph:
def __init__(self, N, string):
self.vertices = dict()
self.string = ' ' + string
self.roots = set()
self.max = 0
self.N = N
for i in range(1, N + 1):
self.vertices[i] = Node()
def add_edge(self, a, b):
self.vertices[a].adjacent_out += [b]
self.vertices[b].indegree += 1
if b in self.roots:
self.roots.remove(b)
if not self.vertices[a].indegree:
self.roots.add(a)
def topo_sort(self, queue=deque()):
for i in self.roots: # база
self.vertices[i].lengths[self.string[i]] += 1
proceeded = 0
while queue:
q = queue.popleft()
proceeded += 1
for neighbour in self.vertices[q].adjacent_out:
self.vertices[neighbour].indegree -= 1
self.vertices[neighbour].lengths = copy.deepcopy(self.vertices[q].lengths)
self.vertices[neighbour].lengths[self.string[neighbour]] += 1
if self.max < self.vertices[neighbour].lengths[self.string[neighbour]]:
self.max = self.vertices[neighbour].lengths[self.string[neighbour]]
if not self.vertices[neighbour].indegree:
queue.append(neighbour)
if proceeded != self.N:
self.max = -1
def get_max(self):
self.topo_sort(deque(self.roots))
return self.max
N, M = map(int, input().split())
string = input()
graph = Graph(N, string)
for _ in range(M):
graph.add_edge(*list(map(int, input().split())))
print(graph.get_max()) | Title: Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a graph with $n$ nodes and $m$ directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is $3$. Your task is find a path whose value is the largest.
Input Specification:
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), denoting that the graph has $n$ nodes and $m$ directed edges.
The second line contains a string $s$ with only lowercase English letters. The $i$-th character is the letter assigned to the $i$-th node.
Then $m$ lines follow. Each line contains two integers $x, y$ ($1 \leq x, y \leq n$), describing a directed edge from $x$ to $y$. Note that $x$ can be equal to $y$ and there can be multiple edges between $x$ and $y$. Also the graph can be not connected.
Output Specification:
Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.
Demo Input:
['5 4\nabaca\n1 2\n1 3\n3 4\n4 5\n', '6 6\nxzyabc\n1 2\n3 1\n2 3\n5 4\n4 3\n6 4\n', '10 14\nxzyzyzyzqx\n1 2\n2 4\n3 5\n4 5\n2 6\n6 8\n6 5\n2 10\n3 9\n10 9\n4 6\n1 10\n2 8\n3 7\n']
Demo Output:
['3\n', '-1\n', '4\n']
Note:
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times. | ```python
"""
Дан граф из n вершин и m ориентированных ребер. В каждой вершине записана некоторая строчная
латинская буква. Определим величину пути как наибольшее количество раз, которое какая-то буква
встречалась на этом пути. Например, если буквы на пути образуют строку «abaca», то величина
этого пути равна 3. Ваша задача — найти путь с наибольшей величиной.
"""
from collections import deque
import copy
class Node:
def __init__(self):
self.adjacent_out = list()
self.indegree = 0
self.lengths = dict()
for i in 'qwertyuiopasdfghjklzxcvbnm':
self.lengths[i] = 0
class Graph:
def __init__(self, N, string):
self.vertices = dict()
self.string = ' ' + string
self.roots = set()
self.max = 0
self.N = N
for i in range(1, N + 1):
self.vertices[i] = Node()
def add_edge(self, a, b):
self.vertices[a].adjacent_out += [b]
self.vertices[b].indegree += 1
if b in self.roots:
self.roots.remove(b)
if not self.vertices[a].indegree:
self.roots.add(a)
def topo_sort(self, queue=deque()):
for i in self.roots: # база
self.vertices[i].lengths[self.string[i]] += 1
proceeded = 0
while queue:
q = queue.popleft()
proceeded += 1
for neighbour in self.vertices[q].adjacent_out:
self.vertices[neighbour].indegree -= 1
self.vertices[neighbour].lengths = copy.deepcopy(self.vertices[q].lengths)
self.vertices[neighbour].lengths[self.string[neighbour]] += 1
if self.max < self.vertices[neighbour].lengths[self.string[neighbour]]:
self.max = self.vertices[neighbour].lengths[self.string[neighbour]]
if not self.vertices[neighbour].indegree:
queue.append(neighbour)
if proceeded != self.N:
self.max = -1
def get_max(self):
self.topo_sort(deque(self.roots))
return self.max
N, M = map(int, input().split())
string = input()
graph = Graph(N, string)
for _ in range(M):
graph.add_edge(*list(map(int, input().split())))
print(graph.get_max())
``` | 0 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4... | 1,620,669,481 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 14,438,400 | n=int(input())
l=list(map(int,input().split()))
q=int(input())
ql=list(map(int,input().split()))
a,b=0,0
for j in ql:
for i in l:
a+=1
if j==i:
break
l.reverse()
for j in ql:
for i in l:
b+=1
if j==i:
break
print(a,'',b)
| Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
n=int(input())
l=list(map(int,input().split()))
q=int(input())
ql=list(map(int,input().split()))
a,b=0,0
for j in ql:
for i in l:
a+=1
if j==i:
break
l.reverse()
for j in ql:
for i in l:
b+=1
if j==i:
break
print(a,'',b)
``` | 0 | |
109 | A | Lucky Sum of Digits | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | A. Lucky Sum of Digits | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. | The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. | Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. | [
"11\n",
"10\n"
] | [
"47\n",
"-1\n"
] | none | 500 | [
{
"input": "11",
"output": "47"
},
{
"input": "10",
"output": "-1"
},
{
"input": "64",
"output": "4477777777"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "7"
},
{
"input": "12",
"outpu... | 1,631,427,441 | 2,147,483,647 | PyPy 3 | OK | TESTS | 51 | 404 | 29,184,000 | # Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout
# -------------- INPUT FUNCTIONS ------------------
def get_ints_in_variables(): return map(
int, sys.stdin.readline().strip().split())
def get_int(): return int(sys.stdin.readline())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
# -------- SOME CUSTOMIZED FUNCTIONS-----------
def myceil(x, y): return (x + y - 1) // y
# -------------- SOLUTION FUNCTION ------------------
def Solution(n):
# Write Your Code Here
sevens = n//7
sevenRem = n % 7
fours = sevenRem//4
foursRem = sevenRem % 4
if foursRem <= sevens:
sevens -= foursRem
fours += (2*foursRem)
for _ in range(fours):
print(4, end="")
for _ in range(sevens):
print(7, end="")
print()
else:
print(-1)
def main():
# Take input Here and Call solution function
Solution(get_int())
# calling main Function
if __name__ == '__main__':
main()
| Title: Lucky Sum of Digits
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Output Specification:
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
Demo Input:
['11\n', '10\n']
Demo Output:
['47\n', '-1\n']
Note:
none | ```python
# Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout
# -------------- INPUT FUNCTIONS ------------------
def get_ints_in_variables(): return map(
int, sys.stdin.readline().strip().split())
def get_int(): return int(sys.stdin.readline())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
# -------- SOME CUSTOMIZED FUNCTIONS-----------
def myceil(x, y): return (x + y - 1) // y
# -------------- SOLUTION FUNCTION ------------------
def Solution(n):
# Write Your Code Here
sevens = n//7
sevenRem = n % 7
fours = sevenRem//4
foursRem = sevenRem % 4
if foursRem <= sevens:
sevens -= foursRem
fours += (2*foursRem)
for _ in range(fours):
print(4, end="")
for _ in range(sevens):
print(7, end="")
print()
else:
print(-1)
def main():
# Take input Here and Call solution function
Solution(get_int())
# calling main Function
if __name__ == '__main__':
main()
``` | 3.844641 |
161 | A | Dress'em in Vests! | PROGRAMMING | 1,300 | [
"binary search",
"brute force",
"greedy",
"two pointers"
] | null | null | The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*. | The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests. | In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them. | [
"5 3 0 0\n1 2 3 3 4\n1 3 5\n",
"3 3 2 2\n1 5 9\n3 5 7\n"
] | [
"2\n1 1\n3 2\n",
"3\n1 1\n2 2\n3 3\n"
] | In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped. | 1,000 | [
{
"input": "5 3 0 0\n1 2 3 3 4\n1 3 5",
"output": "2\n1 1\n3 2"
},
{
"input": "3 3 2 2\n1 5 9\n3 5 7",
"output": "3\n1 1\n2 2\n3 3"
},
{
"input": "1 1 0 0\n1\n1",
"output": "1\n1 1"
},
{
"input": "1 1 0 0\n1\n2",
"output": "0"
},
{
"input": "2 3 1 4\n1 5\n1 2 2",
... | 1,580,887,576 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 3,000 | 10,137,600 | class Pair:
def __init__(self, u, v):
self.u = u
self.v = v
def printData(self):
print("{0} {1}".format(self.u, self.v))
def solution():
n, m, x, y = map(int, input().split())
s = list(map(int, input().split()))
a = list(map(int, input().split()))
result = []
sIndex = 0
aIndex = 0
previousIndex = 0
while(sIndex < n):
u = s[sIndex]
while(aIndex < m):
v = a[aIndex]
if(u - x <= v and u + y >= v):
result.append(Pair(sIndex + 1, aIndex + 1))
previousIndex = aIndex
aIndex += 1
break
else:
aIndex += 1
if(aIndex == m):
aIndex = previousIndex + 1
if(len(result) == m):
break
sIndex += 1
l = len(result)
print(l)
for i in range(l):
result[i].printData()
solution()
| Title: Dress'em in Vests!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
Input Specification:
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
Output Specification:
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
Demo Input:
['5 3 0 0\n1 2 3 3 4\n1 3 5\n', '3 3 2 2\n1 5 9\n3 5 7\n']
Demo Output:
['2\n1 1\n3 2\n', '3\n1 1\n2 2\n3 3\n']
Note:
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped. | ```python
class Pair:
def __init__(self, u, v):
self.u = u
self.v = v
def printData(self):
print("{0} {1}".format(self.u, self.v))
def solution():
n, m, x, y = map(int, input().split())
s = list(map(int, input().split()))
a = list(map(int, input().split()))
result = []
sIndex = 0
aIndex = 0
previousIndex = 0
while(sIndex < n):
u = s[sIndex]
while(aIndex < m):
v = a[aIndex]
if(u - x <= v and u + y >= v):
result.append(Pair(sIndex + 1, aIndex + 1))
previousIndex = aIndex
aIndex += 1
break
else:
aIndex += 1
if(aIndex == m):
aIndex = previousIndex + 1
if(len(result) == m):
break
sIndex += 1
l = len(result)
print(l)
for i in range(l):
result[i].printData()
solution()
``` | 0 | |
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": "... | 1,699,933,502 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 46 | 0 | xa, ya, xb, yb = [int(x) for x in input().split(' ')]
xv = xa-xb if xa > xb else xb-xa
yv = ya - yb if ya > yb else yb - ya
# def torr():
if xa == xb or ya == yb:
print(1, end=' ')
else:
print(2, end=' ')
# def alf():
if ((xa + ya)%2 == 0) != ((xb + yb)%2 == 0):
print(0, end=' ')
# return
elif xv == yv:
print(1, end=' ')
else:
print(2, end=' ')
# def king():
print(max(xv, yv), end='')
| 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
xa, ya, xb, yb = [int(x) for x in input().split(' ')]
xv = xa-xb if xa > xb else xb-xa
yv = ya - yb if ya > yb else yb - ya
# def torr():
if xa == xb or ya == yb:
print(1, end=' ')
else:
print(2, end=' ')
# def alf():
if ((xa + ya)%2 == 0) != ((xb + yb)%2 == 0):
print(0, end=' ')
# return
elif xv == yv:
print(1, end=' ')
else:
print(2, end=' ')
# def king():
print(max(xv, yv), end='')
``` | 3 | |
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... | 1,554,622,584 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 218 | 0 | s = input()
s1 = ''
k = 0
for c in s:
if 'A' <= c <= 'Z':
k += 1
if k <= len(s) - k:
for i in range(len(s)):
if 'A' <= s[i] <= 'Z':
s1 += s[i].lower()
else:
s1 += s[i]
else:
for i in range(len(s)):
if 'a' <= s[i] <= 'z':
s1 += s[i].upper()
else:
s1 += s[i]
print(s1) | 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()
s1 = ''
k = 0
for c in s:
if 'A' <= c <= 'Z':
k += 1
if k <= len(s) - k:
for i in range(len(s)):
if 'A' <= s[i] <= 'Z':
s1 += s[i].lower()
else:
s1 += s[i]
else:
for i in range(len(s)):
if 'a' <= s[i] <= 'z':
s1 += s[i].upper()
else:
s1 += s[i]
print(s1)
``` | 3.9455 |
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... | 1,569,233,416 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 748 | 29,286,400 | import heapq as hq
Inf = float("inf")
def Dijkstra(n, adj, src):
pq=[]
d=[Inf]*(n+1)
parent=[None]*(n+1)
d[src]=0
hq.heappush(pq,(d[src],src))
while pq:
_,u=hq.heappop(pq)
for v,w in adj[u]:
if d[v]>d[u]+w:
d[v]=d[u]+w
parent[v]=u
hq.heappush(pq,(d[v],v))
return d,parent
n,m=map(int,input().split())
adj={}
for i in range(1,n+1):
adj[i]=[]
for _ in range(m):
u,v,w=map(int,input().split())
adj[u].append([v,w])
adj[v].append([u,w])
d,parent=Dijkstra(n,adj,1)
if d[n]==Inf:
print ("-1")
else:
path=[]
x=n
while x is not None:
path.append(x)
x=parent[x]
l= len(path)
for i in range(l):
print(path[l-i-1],end=" ")
| 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
import heapq as hq
Inf = float("inf")
def Dijkstra(n, adj, src):
pq=[]
d=[Inf]*(n+1)
parent=[None]*(n+1)
d[src]=0
hq.heappush(pq,(d[src],src))
while pq:
_,u=hq.heappop(pq)
for v,w in adj[u]:
if d[v]>d[u]+w:
d[v]=d[u]+w
parent[v]=u
hq.heappush(pq,(d[v],v))
return d,parent
n,m=map(int,input().split())
adj={}
for i in range(1,n+1):
adj[i]=[]
for _ in range(m):
u,v,w=map(int,input().split())
adj[u].append([v,w])
adj[v].append([u,w])
d,parent=Dijkstra(n,adj,1)
if d[n]==Inf:
print ("-1")
else:
path=[]
x=n
while x is not None:
path.append(x)
x=parent[x]
l= len(path)
for i in range(l):
print(path[l-i-1],end=" ")
``` | 3.407799 |
450 | A | Jzzhu and Children | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order? | The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | Output a single integer, representing the number of the last child. | [
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] | [
"4\n",
"6\n"
] | Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | 500 | [
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18... | 1,644,496,769 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 31 | 0 | x,y=map(int,input().split())
sozluk={k+1:w for k,w in enumerate(map(int,input().split()))}
en_buyuk=sozluk[1]/y
konum=1
for i in range(2,x+1):
if(sozluk[i]/y>en_buyuk):
en_buyuk=sozluk[i]/y
konum=i
else:
if(en_buyuk==int(en_buyuk) and sozluk[i]/y>en_buyuk-1):
en_buyuk = sozluk[i] / y
konum = i
elif(en_buyuk!=int(en_buyuk) and (sozluk[i]/y)>int(en_buyuk)):
en_buyuk = sozluk[i] / y
konum = i
print(konum) | Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | ```python
x,y=map(int,input().split())
sozluk={k+1:w for k,w in enumerate(map(int,input().split()))}
en_buyuk=sozluk[1]/y
konum=1
for i in range(2,x+1):
if(sozluk[i]/y>en_buyuk):
en_buyuk=sozluk[i]/y
konum=i
else:
if(en_buyuk==int(en_buyuk) and sozluk[i]/y>en_buyuk-1):
en_buyuk = sozluk[i] / y
konum = i
elif(en_buyuk!=int(en_buyuk) and (sozluk[i]/y)>int(en_buyuk)):
en_buyuk = sozluk[i] / y
konum = i
print(konum)
``` | 3 | |
934 | A | A Compatible Pair | PROGRAMMING | 1,400 | [
"brute force",
"games"
] | null | null | Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109. | Print a single integer — the brightness of the chosen pair. | [
"2 2\n20 18\n2 14\n",
"5 3\n-1 0 1 2 3\n-1 0 1\n"
] | [
"252\n",
"2\n"
] | In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself. | 500 | [
{
"input": "2 2\n20 18\n2 14",
"output": "252"
},
{
"input": "5 3\n-1 0 1 2 3\n-1 0 1",
"output": "2"
},
{
"input": "10 2\n1 6 2 10 2 3 2 10 6 4\n5 7",
"output": "70"
},
{
"input": "50 50\n1 6 2 10 2 3 2 10 6 4 5 0 3 1 7 3 2 4 4 2 1 5 0 6 10 1 8 0 10 9 0 4 10 5 5 7 4 9 9 5 5 ... | 1,518,627,952 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 62 | 5,632,000 | n,m=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
a.remove(max(a))
z=max(b)
print(z*max(a))
| Title: A Compatible Pair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109.
