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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
216 | B | Forming Teams | PROGRAMMING | 1,700 | [
"dfs and similar",
"implementation"
] | null | null | One day *n* students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.
We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student *A* is an archenemy to student *B*, then student *B* is an archenemy to student *A*.
The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.
Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100) — the number of students and the number of pairs of archenemies correspondingly.
Next *m* lines describe enmity between students. Each enmity is described as two numbers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.
You can consider the students indexed in some manner with distinct integers from 1 to *n*. | Print a single integer — the minimum number of students you will have to send to the bench in order to start the game. | [
"5 4\n1 2\n2 4\n5 3\n1 4\n",
"6 2\n1 4\n3 4\n",
"6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4\n"
] | [
"1",
"0",
"2"
] | none | 1,500 | [
{
"input": "5 4\n1 2\n2 4\n5 3\n1 4",
"output": "1"
},
{
"input": "6 2\n1 4\n3 4",
"output": "0"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "2"
},
{
"input": "5 1\n1 2",
"output": "1"
},
{
"input": "8 8\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 1",
... | 1,679,685,449 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 23 | 92 | 0 | a,b=map(int,input().split())
d={}
li=[]
r=0
for i in range(b):
x,y=map(int,input().split())
d[x]=y
d[y]=x
li+=[(x,y)]
for j,h in li:
if d[j]==d[h]:
r+=1
if (a-r)%2==0:
print(r)
else:
print(r+1) | Title: Forming Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.
We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student *A* is an archenemy to student *B*, then student *B* is an archenemy to student *A*.
The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.
Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100) — the number of students and the number of pairs of archenemies correspondingly.
Next *m* lines describe enmity between students. Each enmity is described as two numbers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.
You can consider the students indexed in some manner with distinct integers from 1 to *n*.
Output Specification:
Print a single integer — the minimum number of students you will have to send to the bench in order to start the game.
Demo Input:
['5 4\n1 2\n2 4\n5 3\n1 4\n', '6 2\n1 4\n3 4\n', '6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4\n']
Demo Output:
['1', '0', '2']
Note:
none | ```python
a,b=map(int,input().split())
d={}
li=[]
r=0
for i in range(b):
x,y=map(int,input().split())
d[x]=y
d[y]=x
li+=[(x,y)]
for j,h in li:
if d[j]==d[h]:
r+=1
if (a-r)%2==0:
print(r)
else:
print(r+1)
``` | 0 | |
761 | C | Dasha and Password | PROGRAMMING | 1,500 | [
"brute force",
"dp",
"implementation"
] | null | null | After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length *n* which satisfies the following requirements:
- There is at least one digit in the string, - There is at least one lowercase (small) letter of the Latin alphabet in the string, - There is at least one of three listed symbols in the string: '#', '*', '&'.
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length *m*, on each of these *n* strings there is a pointer on some character. The *i*-th character displayed on the screen is the pointed character in the *i*-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index *m*, and when we move it to the right from the position *m* it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password. | The first line contains two integers *n*, *m* (3<=≤<=*n*<=≤<=50,<=1<=≤<=*m*<=≤<=50) — the length of the password and the length of strings which are assigned to password symbols.
Each of the next *n* lines contains the string which is assigned to the *i*-th symbol of the password string. Its length is *m*, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password. | Print one integer — the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password. | [
"3 4\n1**2\na3*0\nc4**\n",
"5 5\n#*&#*\n*a1c&\n&q2w*\n#a3c#\n*&#*&\n"
] | [
"1\n",
"3\n"
] | In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
In the second test one of possible algorithms will be:
- to move the pointer of the second symbol once to the right. - to move the pointer of the third symbol twice to the right. | 1,500 | [
{
"input": "3 4\n1**2\na3*0\nc4**",
"output": "1"
},
{
"input": "5 5\n#*&#*\n*a1c&\n&q2w*\n#a3c#\n*&#*&",
"output": "3"
},
{
"input": "5 2\n&l\n*0\n*9\n*#\n#o",
"output": "2"
},
{
"input": "25 16\nvza**ooxkmd#*ywa\ndip#*#&ef&z&&&pv\nwggob&&72#*&&nku\nrsb##*&jm&#u**te\nzif#lu#... | 1,518,855,352 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 62 | 19,968,000 | n,leng=map(int,input().split())
a=leng
d=leng
c=leng
from re import search
e=[[],[],[],[]]
for i in range(n):
s=input()
sr=s[::-1]
let=search('[a-z]',s)
dig=search('\d', s)
sig=search('\*|#|&', s)
if let:
let=min(let.start(),search('[a-z]',sr).start()+1)
e[1].append([let, i])
if dig:
dig = min(dig.start(), search('\d', sr).start() + 1)
e[2].append([dig, i])
if sig:
sig = min(sig.start(), search('\*|#|&', sr).start() + 1)
e[3].append([sig, i])
if len(e[1])>3:
e[1].sort()
e[1].pop()
if len(e[2])>3:
e[2].sort()
e[2].pop()
if len(e[3])>3:
e[3].sort()
e[3].pop()
ans=leng*3