Output Specification:
Print a single integer — the brightness of the chosen pair.
Demo Input:
['2 2\n20 18\n2 14\n', '5 3\n-1 0 1 2 3\n-1 0 1\n']
Demo Output:
['252\n', '2\n']
Note:
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself. | ```python
n,m=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
a.remove(max(a))
z=max(b)
print(z*max(a))
``` | 0 | |
621 | D | Rat Kwesh and Cheese | PROGRAMMING | 2,400 | [
"brute force",
"constructive algorithms",
"math"
] | null | null | Wet Shark asked Rat Kwesh to generate three positive real numbers *x*, *y* and *z*, from 0.1 to 200.0, inclusive. Wet Krash wants to impress Wet Shark, so all generated numbers will have exactly one digit after the decimal point.
Wet Shark knows Rat Kwesh will want a lot of cheese. So he will give the Rat an opportunity to earn a lot of cheese. He will hand the three numbers *x*, *y* and *z* to Rat Kwesh, and Rat Kwesh will pick one of the these twelve options:
1. *a*1<==<=*x**y**z*; 1. *a*2<==<=*x**z**y*; 1. *a*3<==<=(*x**y*)*z*; 1. *a*4<==<=(*x**z*)*y*; 1. *a*5<==<=*y**x**z*; 1. *a*6<==<=*y**z**x*; 1. *a*7<==<=(*y**x*)*z*; 1. *a*8<==<=(*y**z*)*x*; 1. *a*9<==<=*z**x**y*; 1. *a*10<==<=*z**y**x*; 1. *a*11<==<=(*z**x*)*y*; 1. *a*12<==<=(*z**y*)*x*.
Let *m* be the maximum of all the *a**i*, and *c* be the smallest index (from 1 to 12) such that *a**c*<==<=*m*. Rat's goal is to find that *c*, and he asks you to help him. Rat Kwesh wants to see how much cheese he gets, so he you will have to print the expression corresponding to that *a**c*. | The only line of the input contains three space-separated real numbers *x*, *y* and *z* (0.1<=≤<=*x*,<=*y*,<=*z*<=≤<=200.0). Each of *x*, *y* and *z* is given with exactly one digit after the decimal point. | Find the maximum value of expression among *x**y**z*, *x**z**y*, (*x**y*)*z*, (*x**z*)*y*, *y**x**z*, *y**z**x*, (*y**x*)*z*, (*y**z*)*x*, *z**x**y*, *z**y**x*, (*z**x*)*y*, (*z**y*)*x* and print the corresponding expression. If there are many maximums, print the one that comes first in the list.
*x**y**z* should be outputted as x^y^z (without brackets), and (*x**y*)*z* should be outputted as (x^y)^z (quotes for clarity). | [
"1.1 3.4 2.5\n",
"2.0 2.0 2.0\n",
"1.9 1.8 1.7\n"
] | [
"z^y^x\n",
"x^y^z\n",
"(x^y)^z\n"
] | none | 2,000 | [
{
"input": "1.1 3.4 2.5",
"output": "z^y^x"
},
{
"input": "2.0 2.0 2.0",
"output": "x^y^z"
},
{
"input": "1.9 1.8 1.7",
"output": "(x^y)^z"
},
{
"input": "2.0 2.1 2.2",
"output": "x^z^y"
},
{
"input": "1.5 1.7 2.5",
"output": "(z^x)^y"
},
{
"input": "1... | 1,536,270,998 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
#define ld long double
#define ll long long int
using namespace std;
int main(){
ld x, y, z;
cin >> x >> y >> z;
ld ans[15] = {0.0};
ans[1] = pow(y, z)*log(x);
ans[2] = pow(z, y)*log(x); //c*(log(pow(a, b)));
ans[3] = z*log(pow(x, y));
ans[4] = y*log(pow(x, z));
ans[5] = pow(x, z)*log(y);
ans[6] = pow(z, x)*log(y);
ans[7] = z*log(pow(y, x));
ans[8] = x*log(pow(y, z));
ans[9] = pow(x, y)*log(z);
ans[10] = pow(y, x)*log(z);
ans[11] = y*log(pow(z, x));
ans[12] = x*log(pow(z, y));
//cout << ans1 << " " << ans2 << " " << ans3 << " " << ans4 << " " << ans5 << " " << ans6 << " " << ans7 << " " << ans8 << " " << ans9 << " " << ans10 << " " << ans11 << " " << ans12 << endl;
ld mx = ans[1];//// = max({ans1, ans2, ans3, ans4, ans5, ans6, ans7, ans8, ans9, ans10, ans11, ans12});
// for(int i = 1; i <= 12; ++i)cout << ans[i] << " ";
// cout << endl;
for(int i = 1; i <= 12; ++i)
mx = max(mx, ans[i]);
// cout << mx << endl;
for(int i = 1; i <= 12; ++i){
if(abs(mx - ans[i]) < 1e-9){
if(i == 1){
cout << "x^y^z" << endl;
}
else if(i == 2){
cout << "x^z^y" << endl;
}
else if(i == 3){
cout << "(x^y)^z" << endl;
}
else if(i == 4){
cout << "(x^z)^y" << endl;
}
else if(i == 5){
cout << "y^x^z" << endl;
}
else if(i == 6){
cout << "y^z^x" << endl;
}
else if(i == 7){
cout << "(y^x)^z" << endl;
}
else if(i == 8){
cout << "(y^z)^x" << endl;
}
else if(i == 9){
cout << "z^x^y" << endl;
}
else if(i == 10){
cout << "z^y^x" << endl;
}
else if(i == 11){
cout << "(z^x)^y" << endl;
}
else if(i == 12){
cout << "(z^y)^x" << endl;
}
return 0;
}
}
return 0;
} | Title: Rat Kwesh and Cheese
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Wet Shark asked Rat Kwesh to generate three positive real numbers *x*, *y* and *z*, from 0.1 to 200.0, inclusive. Wet Krash wants to impress Wet Shark, so all generated numbers will have exactly one digit after the decimal point.
Wet Shark knows Rat Kwesh will want a lot of cheese. So he will give the Rat an opportunity to earn a lot of cheese. He will hand the three numbers *x*, *y* and *z* to Rat Kwesh, and Rat Kwesh will pick one of the these twelve options:
1. *a*1<==<=*x**y**z*; 1. *a*2<==<=*x**z**y*; 1. *a*3<==<=(*x**y*)*z*; 1. *a*4<==<=(*x**z*)*y*; 1. *a*5<==<=*y**x**z*; 1. *a*6<==<=*y**z**x*; 1. *a*7<==<=(*y**x*)*z*; 1. *a*8<==<=(*y**z*)*x*; 1. *a*9<==<=*z**x**y*; 1. *a*10<==<=*z**y**x*; 1. *a*11<==<=(*z**x*)*y*; 1. *a*12<==<=(*z**y*)*x*.
Let *m* be the maximum of all the *a**i*, and *c* be the smallest index (from 1 to 12) such that *a**c*<==<=*m*. Rat's goal is to find that *c*, and he asks you to help him. Rat Kwesh wants to see how much cheese he gets, so he you will have to print the expression corresponding to that *a**c*.
Input Specification:
The only line of the input contains three space-separated real numbers *x*, *y* and *z* (0.1<=≤<=*x*,<=*y*,<=*z*<=≤<=200.0). Each of *x*, *y* and *z* is given with exactly one digit after the decimal point.
Output Specification:
Find the maximum value of expression among *x**y**z*, *x**z**y*, (*x**y*)*z*, (*x**z*)*y*, *y**x**z*, *y**z**x*, (*y**x*)*z*, (*y**z*)*x*, *z**x**y*, *z**y**x*, (*z**x*)*y*, (*z**y*)*x* and print the corresponding expression. If there are many maximums, print the one that comes first in the list.
*x**y**z* should be outputted as x^y^z (without brackets), and (*x**y*)*z* should be outputted as (x^y)^z (quotes for clarity).
Demo Input:
['1.1 3.4 2.5\n', '2.0 2.0 2.0\n', '1.9 1.8 1.7\n']
Demo Output:
['z^y^x\n', 'x^y^z\n', '(x^y)^z\n']
Note:
none | ```python
#include <bits/stdc++.h>
#define ld long double
#define ll long long int
using namespace std;
int main(){
ld x, y, z;
cin >> x >> y >> z;
ld ans[15] = {0.0};
ans[1] = pow(y, z)*log(x);
ans[2] = pow(z, y)*log(x); //c*(log(pow(a, b)));
ans[3] = z*log(pow(x, y));
ans[4] = y*log(pow(x, z));
ans[5] = pow(x, z)*log(y);
ans[6] = pow(z, x)*log(y);
ans[7] = z*log(pow(y, x));
ans[8] = x*log(pow(y, z));
ans[9] = pow(x, y)*log(z);
ans[10] = pow(y, x)*log(z);
ans[11] = y*log(pow(z, x));
ans[12] = x*log(pow(z, y));
//cout << ans1 << " " << ans2 << " " << ans3 << " " << ans4 << " " << ans5 << " " << ans6 << " " << ans7 << " " << ans8 << " " << ans9 << " " << ans10 << " " << ans11 << " " << ans12 << endl;
ld mx = ans[1];//// = max({ans1, ans2, ans3, ans4, ans5, ans6, ans7, ans8, ans9, ans10, ans11, ans12});
// for(int i = 1; i <= 12; ++i)cout << ans[i] << " ";
// cout << endl;
for(int i = 1; i <= 12; ++i)
mx = max(mx, ans[i]);
// cout << mx << endl;
for(int i = 1; i <= 12; ++i){
if(abs(mx - ans[i]) < 1e-9){
if(i == 1){
cout << "x^y^z" << endl;
}
else if(i == 2){
cout << "x^z^y" << endl;
}
else if(i == 3){
cout << "(x^y)^z" << endl;
}
else if(i == 4){
cout << "(x^z)^y" << endl;
}
else if(i == 5){
cout << "y^x^z" << endl;
}
else if(i == 6){
cout << "y^z^x" << endl;
}
else if(i == 7){
cout << "(y^x)^z" << endl;
}
else if(i == 8){
cout << "(y^z)^x" << endl;
}
else if(i == 9){
cout << "z^x^y" << endl;
}
else if(i == 10){
cout << "z^y^x" << endl;
}
else if(i == 11){
cout << "(z^x)^y" << endl;
}
else if(i == 12){
cout << "(z^y)^x" << endl;
}
return 0;
}
}
return 0;
}
``` | -1 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,651,076,722 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 46 | 0 | i=input()
j=input()
for x,y in zip(i,j):
if x==y:
print('0',end='')
else:
print('1',end='') | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
i=input()
j=input()
for x,y in zip(i,j):
if x==y:
print('0',end='')
else:
print('1',end='')
``` | 3.9885 |
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 ... | 1,686,205,865 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | global grid
global n
def printi():
for i in range(n):
for j in range(n):
print(f"{grid[i][j]}", end=" ")
# print(f"{grid[i][j]:{len(str(n*n))}}", end=" ")
print("")
def walk_grid(x, y, value):
x_walk = x
y_walk = y
# Left
while x_walk > y:
grid[y_walk][x_walk] = value
grid[n - 1 - y_walk][n - 1 - x_walk] = n * n - value + 1
value += 1
x_walk -= 1
# Down
while y_walk < n - y:
grid[y_walk][x_walk] = value
grid[n - 1 - y_walk][n - 1 - x_walk] = n * n - value + 1
value += 1
y_walk += 1
# Right
y_walk -= 1
x_walk += 1
while x_walk < x - y:
grid[y_walk][x_walk] = value
grid[n - 1 - y_walk][n - 1 - x_walk] = n * n - value + 1
value += 1
x_walk += 1
return value
n = int(input())
mid = n//2
grid = [[0]*n for _ in range(n)]
grid[mid][mid] = n * n // 2 + 1
value = 1
for i in range(0, mid):
y = i
value = walk_grid(mid, y, value)
printi()
| 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
global grid
global n
def printi():
for i in range(n):
for j in range(n):
print(f"{grid[i][j]}", end=" ")
# print(f"{grid[i][j]:{len(str(n*n))}}", end=" ")
print("")
def walk_grid(x, y, value):
x_walk = x
y_walk = y
# Left
while x_walk > y:
grid[y_walk][x_walk] = value
grid[n - 1 - y_walk][n - 1 - x_walk] = n * n - value + 1
value += 1
x_walk -= 1
# Down
while y_walk < n - y:
grid[y_walk][x_walk] = value
grid[n - 1 - y_walk][n - 1 - x_walk] = n * n - value + 1
value += 1
y_walk += 1
# Right
y_walk -= 1
x_walk += 1
while x_walk < x - y:
grid[y_walk][x_walk] = value
grid[n - 1 - y_walk][n - 1 - x_walk] = n * n - value + 1
value += 1
x_walk += 1
return value
n = int(input())
mid = n//2
grid = [[0]*n for _ in range(n)]
grid[mid][mid] = n * n // 2 + 1
value = 1
for i in range(0, mid):
y = i
value = walk_grid(mid, y, value)
printi()
``` | 0 | |
452 | E | Three strings | PROGRAMMING | 2,400 | [
"data structures",
"dsu",
"string suffix structures",
"strings"
] | null | null | You are given three strings (*s*1,<=*s*2,<=*s*3). For each integer *l* (1<=≤<=*l*<=≤<=*min*(|*s*1|,<=|*s*2|,<=|*s*3|) you need to find how many triples (*i*1,<=*i*2,<=*i*3) exist such that three strings *s**k*[*i**k*... *i**k*<=+<=*l*<=-<=1] (*k*<==<=1,<=2,<=3) are pairwise equal. Print all found numbers modulo 1000000007 (109<=+<=7).
See notes if you are not sure about some of the denotions used in the statement. | First three lines contain three non-empty input strings. The sum of lengths of all strings is no more than 3·105. All strings consist only of lowercase English letters. | You need to output *min*(|*s*1|,<=|*s*2|,<=|*s*3|) numbers separated by spaces — answers for the problem modulo 1000000007 (109<=+<=7). | [
"abc\nbc\ncbc\n",
"abacaba\nabac\nabcd\n"
] | [
"3 1 \n",
"11 2 0 0 \n"
] | Consider a string *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">|*t*|</sub>, where *t*<sub class="lower-index">*i*</sub> denotes the *i*-th character of the string, and |*t*| denotes the length of the string.
Then *t*[*i*... *j*] (1 ≤ *i* ≤ *j* ≤ |*t*|) represents the string *t*<sub class="lower-index">*i*</sub>*t*<sub class="lower-index">*i* + 1</sub>... *t*<sub class="lower-index">*j*</sub> (substring of *t* from position *i* to position *j* inclusive). | 2,500 | [] | 1,693,396,935 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1693396935.2383485")# 1693396935.2383678 | Title: Three strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given three strings (*s*1,<=*s*2,<=*s*3). For each integer *l* (1<=≤<=*l*<=≤<=*min*(|*s*1|,<=|*s*2|,<=|*s*3|) you need to find how many triples (*i*1,<=*i*2,<=*i*3) exist such that three strings *s**k*[*i**k*... *i**k*<=+<=*l*<=-<=1] (*k*<==<=1,<=2,<=3) are pairwise equal. Print all found numbers modulo 1000000007 (109<=+<=7).
See notes if you are not sure about some of the denotions used in the statement.
Input Specification:
First three lines contain three non-empty input strings. The sum of lengths of all strings is no more than 3·105. All strings consist only of lowercase English letters.
Output Specification:
You need to output *min*(|*s*1|,<=|*s*2|,<=|*s*3|) numbers separated by spaces — answers for the problem modulo 1000000007 (109<=+<=7).
Demo Input:
['abc\nbc\ncbc\n', 'abacaba\nabac\nabcd\n']
Demo Output:
['3 1 \n', '11 2 0 0 \n']
Note:
Consider a string *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">|*t*|</sub>, where *t*<sub class="lower-index">*i*</sub> denotes the *i*-th character of the string, and |*t*| denotes the length of the string.
Then *t*[*i*... *j*] (1 ≤ *i* ≤ *j* ≤ |*t*|) represents the string *t*<sub class="lower-index">*i*</sub>*t*<sub class="lower-index">*i* + 1</sub>... *t*<sub class="lower-index">*j*</sub> (substring of *t* from position *i* to position *j* inclusive). | ```python
print("_RANDOM_GUESS_1693396935.2383485")# 1693396935.2383678
``` | 0 | |
952 | A | Quirky Quantifiers | PROGRAMMING | 800 | [
"math"
] | null | null | The input contains a single integer *a* (10<=≤<=*a*<=≤<=999).
Output 0 or 1. | The input contains a single integer *a* (10<=≤<=*a*<=≤<=999). | Output 0 or 1. | [
"13\n",
"927\n",
"48\n"
] | [
"1\n",
"1\n",
"0\n"
] | none | 0 | [
{
"input": "13",
"output": "1"
},
{
"input": "927",
"output": "1"
},
{
"input": "48",
"output": "0"
},
{
"input": "10",
"output": "0"
},
{
"input": "999",
"output": "1"
},
{
"input": "142",
"output": "0"
},
{
"input": "309",
"output": "... | 1,590,758,594 | 2,147,483,647 | PyPy 3 | OK | TESTS | 15 | 140 | 0 | a=int(input())
print(1 if a%2!=0 else 0) | Title: Quirky Quantifiers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains a single integer *a* (10<=≤<=*a*<=≤<=999).
Output 0 or 1.
Input Specification:
The input contains a single integer *a* (10<=≤<=*a*<=≤<=999).
Output Specification:
Output 0 or 1.
Demo Input:
['13\n', '927\n', '48\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
none | ```python
a=int(input())
print(1 if a%2!=0 else 0)
``` | 3 | |
476 | A | Dreamoon and Stairs | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition? | The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10). | Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead. | [
"10 2\n",
"3 5\n"
] | [
"6\n",
"-1\n"
] | For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | 500 | [
{
"input": "10 2",
"output": "6"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "29 7",
"output": "21"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "10000 2",
"output": "5000"
},
{
"input": "10000 ... | 1,654,679,523 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 46 | 0 | n,m=map(int,input().split())
if m>n:
print(-1)
else:
res=0
if n%2==0:
res+=n/2
else:
res+=(n//2)+1
while res%m != 0:
res+=1
print(int(res)) | Title: Dreamoon and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
Input Specification:
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
Output Specification:
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
Demo Input:
['10 2\n', '3 5\n']
Demo Output:
['6\n', '-1\n']
Note:
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | ```python
n,m=map(int,input().split())
if m>n:
print(-1)
else:
res=0
if n%2==0:
res+=n/2
else:
res+=(n//2)+1
while res%m != 0:
res+=1
print(int(res))
``` | 3 | |
815 | A | Karen and Game | PROGRAMMING | 1,700 | [
"brute force",
"greedy",
"implementation"
] | null | null | On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with *n* rows and *m* columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To win the level, after all the moves, the number in the cell at the *i*-th row and *j*-th column should be equal to *g**i*,<=*j*.
Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task! | The first line of input contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of rows and the number of columns in the grid, respectively.
The next *n* lines each contain *m* integers. In particular, the *j*-th integer in the *i*-th of these rows contains *g**i*,<=*j* (0<=≤<=*g**i*,<=*j*<=≤<=500). | If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer *k*, the minimum number of moves necessary to beat the level.
The next *k* lines should each contain one of the following, describing the moves in the order they must be done:
- row *x*, (1<=≤<=*x*<=≤<=*n*) describing a move of the form "choose the *x*-th row". - col *x*, (1<=≤<=*x*<=≤<=*m*) describing a move of the form "choose the *x*-th column".
If there are multiple optimal solutions, output any one of them. | [
"3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1\n",
"3 3\n0 0 0\n0 1 0\n0 0 0\n",
"3 3\n1 1 1\n1 1 1\n1 1 1\n"
] | [
"4\nrow 1\nrow 1\ncol 4\nrow 3\n",
"-1\n",
"3\nrow 1\nrow 2\nrow 3\n"
] | In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.
In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3. | 500 | [
{
"input": "3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1",
"output": "4\nrow 1\nrow 1\ncol 4\nrow 3"
},
{
"input": "3 3\n0 0 0\n0 1 0\n0 0 0",
"output": "-1"
},
{
"input": "3 3\n1 1 1\n1 1 1\n1 1 1",
"output": "3\nrow 1\nrow 2\nrow 3"
},
{
"input": "3 5\n2 4 2 2 3\n0 2 0 0 1\n1 3 1 1... | 1,498,493,864 | 2,147,483,647 | Python 3 | OK | TESTS | 177 | 311 | 6,348,800 | n, m = list(map(int, input().split()))
a = [list(map(int, input().split())) for _ in range(n)]
r = [a[i][0] - a[0][0] for i in range(n)]
c = [a[0][i] - a[0][0] for i in range(m)]
t = min(r)
r = [x - t for x in r]
t = min(c)
c = [x - t for x in c]
p = a[0][0] - r[0] - c[0]
if n < m:
r = [x + p for x in r]
else:
c = [x + p for x in c]
for i in range(n):
for j in range(m):
if r[i] + c[j] != a[i][j]:
print (-1)
quit()
print (sum(r) + sum(c))
for i, x in enumerate(r):
for j in range(x):
print ("row", i + 1)
for i, x in enumerate(c):
for j in range(x):
print ("col", i + 1)
| Title: Karen and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with *n* rows and *m* columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To win the level, after all the moves, the number in the cell at the *i*-th row and *j*-th column should be equal to *g**i*,<=*j*.
Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!
Input Specification:
The first line of input contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of rows and the number of columns in the grid, respectively.
The next *n* lines each contain *m* integers. In particular, the *j*-th integer in the *i*-th of these rows contains *g**i*,<=*j* (0<=≤<=*g**i*,<=*j*<=≤<=500).
Output Specification:
If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer *k*, the minimum number of moves necessary to beat the level.
The next *k* lines should each contain one of the following, describing the moves in the order they must be done:
- row *x*, (1<=≤<=*x*<=≤<=*n*) describing a move of the form "choose the *x*-th row". - col *x*, (1<=≤<=*x*<=≤<=*m*) describing a move of the form "choose the *x*-th column".
If there are multiple optimal solutions, output any one of them.
Demo Input:
['3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1\n', '3 3\n0 0 0\n0 1 0\n0 0 0\n', '3 3\n1 1 1\n1 1 1\n1 1 1\n']
Demo Output:
['4\nrow 1\nrow 1\ncol 4\nrow 3\n', '-1\n', '3\nrow 1\nrow 2\nrow 3\n']
Note:
In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.
In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3. | ```python
n, m = list(map(int, input().split()))
a = [list(map(int, input().split())) for _ in range(n)]
r = [a[i][0] - a[0][0] for i in range(n)]
c = [a[0][i] - a[0][0] for i in range(m)]
t = min(r)
r = [x - t for x in r]
t = min(c)
c = [x - t for x in c]
p = a[0][0] - r[0] - c[0]
if n < m:
r = [x + p for x in r]
else:
c = [x + p for x in c]
for i in range(n):
for j in range(m):
if r[i] + c[j] != a[i][j]:
print (-1)
quit()
print (sum(r) + sum(c))
for i, x in enumerate(r):
for j in range(x):
print ("row", i + 1)
for i, x in enumerate(c):
for j in range(x):
print ("col", i + 1)
``` | 3 | |
758 | A | Holiday Of Equality | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. | In the only line print the integer *S* — the minimum number of burles which are had to spend. | [
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] | [
"10",
"1",
"4",
"0"
] | In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | 500 | [
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853... | 1,669,629,593 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 46 | 0 | c = int(input())
m = list(map(int, input().split()))
n = max(m)
t = 0
for i in m:
t += n-i
print(t) | Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | ```python
c = int(input())
m = list(map(int, input().split()))
n = max(m)
t = 0
for i in m:
t += n-i
print(t)
``` | 3 | |
926 | E | Merge Equal Elements | PROGRAMMING | 1,900 | [
"constructive algorithms",
"data structures"
] | null | null | You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*.
While possible, you perform the following operation: find a pair of equal consecutive elements. If there are more than one such pair, find the leftmost (with the smallest indices of elements). If the two integers are equal to *x*, delete both and insert a single integer *x*<=+<=1 on their place. This way the number of elements in the sequence is decreased by 1 on each step.
You stop performing the operation when there is no pair of equal consecutive elements.
For example, if the initial sequence is [5,<=2,<=1,<=1,<=2,<=2], then after the first operation you get [5,<=2,<=2,<=2,<=2], after the second — [5,<=3,<=2,<=2], after the third — [5,<=3,<=3], and finally after the fourth you get [5,<=4]. After that there are no equal consecutive elements left in the sequence, so you stop the process.
Determine the final sequence after you stop performing the operation. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of elements in the sequence.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | In the first line print a single integer *k* — the number of elements in the sequence after you stop performing the operation.
In the second line print *k* integers — the sequence after you stop performing the operation. | [
"6\n5 2 1 1 2 2\n",
"4\n1000000000 1000000000 1000000000 1000000000\n",
"7\n4 10 22 11 12 5 6\n"
] | [
"2\n5 4 ",
"1\n1000000002 ",
"7\n4 10 22 11 12 5 6 "
] | The first example is described in the statements.
In the second example the initial sequence is [1000000000, 1000000000, 1000000000, 1000000000]. After the first operation the sequence is equal to [1000000001, 1000000000, 1000000000]. After the second operation the sequence is [1000000001, 1000000001]. After the third operation the sequence is [1000000002].
In the third example there are no two equal consecutive elements initially, so the sequence does not change. | 0 | [
{
"input": "6\n5 2 1 1 2 2",
"output": "2\n5 4 "
},
{
"input": "4\n1000000000 1000000000 1000000000 1000000000",
"output": "1\n1000000002 "
},
{
"input": "7\n4 10 22 11 12 5 6",
"output": "7\n4 10 22 11 12 5 6 "
},
{
"input": "2\n1 1",
"output": "1\n2 "
},
{
"inpu... | 1,553,771,308 | 2,147,483,647 | Python 3 | OK | TESTS | 75 | 716 | 17,817,600 |
n=int(input())
x=[int(i) for i in input().split()]
otv=[]
i=0
while i<n:
if len(otv)<2:
otv.append(x[i])
i+=1
continue
if otv[len(otv)-1]==otv[len(otv)-2]:
otv.pop()
otv[len(otv)-1]+=1
else:
otv.append(x[i])
i+=1
while len(otv)>=2 and otv[len(otv)-1]==otv[len(otv)-2]:
otv.pop()
otv[len(otv)-1]+=1
print(len(otv))
for i in otv:
print(i,end=" ")
| Title: Merge Equal Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*.
While possible, you perform the following operation: find a pair of equal consecutive elements. If there are more than one such pair, find the leftmost (with the smallest indices of elements). If the two integers are equal to *x*, delete both and insert a single integer *x*<=+<=1 on their place. This way the number of elements in the sequence is decreased by 1 on each step.
You stop performing the operation when there is no pair of equal consecutive elements.
For example, if the initial sequence is [5,<=2,<=1,<=1,<=2,<=2], then after the first operation you get [5,<=2,<=2,<=2,<=2], after the second — [5,<=3,<=2,<=2], after the third — [5,<=3,<=3], and finally after the fourth you get [5,<=4]. After that there are no equal consecutive elements left in the sequence, so you stop the process.
Determine the final sequence after you stop performing the operation.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of elements in the sequence.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
In the first line print a single integer *k* — the number of elements in the sequence after you stop performing the operation.
In the second line print *k* integers — the sequence after you stop performing the operation.
Demo Input:
['6\n5 2 1 1 2 2\n', '4\n1000000000 1000000000 1000000000 1000000000\n', '7\n4 10 22 11 12 5 6\n']
Demo Output:
['2\n5 4 ', '1\n1000000002 ', '7\n4 10 22 11 12 5 6 ']
Note:
The first example is described in the statements.
In the second example the initial sequence is [1000000000, 1000000000, 1000000000, 1000000000]. After the first operation the sequence is equal to [1000000001, 1000000000, 1000000000]. After the second operation the sequence is [1000000001, 1000000001]. After the third operation the sequence is [1000000002].
In the third example there are no two equal consecutive elements initially, so the sequence does not change. | ```python
n=int(input())
x=[int(i) for i in input().split()]
otv=[]
i=0
while i<n:
if len(otv)<2:
otv.append(x[i])
i+=1
continue
if otv[len(otv)-1]==otv[len(otv)-2]:
otv.pop()
otv[len(otv)-1]+=1
else:
otv.append(x[i])
i+=1
while len(otv)>=2 and otv[len(otv)-1]==otv[len(otv)-2]:
otv.pop()
otv[len(otv)-1]+=1
print(len(otv))
for i in otv:
print(i,end=" ")
``` | 3 | |
876 | B | Divisiblity of Differences | PROGRAMMING | 1,300 | [
"implementation",
"math",
"number theory"
] | null | null | You are given a multiset of *n* integers. You should select exactly *k* of them in a such way that the difference between any two of them is divisible by *m*, or tell that it is impossible.
Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset. | First line contains three integers *n*, *k* and *m* (2<=≤<=*k*<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the numbers in the multiset. | If it is not possible to select *k* numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print *k* integers *b*1,<=*b*2,<=...,<=*b**k* — the selected numbers. If there are multiple possible solutions, print any of them. | [
"3 2 3\n1 8 4\n",
"3 3 3\n1 8 4\n",
"4 3 5\n2 7 7 7\n"
] | [
"Yes\n1 4 ",
"No",
"Yes\n2 7 7 "
] | none | 1,000 | [
{
"input": "3 2 3\n1 8 4",
"output": "Yes\n1 4 "
},
{
"input": "3 3 3\n1 8 4",
"output": "No"
},
{
"input": "4 3 5\n2 7 7 7",
"output": "Yes\n2 7 7 "
},
{
"input": "9 9 5\n389149775 833127990 969340400 364457730 48649145 316121525 640054660 924273385 973207825",
"output":... | 1,508,865,628 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 5,529,600 | n, k, m = map(int, input().split())
a = list(map(int, input().split()))
c = [0]*m
l = 0
for i in range(n):
c[a[i] % m] += 1
if c[a[i] % m] == k:
print('Yes')
for j in range(n):
if a[j] % m == i:
print(a[j], end=' ')
l += 1
if l == k:
exit()
exit()
t = []
if k == 2:
for i in range(n-1):
for j in range(i+1, n):
if abs(a[j] - a[i]) % m == 0:
print('Yes')
print(a[i], a[j])
exit()
print('No')
else:
if m % 2 == 0:
middle = m//2 - 1
else:
middle = m//2
for i in range(m):
if c[i] == k:
print('Yes')
for j in range(n):
if a[j] % m == i:
print(a[i], end=' ')
exit()
if i < middle:
if c[i] + c[m-i-1] >= k:
print('Yes')
l = 0
for j in range(n):
if a[i] % m == i or a[i] % m == m - i:
print(a[i], end=' ')
l += 1
if l == k:
exit()
print('No') | Title: Divisiblity of Differences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a multiset of *n* integers. You should select exactly *k* of them in a such way that the difference between any two of them is divisible by *m*, or tell that it is impossible.
Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.
Input Specification:
First line contains three integers *n*, *k* and *m* (2<=≤<=*k*<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the numbers in the multiset.
Output Specification:
If it is not possible to select *k* numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print *k* integers *b*1,<=*b*2,<=...,<=*b**k* — the selected numbers. If there are multiple possible solutions, print any of them.
Demo Input:
['3 2 3\n1 8 4\n', '3 3 3\n1 8 4\n', '4 3 5\n2 7 7 7\n']
Demo Output:
['Yes\n1 4 ', 'No', 'Yes\n2 7 7 ']
Note:
none | ```python
n, k, m = map(int, input().split())
a = list(map(int, input().split()))
c = [0]*m
l = 0
for i in range(n):
c[a[i] % m] += 1
if c[a[i] % m] == k:
print('Yes')
for j in range(n):
if a[j] % m == i:
print(a[j], end=' ')
l += 1
if l == k:
exit()
exit()
t = []
if k == 2:
for i in range(n-1):
for j in range(i+1, n):
if abs(a[j] - a[i]) % m == 0:
print('Yes')
print(a[i], a[j])
exit()
print('No')
else:
if m % 2 == 0:
middle = m//2 - 1
else:
middle = m//2
for i in range(m):
if c[i] == k:
print('Yes')
for j in range(n):
if a[j] % m == i:
print(a[i], end=' ')
exit()
if i < middle:
if c[i] + c[m-i-1] >= k:
print('Yes')
l = 0
for j in range(n):
if a[i] % m == i or a[i] % m == m - i:
print(a[i], end=' ')
l += 1
if l == k:
exit()
print('No')
``` | 0 | |
508 | D | Tanya and Password | PROGRAMMING | 2,500 | [
"dfs and similar",
"graphs"
] | null | null | While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of *n*<=+<=2 characters. She has written all the possible *n* three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with *n* pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct. | The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105), the number of three-letter substrings Tanya got.
Next *n* lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit. | If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option. | [
"5\naca\naba\naba\ncab\nbac\n",
"4\nabc\nbCb\ncb1\nb13\n",
"7\naaa\naaa\naaa\naaa\naaa\naaa\naaa\n"
] | [
"YES\nabacaba\n",
"NO\n",
"YES\naaaaaaaaa\n"
] | none | 2,000 | [
{
"input": "5\naca\naba\naba\ncab\nbac",
"output": "YES\nabacaba"
},
{
"input": "4\nabc\nbCb\ncb1\nb13",
"output": "NO"
},
{
"input": "7\naaa\naaa\naaa\naaa\naaa\naaa\naaa",
"output": "YES\naaaaaaaaa"
},
{
"input": "1\nabc",
"output": "YES\nabc"
},
{
"input": "2\n... | 1,677,741,976 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
#define ll long long
using namespace std;
//const ll MAXN = 62;
const ll MAXM = 200001;
//vector < vector<pair<ll, ll> > > g(MAXN);
//vector < vector<int> > g (MAXN, vector<int> (MAXN));
//vector<pair<int,int> > g[MAXN];
map<string, pair<vector<string>, int> > g;
vector<bool> used(MAXM, false);
//vector<bool> visited_n(MAXN, false);
vector<string> order;
//vector<vector<ll> > components;
//vector<ll> component;
void dfs1 (string init) {
while(g[init].first.size()){
string v = g[init].first.back();
g[init].first.pop_back();
used[ g[init].second ] = 1;
dfs1(v);
}
order.push_back(init);
// visited_n[init] = true; // Lecture code - Time Limit
// for (size_t i=0; i<g[init].size(); ++i){
// if (!used[ g[init][i].second ] && !visited_n[g[init][i].first]){
// used[ g[init][i].second ] = 1;
// dfs1(g[init][i].first);
// }
// }
// order.push_back(init);
//-------------------
// stack<int> st;
// st.push(init);
// while(!st.empty()){ # Emax Modified - Time Limit
// int v = st.top();
// int i;
// // for (i=0; i<g[v].size(); ++i){
// // if (!used[ g[v][i].second ]){
// // break;
// // }
// // }
// // if(i == g[v].size()){
// // order.push_back(v);
// // st.pop();
// // }else{
// // used[ g[v][i].second ] = 1;
// // st.push(g[v][i].first);
// // }
// for (i=0; i<n; ++i) # Emax, gives mistake on adj NxN matrix
// if (g[v][i])
// break;
// if (i == n)
// {
// order.push_back (v);
// st.pop();
// }
// else
// {
// --g[v][i];
// st.push(i);
// }
// }
}
pair<int,int> transform(string s){
char v1 = s[0] - '0';
char v2 = s[1] - '0';
if(v1 >= 'A' && v1 <='Z'){
v1 -= 7;
}
if(v2 >= 'A' && v2 <='Z'){
v2 -= 7;
}
if(v1 >= 'a' && v1 <='z'){
v1 -= 7;
v1 -= 6;
}
if(v2 >= 'a' && v2 <='z'){
v2 -= 7;
v2 -= 6;
}
return make_pair(v1, v2);
}
char transform_back(int ord){
if(ord >= 10 && ord <=35){
ord += 7;
}
if(ord >= 36){
ord += 7 + 6;
}
ord += '0';
return ord;
}
void solution(ll q){
//vector<int> in(MAXN+1, 0), out(MAXN+1, 0);
map<string, ll> in, out;
set<string> st;
for(ll i = 0; i<q; i++){
string s; cin>>s;
g[s.substr(0, 2)].first.push_back(s.substr(1, 2));
g[s.substr(0, 2)].second = i;
out[s.substr(0, 2)]++;
in[s.substr(1, 2)]++;
st.insert(s.substr(0, 2));
st.insert(s.substr(1, 2));
// pair<int,int> p1 = transform(s.substr(0, 2));
// g[p1.first].push_back(make_pair(p1.second, i));
// out[p1.first]++;
// in[p1.second]++;
// pair<int,int> p2 = transform(s.substr(1, 2));
// g[p2.first].push_back(make_pair(p2.second, i+1));
// out[p2.first]++;
// in[p2.second]++;
}
int condition1 = 0, condition2 = 0;
for(string c : st){
if(in[c] - out[c] == 1)
condition1++;
else if(in[c] - out[c] == -1){
condition2++;
}else if(in[c]!=out[c]){// || (in[i]%2!=0 && out[i]%2!=0 && in[i]!=0 && out[i]!=0)){
cout<<"NO\n";
return;
}
}
// if(!(condition1 == condition2 && condition1 == 1)){
// cout<<"2NO\n";
// return;
// }
// for(ll i=0; i<MAXN; ++i){
// if(in[i] - out[i] == 1)
// condition1++;
// else if(in[i] - out[i] == -1){
// condition2++;
// }else if(in[i]!=out[i]){// || (in[i]%2!=0 && out[i]%2!=0 && in[i]!=0 && out[i]!=0)){
// cout<<"1NO\n";
// return;
// }
// }
// if(!(condition1 == condition2 && condition1 == 1)){
// cout<<"2NO\n";
// return;
// }
dfs1(*st.begin());
// for(int i = 0; i<62; i++)
// if(!visited_n[i])
// dfs1(i);
// for(ll i=0; i<2*q; ++i){
// if(!used[i]){
// cout<<"3NO\n";
// return;
// }
// }
if((ll)order.size() != q+1){ //if(!visited_n[n-1]){ // order[0] != n-1 ||
cout<<"NO\n";
return;
}
// for(vector<ll>::reverse_iterator it = order.rbegin(); it!=order.rend(); it++)
// cout<<(*it)+1<<" ";
cout<<"YES\n";
for(int i = order.size()-1; i>=0; i--){
//char w = transform_back(order[i]);
//cout<<w<<" ";
if(i == order.size()-1)
cout<<order[i]<<"";
else
cout<<order[i][1]<<"";
}
}
int main(){
// std::freopen("condense2.in", "r", stdin);
// std::freopen("condense2.out", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin>>n;
solution(n);
return 0;
} | Title: Tanya and Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of *n*<=+<=2 characters. She has written all the possible *n* three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with *n* pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105), the number of three-letter substrings Tanya got.
Next *n* lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output Specification:
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Demo Input:
['5\naca\naba\naba\ncab\nbac\n', '4\nabc\nbCb\ncb1\nb13\n', '7\naaa\naaa\naaa\naaa\naaa\naaa\naaa\n']
Demo Output:
['YES\nabacaba\n', 'NO\n', 'YES\naaaaaaaaa\n']
Note:
none | ```python
#include <bits/stdc++.h>
#define ll long long
using namespace std;
//const ll MAXN = 62;
const ll MAXM = 200001;
//vector < vector<pair<ll, ll> > > g(MAXN);
//vector < vector<int> > g (MAXN, vector<int> (MAXN));
//vector<pair<int,int> > g[MAXN];
map<string, pair<vector<string>, int> > g;
vector<bool> used(MAXM, false);
//vector<bool> visited_n(MAXN, false);
vector<string> order;
//vector<vector<ll> > components;
//vector<ll> component;
void dfs1 (string init) {
while(g[init].first.size()){
string v = g[init].first.back();
g[init].first.pop_back();
used[ g[init].second ] = 1;
dfs1(v);
}
order.push_back(init);
// visited_n[init] = true; // Lecture code - Time Limit
// for (size_t i=0; i<g[init].size(); ++i){
// if (!used[ g[init][i].second ] && !visited_n[g[init][i].first]){
// used[ g[init][i].second ] = 1;
// dfs1(g[init][i].first);
// }
// }
// order.push_back(init);
//-------------------
// stack<int> st;
// st.push(init);
// while(!st.empty()){ # Emax Modified - Time Limit
// int v = st.top();
// int i;
// // for (i=0; i<g[v].size(); ++i){
// // if (!used[ g[v][i].second ]){
// // break;
// // }
// // }
// // if(i == g[v].size()){
// // order.push_back(v);
// // st.pop();
// // }else{
// // used[ g[v][i].second ] = 1;
// // st.push(g[v][i].first);
// // }
// for (i=0; i<n; ++i) # Emax, gives mistake on adj NxN matrix
// if (g[v][i])
// break;
// if (i == n)
// {
// order.push_back (v);
// st.pop();
// }
// else
// {
// --g[v][i];
// st.push(i);
// }
// }
}
pair<int,int> transform(string s){
char v1 = s[0] - '0';
char v2 = s[1] - '0';
if(v1 >= 'A' && v1 <='Z'){
v1 -= 7;
}
if(v2 >= 'A' && v2 <='Z'){
v2 -= 7;
}
if(v1 >= 'a' && v1 <='z'){
v1 -= 7;
v1 -= 6;
}
if(v2 >= 'a' && v2 <='z'){
v2 -= 7;
v2 -= 6;
}
return make_pair(v1, v2);
}
char transform_back(int ord){
if(ord >= 10 && ord <=35){
ord += 7;
}
if(ord >= 36){
ord += 7 + 6;
}
ord += '0';
return ord;
}
void solution(ll q){
//vector<int> in(MAXN+1, 0), out(MAXN+1, 0);
map<string, ll> in, out;
set<string> st;
for(ll i = 0; i<q; i++){
string s; cin>>s;
g[s.substr(0, 2)].first.push_back(s.substr(1, 2));
g[s.substr(0, 2)].second = i;
out[s.substr(0, 2)]++;
in[s.substr(1, 2)]++;
st.insert(s.substr(0, 2));
st.insert(s.substr(1, 2));
// pair<int,int> p1 = transform(s.substr(0, 2));
// g[p1.first].push_back(make_pair(p1.second, i));
// out[p1.first]++;
// in[p1.second]++;
// pair<int,int> p2 = transform(s.substr(1, 2));
// g[p2.first].push_back(make_pair(p2.second, i+1));
// out[p2.first]++;
// in[p2.second]++;
}
int condition1 = 0, condition2 = 0;
for(string c : st){
if(in[c] - out[c] == 1)
condition1++;
else if(in[c] - out[c] == -1){
condition2++;
}else if(in[c]!=out[c]){// || (in[i]%2!=0 && out[i]%2!=0 && in[i]!=0 && out[i]!=0)){
cout<<"NO\n";
return;
}
}
// if(!(condition1 == condition2 && condition1 == 1)){
// cout<<"2NO\n";
// return;
// }
// for(ll i=0; i<MAXN; ++i){
// if(in[i] - out[i] == 1)
// condition1++;
// else if(in[i] - out[i] == -1){
// condition2++;
// }else if(in[i]!=out[i]){// || (in[i]%2!=0 && out[i]%2!=0 && in[i]!=0 && out[i]!=0)){
// cout<<"1NO\n";
// return;
// }
// }
// if(!(condition1 == condition2 && condition1 == 1)){
// cout<<"2NO\n";
// return;
// }
dfs1(*st.begin());
// for(int i = 0; i<62; i++)
// if(!visited_n[i])
// dfs1(i);
// for(ll i=0; i<2*q; ++i){
// if(!used[i]){
// cout<<"3NO\n";
// return;
// }
// }
if((ll)order.size() != q+1){ //if(!visited_n[n-1]){ // order[0] != n-1 ||
cout<<"NO\n";
return;
}
// for(vector<ll>::reverse_iterator it = order.rbegin(); it!=order.rend(); it++)
// cout<<(*it)+1<<" ";
cout<<"YES\n";
for(int i = order.size()-1; i>=0; i--){
//char w = transform_back(order[i]);
//cout<<w<<" ";
if(i == order.size()-1)
cout<<order[i]<<"";
else
cout<<order[i][1]<<"";
}
}
int main(){
// std::freopen("condense2.in", "r", stdin);
// std::freopen("condense2.out", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin>>n;
solution(n);
return 0;
}
``` | -1 | |
476 | B | Dreamoon and WiFi | PROGRAMMING | 1,300 | [
"bitmasks",
"brute force",
"combinatorics",
"dp",
"math",
"probabilities"
] | null | null | Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.
Each command is one of the following two types:
1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-'
But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5).
You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands? | The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.
The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.
Lengths of two strings are equal and do not exceed 10. | Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9. | [
"++-+-\n+-+-+\n",
"+-+-\n+-??\n",
"+++\n??-\n"
] | [
"1.000000000000\n",
"0.500000000000\n",
"0.000000000000\n"
] | For the first sample, both *s*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position + 1.
For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5.
For the third sample, *s*<sub class="lower-index">2</sub> could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0. | 1,500 | [
{
"input": "++-+-\n+-+-+",
"output": "1.000000000000"
},
{
"input": "+-+-\n+-??",
"output": "0.500000000000"
},
{
"input": "+++\n??-",
"output": "0.000000000000"
},
{
"input": "++++++++++\n+++??++?++",
"output": "0.125000000000"
},
{
"input": "--+++---+-\n????????... | 1,676,633,887 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | import sys,random,bisect
from collections import deque,defaultdict,Counter
from heapq import heapify,heappop,heappush
from math import gcd
mod = int(1e9 + 7)
inf = int(1e20)
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
ii = lambda :int(input())
py = lambda :print("YES")
pn = lambda :print("NO")
s1=input()
s2=input()
ans=0
for i in s1:
ans+= 1 if i=="+" else -1
res=0
m=0
for i in s2:
if i=="+":
res+=1
elif i=="-":
res-=1
else:
m+=1
cnt=0
for tmp in range(1<<m):
total=0
while tmp:
if tmp&1==1:
total+=1
else:
total-=1
tmp>>=1
if total==ans-res:
cnt+=1
print(cnt/(1<<m))
| Title: Dreamoon and WiFi
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.
Each command is one of the following two types:
1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-'
But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5).
You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?
Input Specification:
The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.
The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.
Lengths of two strings are equal and do not exceed 10.
Output Specification:
Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9.
Demo Input:
['++-+-\n+-+-+\n', '+-+-\n+-??\n', '+++\n??-\n']
Demo Output:
['1.000000000000\n', '0.500000000000\n', '0.000000000000\n']
Note:
For the first sample, both *s*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position + 1.
For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5.
For the third sample, *s*<sub class="lower-index">2</sub> could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0. | ```python
import sys,random,bisect
from collections import deque,defaultdict,Counter
from heapq import heapify,heappop,heappush
from math import gcd
mod = int(1e9 + 7)
inf = int(1e20)
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
ii = lambda :int(input())
py = lambda :print("YES")
pn = lambda :print("NO")
s1=input()
s2=input()
ans=0
for i in s1:
ans+= 1 if i=="+" else -1
res=0
m=0
for i in s2:
if i=="+":
res+=1
elif i=="-":
res-=1
else:
m+=1
cnt=0
for tmp in range(1<<m):
total=0
while tmp:
if tmp&1==1:
total+=1
else:
total-=1
tmp>>=1
if total==ans-res:
cnt+=1
print(cnt/(1<<m))
``` | 0 | |
626 | E | Simple Skewness | PROGRAMMING | 2,400 | [
"binary search",
"math",
"ternary search"
] | null | null | Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of *n* (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.
The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200 000) — the number of elements in the list.
The second line contains *n* integers *x**i* (0<=≤<=*x**i*<=≤<=1<=000<=000) — the *i*th element of the list. | In the first line, print a single integer *k* — the size of the subset.
In the second line, print *k* integers — the elements of the subset in any order.
If there are multiple optimal subsets, print any. | [
"4\n1 2 3 12\n",
"4\n1 1 2 2\n",
"2\n1 2\n"
] | [
"3\n1 2 12 \n",
"3\n1 1 2 \n",
"2\n1 2\n"
] | In the first case, the optimal subset is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/04cdbd07a0375de9c557422eca077386392a9349.png" style="max-width: 100.0%;max-height: 100.0%;"/>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.
In the second case, the optimal subset is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/af49670de7c27def20edf0ec421d9bb17d904c94.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that repetition is allowed.
In the last case, any subset has the same median and mean, so all have simple skewness of 0. | 2,000 | [
{
"input": "4\n1 2 3 12",
"output": "3\n1 2 12 "
},
{
"input": "4\n1 1 2 2",
"output": "3\n1 1 2 "
},
{
"input": "2\n1 2",
"output": "2\n1 2"
},
{
"input": "1\n1000000",
"output": "1\n1000000 "
},
{
"input": "20\n999999 999998 999996 999992 999984 999968 999936 99... | 1,675,158,834 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 966 | 13,926,400 | n = int(input())
a = sorted(list(map(int,input().split())))
s = [0] + a
for i in range(1, len(s)):
s[i] += s[i-1]
def skew(x):
m = n-x
avg = (s[m-m//2] + s[-1] - s[m-m//2+x]) / m
if m&1:
med = s[m-m//2] - s[m-m//2-1]
else:
med = (s[m-m//2] - s[m-m//2-1] + s[m-m//2+x+1] - s[m-m//2+x])/2
return avg - med
def skewed(x):
m = n-x
ans = a[:m-m//2] + a[m-m//2+x:]
return ans
mx = -1
x = 0
for i in range(n):
temp = skew(i)
if (temp > mx):
mx = temp
x = i
print(n-x)
for i in skewed(x):
print(end = str(i)+" ")
"""
4
1 2 3 12
7
1 2 3 4 5 6 7
2
1 2
4
1 1 2 2
""" | Title: Simple Skewness
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of *n* (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.
The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200 000) — the number of elements in the list.
The second line contains *n* integers *x**i* (0<=≤<=*x**i*<=≤<=1<=000<=000) — the *i*th element of the list.
Output Specification:
In the first line, print a single integer *k* — the size of the subset.
In the second line, print *k* integers — the elements of the subset in any order.
If there are multiple optimal subsets, print any.
Demo Input:
['4\n1 2 3 12\n', '4\n1 1 2 2\n', '2\n1 2\n']
Demo Output:
['3\n1 2 12 \n', '3\n1 1 2 \n', '2\n1 2\n']
Note:
In the first case, the optimal subset is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/04cdbd07a0375de9c557422eca077386392a9349.png" style="max-width: 100.0%;max-height: 100.0%;"/>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.
In the second case, the optimal subset is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/af49670de7c27def20edf0ec421d9bb17d904c94.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that repetition is allowed.
In the last case, any subset has the same median and mean, so all have simple skewness of 0. | ```python
n = int(input())
a = sorted(list(map(int,input().split())))
s = [0] + a
for i in range(1, len(s)):
s[i] += s[i-1]
def skew(x):
m = n-x
avg = (s[m-m//2] + s[-1] - s[m-m//2+x]) / m
if m&1:
med = s[m-m//2] - s[m-m//2-1]
else:
med = (s[m-m//2] - s[m-m//2-1] + s[m-m//2+x+1] - s[m-m//2+x])/2
return avg - med
def skewed(x):
m = n-x
ans = a[:m-m//2] + a[m-m//2+x:]
return ans
mx = -1
x = 0
for i in range(n):
temp = skew(i)
if (temp > mx):
mx = temp
x = i
print(n-x)
for i in skewed(x):
print(end = str(i)+" ")
"""
4
1 2 3 12
7
1 2 3 4 5 6 7
2
1 2
4
1 1 2 2
"""
``` | 0 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,689,621,015 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 92 | 0 | n=int(input())
sumi=0
l1=[int(x) for x in input().split()]
l2=[int(x) for x in input().split()]
l3=[int(x) for x in input().split()]
for val in l1:
sumi=sumi+val
for val in l2:
sumi=sumi+val
for val in l3:
sumi=sumi+val
if sumi==0:
print("YES")
else:
print("NO") | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n=int(input())
sumi=0
l1=[int(x) for x in input().split()]
l2=[int(x) for x in input().split()]
l3=[int(x) for x in input().split()]
for val in l1:
sumi=sumi+val
for val in l2:
sumi=sumi+val
for val in l3:
sumi=sumi+val
if sumi==0:
print("YES")
else:
print("NO")
``` | 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": ... | 1,688,535,055 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | a, b = map(int, input().split())
h = a
o = 0
while a>=b:
h+=a//b
o += a % b
a=a//b
if a < b:
print(h+(o+a)//b)
else:
print(h+o//b) | 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
a, b = map(int, input().split())
h = a
o = 0
while a>=b:
h+=a//b
o += a % b
a=a//b
if a < b:
print(h+(o+a)//b)
else:
print(h+o//b)
``` | 0 | |
961 | B | Lecture Sleep | PROGRAMMING | 1,200 | [
"data structures",
"dp",
"implementation",
"two pointers"
] | null | null | Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture. | Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | [
"6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n"
] | [
"16\n"
] | In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | 0 | [
{
"input": "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "16"
},
{
"input": "5 3\n1 9999 10000 10000 10000\n0 0 0 0 0",
"output": "30000"
},
{
"input": "3 3\n10 10 10\n1 1 0",
"output": "30"
},
{
"input": "1 1\n423\n0",
"output": "423"
},
{
"input": "6 6\n1 3 5 2 5 4... | 1,600,692,563 | 2,147,483,647 | PyPy 3 | OK | TESTS | 37 | 234 | 12,697,600 | import bisect
n,k=list(map(int,input().split()))
a=list(map(int,input().split()))
t=list(map(int,input().split()))
ans=0
ax,tx=[],[]
for i in range(n):
if t[i]==1:
ans=ans+a[i]
else:
ax.append(a[i])
tx.append(i)
m=0
for i in range(1,len(ax)):
ax[i]+=ax[i-1]
for i in range(len(ax)):
j=bisect.bisect_right(tx,tx[i]+k-1)-1
if i>0:
m=max(m,ax[j]-ax[i-1])
else:
m=max(m,ax[j])
ans=ans+m
print(ans) | Title: Lecture Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input Specification:
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
Output Specification:
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Demo Input:
['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n']
Demo Output:
['16\n']
Note:
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | ```python
import bisect
n,k=list(map(int,input().split()))
a=list(map(int,input().split()))
t=list(map(int,input().split()))
ans=0
ax,tx=[],[]
for i in range(n):
if t[i]==1:
ans=ans+a[i]
else:
ax.append(a[i])
tx.append(i)
m=0
for i in range(1,len(ax)):
ax[i]+=ax[i-1]
for i in range(len(ax)):
j=bisect.bisect_right(tx,tx[i]+k-1)-1
if i>0:
m=max(m,ax[j]-ax[i-1])
else:
m=max(m,ax[j])
ans=ans+m
print(ans)
``` | 3 | |
350 | A | TL | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation"
] | null | null | Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist. | The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds. | If there is a valid TL value, print it. Otherwise, print -1. | [
"3 6\n4 5 2\n8 9 6 10 7 11\n",
"3 1\n3 4 5\n6\n"
] | [
"5",
"-1\n"
] | none | 500 | [
{
"input": "3 6\n4 5 2\n8 9 6 10 7 11",
"output": "5"
},
{
"input": "3 1\n3 4 5\n6",
"output": "-1"
},
{
"input": "2 5\n45 99\n49 41 77 83 45",
"output": "-1"
},
{
"input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43... | 1,589,870,229 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 218 | 0 | n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
v = max(max(a),2*min(a))
if v<min(b):
print(v)
else:
print('-1')
| Title: TL
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
Output Specification:
If there is a valid TL value, print it. Otherwise, print -1.
Demo Input:
['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n']
Demo Output:
['5', '-1\n']
Note:
none | ```python
n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
v = max(max(a),2*min(a))
if v<min(b):
print(v)
else:
print('-1')
``` | 3 | |
617 | A | Elephant | PROGRAMMING | 800 | [
"math"
] | null | null | An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house. | The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house. | Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*. | [
"5\n",
"12\n"
] | [
"1\n",
"3\n"
] | In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | 500 | [
{
"input": "5",
"output": "1"
},
{
"input": "12",
"output": "3"
},
{
"input": "999999",
"output": "200000"
},
{
"input": "41",
"output": "9"
},
{
"input": "1000000",
"output": "200000"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
... | 1,695,111,024 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 0 | a=int(input())
b=int((a/5)+1)
c=int(a/5)
if a<5:
print(1)
elif a%5==0:
print(c)
else:
print(b) | Title: Elephant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
Input Specification:
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Output Specification:
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
Demo Input:
['5\n', '12\n']
Demo Output:
['1\n', '3\n']
Note:
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | ```python
a=int(input())
b=int((a/5)+1)
c=int(a/5)
if a<5:
print(1)
elif a%5==0:
print(c)
else:
print(b)
``` | 3 | |
989 | A | A Blend of Springtime | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | "What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order. | The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively. | Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower). | [
".BAC.\n",
"AA..CB\n"
] | [
"Yes\n",
"No\n"
] | In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | 500 | [
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
... | 1,542,210,045 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 202 | 1,024,000 | import itertools
seq = input()
for subseq in itertools.permutations('ABC'):
if subseq in seq:
print('Yes')
quit()
print('No') | Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | ```python
import itertools
seq = input()
for subseq in itertools.permutations('ABC'):
if subseq in seq:
print('Yes')
quit()
print('No')
``` | -1 | |
1,000 | D | Yet Another Problem On a Subsequence | PROGRAMMING | 1,900 | [
"combinatorics",
"dp"
] | null | null | The sequence of integers $a_1, a_2, \dots, a_k$ is called a good array if $a_1 = k - 1$ and $a_1 > 0$. For example, the sequences $[3, -1, 44, 0], [1, -99]$ are good arrays, and the sequences $[3, 7, 8], [2, 5, 4, 1], [0]$ — are not.
A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences $[2, -3, 0, 1, 4]$, $[1, 2, 3, -3, -9, 4]$ are good, and the sequences $[2, -3, 0, 1]$, $[1, 2, 3, -3 -9, 4, 1]$ — are not.
For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353. | The first line contains the number $n~(1 \le n \le 10^3)$ — the length of the initial sequence. The following line contains $n$ integers $a_1, a_2, \dots, a_n~(-10^9 \le a_i \le 10^9)$ — the sequence itself. | In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353. | [
"3\n2 1 1\n",
"4\n1 1 1 1\n"
] | [
"2\n",
"7\n"
] | In the first test case, two good subsequences — $[a_1, a_2, a_3]$ and $[a_2, a_3]$.
In the second test case, seven good subsequences — $[a_1, a_2, a_3, a_4], [a_1, a_2], [a_1, a_3], [a_1, a_4], [a_2, a_3], [a_2, a_4]$ and $[a_3, a_4]$. | 0 | [
{
"input": "3\n2 1 1",
"output": "2"
},
{
"input": "4\n1 1 1 1",
"output": "7"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
}
] | 1,645,705,944 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 93 | 3,584,000 | import sys
input = sys.stdin.readline
MOD = 998244353
N = 10005
fact = [1]
for i in range(1, N):
fact.append(fact[-1] * i % MOD)
factinv = [0] * N
factinv[N - 1] = pow(fact[N - 1], MOD - 2, MOD)
for j in range(N - 2, -1, -1):
factinv[j] = factinv[j + 1] * (j + 1) % MOD
def nCk(n, k):
if k < 0 or n < k:
return 0
return fact[n] * factinv[n - k] * factinv[k] % MOD
n = int(input())
a = list(map(int, input().split()))
dp = [0] * (n + 1)
dp[n] = 1
for j in range(n - 1, -1, -1):
if a[j] <= 0:
continue
for k in range(j + a[j] + 1, n + 1):
dp[j] += dp[k] * nCk(k - j - 1, a[j]) % MOD
print(sum(dp[:n]))
| Title: Yet Another Problem On a Subsequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The sequence of integers $a_1, a_2, \dots, a_k$ is called a good array if $a_1 = k - 1$ and $a_1 > 0$. For example, the sequences $[3, -1, 44, 0], [1, -99]$ are good arrays, and the sequences $[3, 7, 8], [2, 5, 4, 1], [0]$ — are not.
A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences $[2, -3, 0, 1, 4]$, $[1, 2, 3, -3, -9, 4]$ are good, and the sequences $[2, -3, 0, 1]$, $[1, 2, 3, -3 -9, 4, 1]$ — are not.
For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353.
Input Specification:
The first line contains the number $n~(1 \le n \le 10^3)$ — the length of the initial sequence. The following line contains $n$ integers $a_1, a_2, \dots, a_n~(-10^9 \le a_i \le 10^9)$ — the sequence itself.
Output Specification:
In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353.
Demo Input:
['3\n2 1 1\n', '4\n1 1 1 1\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test case, two good subsequences — $[a_1, a_2, a_3]$ and $[a_2, a_3]$.
In the second test case, seven good subsequences — $[a_1, a_2, a_3, a_4], [a_1, a_2], [a_1, a_3], [a_1, a_4], [a_2, a_3], [a_2, a_4]$ and $[a_3, a_4]$. | ```python
import sys
input = sys.stdin.readline
MOD = 998244353
N = 10005
fact = [1]
for i in range(1, N):
fact.append(fact[-1] * i % MOD)
factinv = [0] * N
factinv[N - 1] = pow(fact[N - 1], MOD - 2, MOD)
for j in range(N - 2, -1, -1):
factinv[j] = factinv[j + 1] * (j + 1) % MOD
def nCk(n, k):
if k < 0 or n < k:
return 0
return fact[n] * factinv[n - k] * factinv[k] % MOD
n = int(input())
a = list(map(int, input().split()))
dp = [0] * (n + 1)
dp[n] = 1
for j in range(n - 1, -1, -1):
if a[j] <= 0:
continue
for k in range(j + a[j] + 1, n + 1):
dp[j] += dp[k] * nCk(k - j - 1, a[j]) % MOD
print(sum(dp[:n]))
``` | 0 | |
1,005 | D | Polycarp and Div 3 | PROGRAMMING | 1,500 | [
"dp",
"greedy",
"number theory"
] | null | null | Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain? | The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0. | Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$. | [
"3121\n",
"6\n",
"1000000000000000000000000000000000\n",
"201920181\n"
] | [
"2\n",
"1\n",
"33\n",
"4\n"
] | In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | 0 | [
{
"input": "3121",
"output": "2"
},
{
"input": "6",
"output": "1"
},
{
"input": "1000000000000000000000000000000000",
"output": "33"
},
{
"input": "201920181",
"output": "4"
},
{
"input": "4",
"output": "0"
},
{
"input": "10",
"output": "1"
},
... | 1,678,459,229 | 1,229 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 92 | 3,072,000 | a = input()
b = len(a)
c = 0
d = 0
q = 0
for i in range(b):
c += int(s[i])
d += 1
if c%3 == 0 or int(a[i]) == 0 or d%3 == 0:
c = 0
d = 0
q+=1
print(q) | Title: Polycarp and Div 3
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
Input Specification:
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
Output Specification:
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
Demo Input:
['3121\n', '6\n', '1000000000000000000000000000000000\n', '201920181\n']
Demo Output:
['2\n', '1\n', '33\n', '4\n']
Note:
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | ```python
a = input()
b = len(a)
c = 0
d = 0
q = 0
for i in range(b):
c += int(s[i])
d += 1
if c%3 == 0 or int(a[i]) == 0 or d%3 == 0:
c = 0
d = 0
q+=1
print(q)
``` | -1 | |
38 | A | Army | PROGRAMMING | 800 | [
"implementation"
] | A. Army | 2 | 256 | The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. | The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated. | Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. | [
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] | [
"5\n",
"11\n"
] | none | 0 | [
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n... | 1,662,433,558 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 92 | 0 | n = int(input())
d = list(input().split())
d = [int(i) for i in d]
a,b = input().split()
a = int(a)
b = int(b)
ans = 0
for i in range(a-1,b-1):
ans += d[i]
print(ans) | Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none | ```python
n = int(input())
d = list(input().split())
d = [int(i) for i in d]
a,b = input().split()
a = int(a)
b = int(b)
ans = 0
for i in range(a-1,b-1):
ans += d[i]
print(ans)
``` | 3.977 |
538 | B | Quasi Binary | PROGRAMMING | 1,400 | [
"constructive algorithms",
"dp",
"greedy",
"implementation"
] | null | null | A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106). | In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them. | [
"9\n",
"32\n"
] | [
"9\n1 1 1 1 1 1 1 1 1 \n",
"3\n10 11 11 \n"
] | none | 1,000 | [
{
"input": "9",
"output": "9\n1 1 1 1 1 1 1 1 1 "
},
{
"input": "32",
"output": "3\n10 11 11 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "415",
"output": "5\n1 101 101 101 111 "
},
{
"input": "10011",
"output": "1\n10011 "
},
{
"input": "10201... | 1,440,331,840 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 108 | 0 | s = input()
m = max(int(c) for c in s)
print(m)
for i in range(m):
print(int(''.join('1' if int(c) > i else '0' for c in s)), end=' ')
| Title: Quasi Binary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
Output Specification:
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Demo Input:
['9\n', '32\n']
Demo Output:
['9\n1 1 1 1 1 1 1 1 1 \n', '3\n10 11 11 \n']
Note:
none | ```python
s = input()
m = max(int(c) for c in s)
print(m)
for i in range(m):
print(int(''.join('1' if int(c) > i else '0' for c in s)), end=' ')
``` | 3 | |
735 | D | Taxes | PROGRAMMING | 1,600 | [
"math",
"number theory"
] | null | null | Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to *n* (*n*<=≥<=2) burles and the amount of tax he has to pay is calculated as the maximum divisor of *n* (not equal to *n*, of course). For example, if *n*<==<=6 then Funt has to pay 3 burles, while for *n*<==<=25 he needs to pay 5 and if *n*<==<=2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial *n* in several parts *n*1<=+<=*n*2<=+<=...<=+<=*n**k*<==<=*n* (here *k* is arbitrary, even *k*<==<=1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition *n**i*<=≥<=2 should hold for all *i* from 1 to *k*.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split *n* in parts. | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·109) — the total year income of mr. Funt. | Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax. | [
"4\n",
"27\n"
] | [
"2\n",
"3\n"
] | none | 1,750 | [
{
"input": "4",
"output": "2"
},
{
"input": "27",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "2000000000",
"output": "2"
},
{
"input": "26",
"output": ... | 1,605,281,356 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 2,000 | 1,024,000 | def checkprime(a:int):
i = 0
for i in range(2,a-i*i):
if(a%i==0):
return False
return True
n = int(input())
ans = 0
if(checkprime(n)): ans = 1
elif(n%2==0 or checkprime(n-2)): ans = 2
else: ans=3
print(ans)
| Title: Taxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to *n* (*n*<=≥<=2) burles and the amount of tax he has to pay is calculated as the maximum divisor of *n* (not equal to *n*, of course). For example, if *n*<==<=6 then Funt has to pay 3 burles, while for *n*<==<=25 he needs to pay 5 and if *n*<==<=2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial *n* in several parts *n*1<=+<=*n*2<=+<=...<=+<=*n**k*<==<=*n* (here *k* is arbitrary, even *k*<==<=1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition *n**i*<=≥<=2 should hold for all *i* from 1 to *k*.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split *n* in parts.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·109) — the total year income of mr. Funt.
Output Specification:
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Demo Input:
['4\n', '27\n']
Demo Output:
['2\n', '3\n']
Note:
none | ```python
def checkprime(a:int):
i = 0
for i in range(2,a-i*i):
if(a%i==0):
return False
return True
n = int(input())
ans = 0
if(checkprime(n)): ans = 1
elif(n%2==0 or checkprime(n-2)): ans = 2
else: ans=3
print(ans)
``` | 0 | |
620 | A | Professor GukiZ's Robot | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position. | The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot. | Print the only integer *d* — the minimal number of steps to get the finish position. | [
"0 0\n4 5\n",
"3 4\n6 1\n"
] | [
"5\n",
"3\n"
] | In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times. | 0 | [
{
"input": "0 0\n4 5",
"output": "5"
},
{
"input": "3 4\n6 1",
"output": "3"
},
{
"input": "0 0\n4 6",
"output": "6"
},
{
"input": "1 1\n-3 -5",
"output": "6"
},
{
"input": "-1 -1\n-10 100",
"output": "101"
},
{
"input": "1 -1\n100 -100",
"output":... | 1,633,262,340 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <string>
#include <bits/stdc++.h>
#include <set>
using namespace std;
int main()
{
int a,b, c, d, ans;
cin >> a;
cin >> b;
cin >> c;
cin >> d;
cout << min(abs(abs(a)-abs(c)),abs(abs(b)-abs(d))) + abs(abs(abs(a)-abs(c))-abs(abs(b)-abs(d))) << endl;
} | Title: Professor GukiZ's Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position.
Input Specification:
The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot.
Output Specification:
Print the only integer *d* — the minimal number of steps to get the finish position.
Demo Input:
['0 0\n4 5\n', '3 4\n6 1\n']
Demo Output:
['5\n', '3\n']
Note:
In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times. | ```python
#include <iostream>
#include <string>
#include <bits/stdc++.h>
#include <set>
using namespace std;
int main()
{
int a,b, c, d, ans;
cin >> a;
cin >> b;
cin >> c;
cin >> d;
cout << min(abs(abs(a)-abs(c)),abs(abs(b)-abs(d))) + abs(abs(abs(a)-abs(c))-abs(abs(b)-abs(d))) << endl;
}
``` | -1 | |
452 | A | Eevee | PROGRAMMING | 1,000 | [
"brute force",
"implementation",
"strings"
] | null | null | You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.
You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it. | First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string.
Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword). | Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter). | [
"7\nj......\n",
"7\n...feon\n",
"7\n.l.r.o.\n"
] | [
"jolteon\n",
"leafeon\n",
"flareon\n"
] | Here's a set of names in a form you can paste into your solution:
["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"} | 500 | [
{
"input": "7\n...feon",
"output": "leafeon"
},
{
"input": "7\n.l.r.o.",
"output": "flareon"
},
{
"input": "6\n.s..o.",
"output": "espeon"
},
{
"input": "7\nglaceon",
"output": "glaceon"
},
{
"input": "8\n.a.o.e.n",
"output": "vaporeon"
},
{
"input": "... | 1,406,480,998 | 598 | Python 3 | RUNTIME_ERROR | PRETESTS | 0 | 46 | 0 | n = int(input())
s = raw_input()
assert len(s) == n
options = ["vaporeon",
"jolteon",
"flareon",
"espeon",
"umbreon",
"leafeon",
"glaceon",
"sylveon"]
for s2 in options:
#print(s2)
#print(len(s2), n)
if len(s2) != n: continue
#print([s[i] == '.' or s[i] == s2[i] for i in range(n)])
if all([s[i] == '.' or s[i] == s2[i] for i in range(n)]):
print(s2)
break
| Title: Eevee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.
You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it.
Input Specification:
First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string.
Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword).
Output Specification:
Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter).
Demo Input:
['7\nj......\n', '7\n...feon\n', '7\n.l.r.o.\n']
Demo Output:
['jolteon\n', 'leafeon\n', 'flareon\n']
Note:
Here's a set of names in a form you can paste into your solution:
["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"} | ```python
n = int(input())
s = raw_input()
assert len(s) == n
options = ["vaporeon",
"jolteon",
"flareon",
"espeon",
"umbreon",
"leafeon",
"glaceon",
"sylveon"]
for s2 in options:
#print(s2)
#print(len(s2), n)
if len(s2) != n: continue
#print([s[i] == '.' or s[i] == s2[i] for i in range(n)])
if all([s[i] == '.' or s[i] == s2[i] for i in range(n)]):
print(s2)
break
``` | -1 | |
828 | A | Restaurant Tables | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to. | The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people. | Print the total number of people the restaurant denies service to. | [
"4 1 2\n1 2 1 1\n",
"4 1 1\n1 1 2 1\n"
] | [
"0\n",
"2\n"
] | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | 500 | [
{
"input": "4 1 2\n1 2 1 1",
"output": "0"
},
{
"input": "4 1 1\n1 1 2 1",
"output": "2"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 1 2\n2 2",
"output": "0"
},
{
"input": "5 1 3\n1 2 2 2 1",
"output": "1"
},
{
"input": "7 6 1\n1 1 1 1 1 1... | 1,547,511,293 | 2,147,483,647 | Python 3 | OK | TESTS | 97 | 187 | 3,072,000 | n, a, b = map(int, input().split())
b2 = 0
deny = 0
for x in map(int, input().split()):
if x == 1:
if a > 0:
a -= 1
elif b > 0:
b -= 1
b2 += 1
elif b2 > 0:
b2 -= 1
else:
deny += 1
else:
if b > 0:
b -= 1
else:
deny += 2
print(deny) | Title: Restaurant Tables
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to.
Input Specification:
The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people.
Output Specification:
Print the total number of people the restaurant denies service to.
Demo Input:
['4 1 2\n1 2 1 1\n', '4 1 1\n1 1 2 1\n']
Demo Output:
['0\n', '2\n']
Note:
In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | ```python
n, a, b = map(int, input().split())
b2 = 0
deny = 0
for x in map(int, input().split()):
if x == 1:
if a > 0:
a -= 1
elif b > 0:
b -= 1
b2 += 1
elif b2 > 0:
b2 -= 1
else:
deny += 1
else:
if b > 0:
b -= 1
else:
deny += 2
print(deny)
``` | 3 | |
678 | D | Iterated Linear Function | PROGRAMMING | 1,700 | [
"math",
"number theory"
] | null | null | Consider a linear function *f*(*x*)<==<=*Ax*<=+<=*B*. Let's define *g*(0)(*x*)<==<=*x* and *g*(*n*)(*x*)<==<=*f*(*g*(*n*<=-<=1)(*x*)) for *n*<=><=0. For the given integer values *A*, *B*, *n* and *x* find the value of *g*(*n*)(*x*) modulo 109<=+<=7. | The only line contains four integers *A*, *B*, *n* and *x* (1<=≤<=*A*,<=*B*,<=*x*<=≤<=109,<=1<=≤<=*n*<=≤<=1018) — the parameters from the problem statement.
Note that the given value *n* can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | Print the only integer *s* — the value *g*(*n*)(*x*) modulo 109<=+<=7. | [
"3 4 1 1\n",
"3 4 2 1\n",
"3 4 3 1\n"
] | [
"7\n",
"25\n",
"79\n"
] | none | 0 | [
{
"input": "3 4 1 1",
"output": "7"
},
{
"input": "3 4 2 1",
"output": "25"
},
{
"input": "3 4 3 1",
"output": "79"
},
{
"input": "1 1 1 1",
"output": "2"
},
{
"input": "3 10 723 6",
"output": "443623217"
},
{
"input": "14 81 51 82",
"output": "908... | 1,697,025,093 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 36 | 77 | 0 | import math
a, b, n, x = [int(i) for i in input().split()]
def modInverse(b, m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def modDivide(a, b, m):
a = pow(a, 1, m)
inv = modInverse(b, m)
if (inv == -1):
return -1
else:
return pow((inv * a), 1, m)
newA = pow(a, n, 1000000007)
if a > 1:
bTerm = modDivide((pow(a, n, 1000000007)-1), (a-1), 1000000007)
else:
bTerm = (n)
print(pow(newA*x + bTerm*b, 1, 1000000007)) | Title: Iterated Linear Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider a linear function *f*(*x*)<==<=*Ax*<=+<=*B*. Let's define *g*(0)(*x*)<==<=*x* and *g*(*n*)(*x*)<==<=*f*(*g*(*n*<=-<=1)(*x*)) for *n*<=><=0. For the given integer values *A*, *B*, *n* and *x* find the value of *g*(*n*)(*x*) modulo 109<=+<=7.
Input Specification:
The only line contains four integers *A*, *B*, *n* and *x* (1<=≤<=*A*,<=*B*,<=*x*<=≤<=109,<=1<=≤<=*n*<=≤<=1018) — the parameters from the problem statement.
Note that the given value *n* can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output Specification:
Print the only integer *s* — the value *g*(*n*)(*x*) modulo 109<=+<=7.
Demo Input:
['3 4 1 1\n', '3 4 2 1\n', '3 4 3 1\n']
Demo Output:
['7\n', '25\n', '79\n']
Note:
none | ```python
import math
a, b, n, x = [int(i) for i in input().split()]
def modInverse(b, m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def modDivide(a, b, m):
a = pow(a, 1, m)
inv = modInverse(b, m)
if (inv == -1):
return -1
else:
return pow((inv * a), 1, m)
newA = pow(a, n, 1000000007)
if a > 1:
bTerm = modDivide((pow(a, n, 1000000007)-1), (a-1), 1000000007)
else:
bTerm = (n)
print(pow(newA*x + bTerm*b, 1, 1000000007))
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,487,728,233 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 186 | 4,710,400 | '''input
3
3 -1 7
-5 2 -4
2 -1 -3
'''
t1, t2, t3 = 0, 0, 0
for _ in range(int(input())):
i = list(map(int, input().split()))
t1 += i[0]
t2 += i[1]
t3 += i[2]
if t1 == t2 == t3 == 0:
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
'''input
3
3 -1 7
-5 2 -4
2 -1 -3
'''
t1, t2, t3 = 0, 0, 0
for _ in range(int(input())):
i = list(map(int, input().split()))
t1 += i[0]
t2 += i[1]
t3 += i[2]
if t1 == t2 == t3 == 0:
print("YES")
else:
print("NO")
``` | 3.944726 |
686 | A | Free Ice Cream | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. | The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. | Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. | [
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] | [
"22 1\n",
"3 2\n"
] | Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | 500 | [
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
... | 1,627,262,983 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 62 | 6,963,200 | n,x=map(int,input().split())
l=[]
s=0
for i in range(n):
l.append(input())
if l[i][0]=="+":
x=x+int(l[i][2:])
elif l[i][0]=="-" and x<int(l[i][2:]):
s=s+1
else:
x = x - int(l[i][2:])
print(x,s)
| Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | ```python
n,x=map(int,input().split())
l=[]
s=0
for i in range(n):
l.append(input())
if l[i][0]=="+":
x=x+int(l[i][2:])
elif l[i][0]=="-" and x<int(l[i][2:]):
s=s+1
else:
x = x - int(l[i][2:])
print(x,s)
``` | 3 | |
686 | A | Free Ice Cream | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. | The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. | Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. | [
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] | [
"22 1\n",
"3 2\n"
] | Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | 500 | [
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
... | 1,670,770,732 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 0 | n=input().split()
a,b,t=int(n[0]),int(n[1]),0
for i in range(a):
c=input().split()
q,w=c[0],int(c[1])
if q=="+":b+=w
else:
if b>=w:b-=w
else :t+=1
print(b,t) | Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | ```python
n=input().split()
a,b,t=int(n[0]),int(n[1]),0
for i in range(a):
c=input().split()
q,w=c[0],int(c[1])
if q=="+":b+=w
else:
if b>=w:b-=w
else :t+=1
print(b,t)
``` | 3 | |
543 | D | Road Improvement | PROGRAMMING | 2,300 | [
"dp",
"trees"
] | null | null | The country has *n* cities and *n*<=-<=1 bidirectional roads, it is possible to get from every city to any other one if you move only along the roads. The cities are numbered with integers from 1 to *n* inclusive.
All the roads are initially bad, but the government wants to improve the state of some roads. We will assume that the citizens are happy about road improvement if the path from the capital located in city *x* to any other city contains at most one bad road.
Your task is — for every possible *x* determine the number of ways of improving the quality of some roads in order to meet the citizens' condition. As those values can be rather large, you need to print each value modulo 1<=000<=000<=007 (109<=+<=7). | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of cities in the country. Next line contains *n*<=-<=1 positive integers *p*2,<=*p*3,<=*p*4,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*i*<=-<=1) — the description of the roads in the country. Number *p**i* means that the country has a road connecting city *p**i* and city *i*. | Print *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sought number of ways to improve the quality of the roads modulo 1<=000<=000<=007 (109<=+<=7), if the capital of the country is at city number *i*. | [
"3\n1 1\n",
"5\n1 2 3 4\n"
] | [
"4 3 3",
"5 8 9 8 5"
] | none | 1,750 | [
{
"input": "3\n1 1",
"output": "4 3 3"
},
{
"input": "5\n1 2 3 4",
"output": "5 8 9 8 5"
},
{
"input": "31\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "73741817 536870913 536870913 536870913 536870913 536870913 536870913 536870913 536870913 536870913 5368... | 1,691,607,544 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | print("_RANDOM_GUESS_1691607544.185782")# 1691607544.1857975 | Title: Road Improvement
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The country has *n* cities and *n*<=-<=1 bidirectional roads, it is possible to get from every city to any other one if you move only along the roads. The cities are numbered with integers from 1 to *n* inclusive.
All the roads are initially bad, but the government wants to improve the state of some roads. We will assume that the citizens are happy about road improvement if the path from the capital located in city *x* to any other city contains at most one bad road.
Your task is — for every possible *x* determine the number of ways of improving the quality of some roads in order to meet the citizens' condition. As those values can be rather large, you need to print each value modulo 1<=000<=000<=007 (109<=+<=7).
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of cities in the country. Next line contains *n*<=-<=1 positive integers *p*2,<=*p*3,<=*p*4,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*i*<=-<=1) — the description of the roads in the country. Number *p**i* means that the country has a road connecting city *p**i* and city *i*.
Output Specification:
Print *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sought number of ways to improve the quality of the roads modulo 1<=000<=000<=007 (109<=+<=7), if the capital of the country is at city number *i*.
Demo Input:
['3\n1 1\n', '5\n1 2 3 4\n']
Demo Output:
['4 3 3', '5 8 9 8 5']
Note:
none | ```python
print("_RANDOM_GUESS_1691607544.185782")# 1691607544.1857975
``` | 0 | |
538 | C | Tourist's Notes | PROGRAMMING | 1,600 | [
"binary search",
"brute force",
"greedy",
"implementation",
"math"
] | null | null | A tourist hiked along the mountain range. The hike lasted for *n* days, during each day the tourist noted height above the sea level. On the *i*-th day height was equal to some integer *h**i*. The tourist pick smooth enough route for his hike, meaning that the between any two consecutive days height changes by at most 1, i.e. for all *i*'s from 1 to *n*<=-<=1 the inequality |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1 holds.
At the end of the route the tourist rafted down a mountain river and some notes in the journal were washed away. Moreover, the numbers in the notes could have been distorted. Now the tourist wonders what could be the maximum height during his hike. Help him restore the maximum possible value of the maximum height throughout the hike or determine that the notes were so much distorted that they do not represent any possible height values that meet limits |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1. | The first line contains two space-separated numbers, *n* and *m* (1<=≤<=*n*<=≤<=108, 1<=≤<=*m*<=≤<=105) — the number of days of the hike and the number of notes left in the journal.
Next *m* lines contain two space-separated integers *d**i* and *h**d**i* (1<=≤<=*d**i*<=≤<=*n*, 0<=≤<=*h**d**i*<=≤<=108) — the number of the day when the *i*-th note was made and height on the *d**i*-th day. It is guaranteed that the notes are given in the chronological order, i.e. for all *i* from 1 to *m*<=-<=1 the following condition holds: *d**i*<=<<=*d**i*<=+<=1. | If the notes aren't contradictory, print a single integer — the maximum possible height value throughout the whole route.
If the notes do not correspond to any set of heights, print a single word 'IMPOSSIBLE' (without the quotes). | [
"8 2\n2 0\n7 0\n",
"8 3\n2 0\n7 0\n8 3\n"
] | [
"2\n",
"IMPOSSIBLE\n"
] | For the first sample, an example of a correct height sequence with a maximum of 2: (0, 0, 1, 2, 1, 1, 0, 1).
In the second sample the inequality between *h*<sub class="lower-index">7</sub> and *h*<sub class="lower-index">8</sub> does not hold, thus the information is inconsistent. | 1,500 | [
{
"input": "8 2\n2 0\n7 0",
"output": "2"
},
{
"input": "8 3\n2 0\n7 0\n8 3",
"output": "IMPOSSIBLE"
},
{
"input": "10 10\n1 0\n2 0\n3 0\n4 0\n5 1\n6 2\n7 3\n8 2\n9 3\n10 4",
"output": "4"
},
{
"input": "50 10\n1 42\n7 36\n16 40\n21 40\n26 39\n30 41\n32 41\n36 40\n44 37\n50 4... | 1,430,092,243 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 31 | 4,710,400 | n, m = map(int, input().split())
l_d, l_h = map(int, input().split())
m_h = l_h + l_d - 1
for i in range(1, m):
n_d, n_h = map(int, input().split())
if abs(n_h - l_h) > n_d - l_d:
print("Impossible")
exit(0)
m_h = max(m_h, max(l_h, n_h) + (n_d - l_d - abs(n_h - l_h)) // 2)
l_d, l_h = n_d, n_h
m_h = max(m_h, l_h + n - l_d)
print(m_h) | Title: Tourist's Notes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tourist hiked along the mountain range. The hike lasted for *n* days, during each day the tourist noted height above the sea level. On the *i*-th day height was equal to some integer *h**i*. The tourist pick smooth enough route for his hike, meaning that the between any two consecutive days height changes by at most 1, i.e. for all *i*'s from 1 to *n*<=-<=1 the inequality |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1 holds.
At the end of the route the tourist rafted down a mountain river and some notes in the journal were washed away. Moreover, the numbers in the notes could have been distorted. Now the tourist wonders what could be the maximum height during his hike. Help him restore the maximum possible value of the maximum height throughout the hike or determine that the notes were so much distorted that they do not represent any possible height values that meet limits |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1.
Input Specification:
The first line contains two space-separated numbers, *n* and *m* (1<=≤<=*n*<=≤<=108, 1<=≤<=*m*<=≤<=105) — the number of days of the hike and the number of notes left in the journal.
Next *m* lines contain two space-separated integers *d**i* and *h**d**i* (1<=≤<=*d**i*<=≤<=*n*, 0<=≤<=*h**d**i*<=≤<=108) — the number of the day when the *i*-th note was made and height on the *d**i*-th day. It is guaranteed that the notes are given in the chronological order, i.e. for all *i* from 1 to *m*<=-<=1 the following condition holds: *d**i*<=<<=*d**i*<=+<=1.
Output Specification:
If the notes aren't contradictory, print a single integer — the maximum possible height value throughout the whole route.
If the notes do not correspond to any set of heights, print a single word 'IMPOSSIBLE' (without the quotes).
Demo Input:
['8 2\n2 0\n7 0\n', '8 3\n2 0\n7 0\n8 3\n']
Demo Output:
['2\n', 'IMPOSSIBLE\n']
Note:
For the first sample, an example of a correct height sequence with a maximum of 2: (0, 0, 1, 2, 1, 1, 0, 1).
In the second sample the inequality between *h*<sub class="lower-index">7</sub> and *h*<sub class="lower-index">8</sub> does not hold, thus the information is inconsistent. | ```python
n, m = map(int, input().split())
l_d, l_h = map(int, input().split())
m_h = l_h + l_d - 1
for i in range(1, m):
n_d, n_h = map(int, input().split())
if abs(n_h - l_h) > n_d - l_d:
print("Impossible")
exit(0)
m_h = max(m_h, max(l_h, n_h) + (n_d - l_d - abs(n_h - l_h)) // 2)
l_d, l_h = n_d, n_h
m_h = max(m_h, l_h + n - l_d)
print(m_h)
``` | 0 | |
733 | B | Parade | PROGRAMMING | 1,100 | [
"math"
] | null | null | Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be *n* columns participating in the parade, the *i*-th column consists of *l**i* soldiers, who start to march from left leg, and *r**i* soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if *L* is the total number of soldiers on the parade who start to march from the left leg, and *R* is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |*L*<=-<=*R*|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index *i* and swap values *l**i* and *r**i*.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=105) — the number of columns.
The next *n* lines contain the pairs of integers *l**i* and *r**i* (1<=≤<=*l**i*,<=*r**i*<=≤<=500) — the number of soldiers in the *i*-th column which start to march from the left or the right leg respectively. | Print single integer *k* — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to *n* in the order they are given in the input data.
If there are several answers, print any of them. | [
"3\n5 6\n8 9\n10 3\n",
"2\n6 5\n5 6\n",
"6\n5 9\n1 3\n4 8\n4 5\n23 54\n12 32\n"
] | [
"3\n",
"1\n",
"0\n"
] | In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | 1,000 | [
{
"input": "3\n5 6\n8 9\n10 3",
"output": "3"
},
{
"input": "2\n6 5\n5 6",
"output": "1"
},
{
"input": "6\n5 9\n1 3\n4 8\n4 5\n23 54\n12 32",
"output": "0"
},
{
"input": "2\n500 499\n500 500",
"output": "0"
},
{
"input": "1\n139 252",
"output": "0"
},
{
... | 1,596,909,479 | 2,147,483,647 | PyPy 3 | OK | TESTS | 40 | 857 | 29,900,800 |
n = int(input())
l = []
r = []
for i in range(n):
x , y = map(int,input().split())
l.append(x)
r.append(y)
r1 = sum(l)
r2 = sum(r)
v = abs(r1 - r2)
s = 0
for i in range(n):
k = abs(r1 - l[i] + r[i] - (r2 - r[i] + l[i]))
if k > v :
v = k
s = i + 1
print(s) | Title: Parade
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be *n* columns participating in the parade, the *i*-th column consists of *l**i* soldiers, who start to march from left leg, and *r**i* soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if *L* is the total number of soldiers on the parade who start to march from the left leg, and *R* is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |*L*<=-<=*R*|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index *i* and swap values *l**i* and *r**i*.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=105) — the number of columns.
The next *n* lines contain the pairs of integers *l**i* and *r**i* (1<=≤<=*l**i*,<=*r**i*<=≤<=500) — the number of soldiers in the *i*-th column which start to march from the left or the right leg respectively.
Output Specification:
Print single integer *k* — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to *n* in the order they are given in the input data.
If there are several answers, print any of them.
Demo Input:
['3\n5 6\n8 9\n10 3\n', '2\n6 5\n5 6\n', '6\n5 9\n1 3\n4 8\n4 5\n23 54\n12 32\n']
Demo Output:
['3\n', '1\n', '0\n']
Note:
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | ```python
n = int(input())
l = []
r = []
for i in range(n):
x , y = map(int,input().split())
l.append(x)
r.append(y)
r1 = sum(l)
r2 = sum(r)
v = abs(r1 - r2)
s = 0
for i in range(n):
k = abs(r1 - l[i] + r[i] - (r2 - r[i] + l[i]))
if k > v :
v = k
s = i + 1
print(s)
``` | 3 | |
897 | B | Chtholly's request | PROGRAMMING | 1,300 | [
"brute force"
] | null | null | — I experienced so many great things.
— You gave me memories like dreams... But I have to leave now...
— One last request, can you...
— Help me solve a Codeforces problem?
— ......
— What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help! | The first line contains two integers *k* and *p* (1<=≤<=*k*<=≤<=105,<=1<=≤<=*p*<=≤<=109). | Output single integer — answer to the problem. | [
"2 100\n",
"5 30\n"
] | [
"33\n",
"15\n"
] | In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,000 | [
{
"input": "2 100",
"output": "33"
},
{
"input": "5 30",
"output": "15"
},
{
"input": "42147 412393322",
"output": "251637727"
},
{
"input": "77809 868097296",
"output": "440411873"
},
{
"input": "5105 443422097",
"output": "363192634"
},
{
"input": "7... | 1,535,032,075 | 2,147,483,647 | PyPy 3 | OK | TESTS | 26 | 186 | 819,200 | k, p = map(int, input().split())
total = 0
for half_zcy in range(1,k+1):
current_zcy = int(str(half_zcy) + str(half_zcy)[::-1])
total += current_zcy
print(total % p)
| Title: Chtholly's request
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
— I experienced so many great things.
— You gave me memories like dreams... But I have to leave now...
— One last request, can you...
— Help me solve a Codeforces problem?
— ......
— What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help!
Input Specification:
The first line contains two integers *k* and *p* (1<=≤<=*k*<=≤<=105,<=1<=≤<=*p*<=≤<=109).
Output Specification:
Output single integer — answer to the problem.
Demo Input:
['2 100\n', '5 30\n']
Demo Output:
['33\n', '15\n']
Note:
In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
k, p = map(int, input().split())
total = 0
for half_zcy in range(1,k+1):
current_zcy = int(str(half_zcy) + str(half_zcy)[::-1])
total += current_zcy
print(total % p)
``` | 3 | |
614 | B | Gena's Code | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case. | The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000. | Print a single number without leading zeroes — the product of the number of tanks presented by each country. | [
"3\n5 10 1\n",
"4\n1 1 10 11\n",
"5\n0 3 1 100 1\n"
] | [
"50",
"110",
"0"
] | In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | 1,000 | [
{
"input": "3\n5 10 1",
"output": "50"
},
{
"input": "4\n1 1 10 11",
"output": "110"
},
{
"input": "5\n0 3 1 100 1",
"output": "0"
},
{
"input": "40\n10 100 10 1 10 10 100 10 10 100 10 100 100 10 1824868942 100 100 1 10 100 100 10 100 100 10 100 10 1 10 100 100 100 10 1 10 1 ... | 1,634,015,388 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 109 | 20,172,800 | n = int(input())
a = list(input().split())
num_of_zeros = 0
has_0 = False
ugly = ''
for i in a:
if i == '0':
has_0 = True
break
all_0 = False
if i[0] == '1':
all_0 = True
for j in i[1:]:
if j != '0':
all_0 = False
break
if not all_0:
ugly = i
else:
num_of_zeros += len(i) - 1
if has_0:
print('0')
else:
print(ugly, end='')
for _ in range(num_of_zeros): print('0', end='') | Title: Gena's Code
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input Specification:
The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000.
Output Specification:
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Demo Input:
['3\n5 10 1\n', '4\n1 1 10 11\n', '5\n0 3 1 100 1\n']
Demo Output:
['50', '110', '0']
Note:
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | ```python
n = int(input())
a = list(input().split())
num_of_zeros = 0
has_0 = False
ugly = ''
for i in a:
if i == '0':
has_0 = True
break
all_0 = False
if i[0] == '1':
all_0 = True
for j in i[1:]:
if j != '0':
all_0 = False
break
if not all_0:
ugly = i
else:
num_of_zeros += len(i) - 1
if has_0:
print('0')
else:
print(ugly, end='')
for _ in range(num_of_zeros): print('0', end='')
``` | 0 | |
253 | A | Boys and Girls | PROGRAMMING | 1,100 | [
"greedy"
] | null | null | There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line. | The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space. | Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them. | [
"3 3\n",
"4 2\n"
] | [
"GBGBGB\n",
"BGBGBB\n"
] | In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal. | 500 | [
{
"input": "3 3",
"output": "GBGBGB"
},
{
"input": "4 2",
"output": "BGBGBB"
},
{
"input": "5 5",
"output": "GBGBGBGBGB"
},
{
"input": "6 4",
"output": "BGBGBGBGBB"
},
{
"input": "100 1",
"output": "BGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB... | 1,628,977,356 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 6,963,200 | file1=open("input.txt")
n,m=file1.readline().split()
n=int(n); m=int(m)
file1.close()
file2=open("output.txt")
t=int(0)
list=[]
if(n<m):
for i in range((n+m)):
if((i%2==0)and(t<n)):
list.append('B')
t=t+1
else:
list.append('G')
elif(m<=n):
for i in range((n+m)):
if((i%2==0)and(t<m)):
list.append('G')
t=t+1
else:
list.append('B')
final=''.join(list)
file2.write(final)
file2.close() | Title: Boys and Girls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line.
Input Specification:
The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them.
Demo Input:
['3 3\n', '4 2\n']
Demo Output:
['GBGBGB\n', 'BGBGBB\n']
Note:
In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal. | ```python
file1=open("input.txt")
n,m=file1.readline().split()
n=int(n); m=int(m)
file1.close()
file2=open("output.txt")
t=int(0)
list=[]
if(n<m):
for i in range((n+m)):
if((i%2==0)and(t<n)):
list.append('B')
t=t+1
else:
list.append('G')
elif(m<=n):
for i in range((n+m)):
if((i%2==0)and(t<m)):
list.append('G')
t=t+1
else:
list.append('B')
final=''.join(list)
file2.write(final)
file2.close()
``` | -1 | |
828 | A | Restaurant Tables | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to. | The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people. | Print the total number of people the restaurant denies service to. | [
"4 1 2\n1 2 1 1\n",
"4 1 1\n1 1 2 1\n"
] | [
"0\n",
"2\n"
] | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | 500 | [
{
"input": "4 1 2\n1 2 1 1",
"output": "0"
},
{
"input": "4 1 1\n1 1 2 1",
"output": "2"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 1 2\n2 2",
"output": "0"
},
{
"input": "5 1 3\n1 2 2 2 1",
"output": "1"
},
{
"input": "7 6 1\n1 1 1 1 1 1... | 1,499,875,080 | 2,147,483,647 | Python 3 | OK | TESTS | 97 | 140 | 8,089,600 | n, a, b = (int(each) for each in input().split())
t_list = [int(each) for each in input().split()]
single_seat = a
two_seat = b
half_seat = 0
result = 0
for t in t_list:
if (t == 1):
if (single_seat > 0):
single_seat -= 1
elif (two_seat > 0):
two_seat -= 1
half_seat += 1
elif (half_seat > 0):
half_seat -= 1
else:
result += 1
elif (t == 2):
if (two_seat > 0):
two_seat -= 1
else:
result += 2
print(result) | Title: Restaurant Tables
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to.
Input Specification:
The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people.
Output Specification:
Print the total number of people the restaurant denies service to.
Demo Input:
['4 1 2\n1 2 1 1\n', '4 1 1\n1 1 2 1\n']
Demo Output:
['0\n', '2\n']
Note:
In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | ```python
n, a, b = (int(each) for each in input().split())
t_list = [int(each) for each in input().split()]
single_seat = a
two_seat = b
half_seat = 0
result = 0
for t in t_list:
if (t == 1):
if (single_seat > 0):
single_seat -= 1
elif (two_seat > 0):
two_seat -= 1
half_seat += 1
elif (half_seat > 0):
half_seat -= 1
else:
result += 1
elif (t == 2):
if (two_seat > 0):
two_seat -= 1
else:
result += 2
print(result)
``` | 3 | |
714 | B | Filya and Homework | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | null | null | Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.
Filya is given an array of non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. First, he pick an integer *x* and then he adds *x* to some elements of the array (no more than once), subtract *x* from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.
Now he wonders if it's possible to pick such integer *x* and change some elements of the array using this *x* in order to make all elements equal. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of integers in the Filya's array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — elements of the array. | If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes). | [
"5\n1 3 3 2 1\n",
"5\n1 2 3 4 5\n"
] | [
"YES\n",
"NO\n"
] | In the first sample Filya should select *x* = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements. | 1,000 | [
{
"input": "5\n1 3 3 2 1",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5",
"output": "NO"
},
{
"input": "2\n1 2",
"output": "YES"
},
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
},
{
"input": "2\n1 1000000000",
"ou... | 1,605,037,774 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 93 | 307,200 | n=int(input())
l=list(input().split())
l=[int(y) for y in l]
d=dict()
for x in range(min(l), max(l)):
poss=True
l1=l.copy()
for i in range(len(l)):
if l1[i] < min(l)+x:
l1[i] += x
elif l1[i] > min(l)+x:
l1[i]-=x
if i>0:
if l1[i] != l1[i-1]:
poss=False
d[x]=poss
if True in d.values():
print('YES')
else:
print('NO') | Title: Filya and Homework
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.
Filya is given an array of non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. First, he pick an integer *x* and then he adds *x* to some elements of the array (no more than once), subtract *x* from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.
Now he wonders if it's possible to pick such integer *x* and change some elements of the array using this *x* in order to make all elements equal.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of integers in the Filya's array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — elements of the array.
Output Specification:
If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).
Demo Input:
['5\n1 3 3 2 1\n', '5\n1 2 3 4 5\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Filya should select *x* = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements. | ```python
n=int(input())
l=list(input().split())
l=[int(y) for y in l]
d=dict()
for x in range(min(l), max(l)):
poss=True
l1=l.copy()
for i in range(len(l)):
if l1[i] < min(l)+x:
l1[i] += x
elif l1[i] > min(l)+x:
l1[i]-=x
if i>0:
if l1[i] != l1[i-1]:
poss=False
d[x]=poss
if True in d.values():
print('YES')
else:
print('NO')
``` | 0 | |
909 | A | Generate Login | PROGRAMMING | 1,000 | [
"brute force",
"greedy",
"sortings"
] | null | null | The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac". | The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive. | Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well. | [
"harry potter\n",
"tom riddle\n"
] | [
"hap\n",
"tomr\n"
] | none | 500 | [
{
"input": "harry potter",
"output": "hap"
},
{
"input": "tom riddle",
"output": "tomr"
},
{
"input": "a qdpinbmcrf",
"output": "aq"
},
{
"input": "wixjzniiub ssdfodfgap",
"output": "wis"
},
{
"input": "z z",
"output": "zz"
},
{
"input": "ertuyivhfg v"... | 1,690,425,935 | 2,147,483,647 | Python 3 | OK | TESTS | 64 | 46 | 0 | a, b = input().split()
v = []
for i in range(len(a)):
v.append(a[:i+1]+b[0])
v.sort()
print(v[0]) | Title: Generate Login
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
Input Specification:
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
Output Specification:
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
Demo Input:
['harry potter\n', 'tom riddle\n']
Demo Output:
['hap\n', 'tomr\n']
Note:
none | ```python
a, b = input().split()
v = []
for i in range(len(a)):
v.append(a[:i+1]+b[0])
v.sort()
print(v[0])
``` | 3 | |
465 | A | inc ARG | PROGRAMMING | 900 | [
"implementation"
] | null | null | Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit. | Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell. | [
"4\n1100\n",
"4\n1111\n"
] | [
"3\n",
"4\n"
] | In the first sample the cell ends up with value 0010, in the second sample — with 0000. | 500 | [
{
"input": "4\n1100",
"output": "3"
},
{
"input": "4\n1111",
"output": "4"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "1"
},
{
"input": "2\n01",
"output": "1"
},
{
"input": "2\n1... | 1,586,002,546 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 108 | 307,200 | n=int(input())
st=input()
num=0
for i in range(n):
if(st[i]=='1'):
num+=(2**i)
num=num+1
b=bin(num).replace("0b","")
b=b[::-1]
ans=0
l=min(n,len(b))
for i in range(l):
if(st[i]!=b[i]):
ans+=1
for i in range(l,n):
if(st[i]!='0'):
ans=ans+1
print(ans)
| Title: inc ARG
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.
Output Specification:
Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.
Demo Input:
['4\n1100\n', '4\n1111\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample the cell ends up with value 0010, in the second sample — with 0000. | ```python
n=int(input())
st=input()
num=0
for i in range(n):
if(st[i]=='1'):
num+=(2**i)
num=num+1
b=bin(num).replace("0b","")
b=b[::-1]
ans=0
l=min(n,len(b))
for i in range(l):
if(st[i]!=b[i]):
ans+=1
for i in range(l,n):
if(st[i]!='0'):
ans=ans+1
print(ans)
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,597,673,539 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 77 | 0 | s=input()
h="hello"
if s==5:
print("NO")
else:
print("YES")
| Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s=input()
h="hello"
if s==5:
print("NO")
else:
print("YES")
``` | 0 |
13 | A | Numbers | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Numbers | 1 | 64 | Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10. | Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000). | Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator. | [
"5\n",
"3\n"
] | [
"7/3\n",
"2/1\n"
] | In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | 0 | [
{
"input": "5",
"output": "7/3"
},
{
"input": "3",
"output": "2/1"
},
{
"input": "1000",
"output": "90132/499"
},
{
"input": "927",
"output": "155449/925"
},
{
"input": "260",
"output": "6265/129"
},
{
"input": "131",
"output": "3370/129"
},
{
... | 1,609,246,197 | 2,147,483,647 | Python 3 | OK | TESTS | 99 | 218 | 0 | def num(r):
import math
x = 0
for a in range(2, r):
temp = r
while (temp != 0):
x += temp%a;temp //= a
t = (math.gcd(x, r-2))
print(str(x//t)+'/'+str((r-2)//t))
r = int(input())
num(r)
| Title: Numbers
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input Specification:
Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).
Output Specification:
Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.
Demo Input:
['5\n', '3\n']
Demo Output:
['7/3\n', '2/1\n']
Note:
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | ```python
def num(r):
import math
x = 0
for a in range(2, r):
temp = r
while (temp != 0):
x += temp%a;temp //= a
t = (math.gcd(x, r-2))
print(str(x//t)+'/'+str((r-2)//t))
r = int(input())
num(r)
``` | 3.891 |
344 | A | Magnets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | On the single line of the output print the number of groups of magnets. | [
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] | [
"3\n",
"2\n"
] | The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | 500 | [
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",... | 1,698,436,024 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 278 | 8,704,000 | n = int(input())
l = list()
for i in range(n):
x = input()
l.append(x)
s = 0
for i in range(n-1):
if l[i+1] != l[i]:
s += 1
print(s+1)
| Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | ```python
n = int(input())
l = list()
for i in range(n):
x = input()
l.append(x)
s = 0
for i in range(n-1):
if l[i+1] != l[i]:
s += 1
print(s+1)
``` | 3 | |
950 | A | Left-handers, Right-handers and Ambidexters | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively. | The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training. | Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players. | [
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] | [
"6\n",
"14\n",
"0\n"
] | In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand. | 500 | [
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
... | 1,530,354,108 | 2,147,483,647 | Python 3 | OK | TESTS | 79 | 109 | 0 | l,r,a= map(int,input().split())
mn = min(l,r)
mx = max(l,r)
if(l== 0 and a == 0 ) or (r==0 and a ==0):print(0)
elif mx != mn :
if(mn+a<=mx):print(2*(mn+a))
else:
rm = (mn+a)-mx
if(rm%2):print(2*mx+rm-1)
else:print(2*mx+rm)
else:
if(l+a+r)%2:print((l+a+r)-1)
else:print((l+a+r))
| Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand. | ```python
l,r,a= map(int,input().split())
mn = min(l,r)
mx = max(l,r)
if(l== 0 and a == 0 ) or (r==0 and a ==0):print(0)
elif mx != mn :
if(mn+a<=mx):print(2*(mn+a))
else:
rm = (mn+a)-mx
if(rm%2):print(2*mx+rm-1)
else:print(2*mx+rm)
else:
if(l+a+r)%2:print((l+a+r)-1)
else:print((l+a+r))
``` | 3 | |
780 | A | Andryusha and Socks | PROGRAMMING | 800 | [
"implementation"
] | null | null | Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time? | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair. | Print single integer — the maximum number of socks that were on the table at the same time. | [
"1\n1 1\n",
"3\n2 1 1 3 2 3\n"
] | [
"1\n",
"2\n"
] | In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | 500 | [
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n2 1 1 3 2 3",
"output": "2"
},
{
"input": "5\n5 1 3 2 4 3 1 2 4 5",
"output": "5"
},
{
"input": "10\n4 2 6 3 4 8 7 1 1 5 2 10 6 8 3 5 10 9 9 7",
"output": "6"
},
{
"input": "50\n30 47 31 38 37 50 36 43 9 23 2 2 ... | 1,579,181,336 | 2,147,483,647 | Python 3 | OK | TESTS | 56 | 233 | 14,028,800 | def solve(n, x):
maxi = 0
c = set()
for i in x:
if i in c:
c.remove(i)
else:
c.add(i)
maxi = max(maxi, len(c))
return maxi
def main():
n = int(input())
x = list(map(int, input().split()))
print(solve(n, x))
main()
| Title: Andryusha and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
Output Specification:
Print single integer — the maximum number of socks that were on the table at the same time.
Demo Input:
['1\n1 1\n', '3\n2 1 1 3 2 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | ```python
def solve(n, x):
maxi = 0
c = set()
for i in x:
if i in c:
c.remove(i)
else:
c.add(i)
maxi = max(maxi, len(c))
return maxi
def main():
n = int(input())
x = list(map(int, input().split()))
print(solve(n, x))
main()
``` | 3 | |
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"
}... | 1,622,854,152 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 14 | 186 | 0 | a = input ()
b = input ()
c1 = str (int (a) + int (b))
while '0' in a or '0' in b:
a = a.replace ('0', '')
b = b.replace ('0', '')
c1 = c1.replace ('0', '')
c2 = str (int (a) + int (b))
if c1 == c2:
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
a = input ()
b = input ()
c1 = str (int (a) + int (b))
while '0' in a or '0' in b:
a = a.replace ('0', '')
b = b.replace ('0', '')
c1 = c1.replace ('0', '')
c2 = str (int (a) + int (b))
if c1 == c2:
print ('YES')
else:
print ('NO')
``` | 0 |
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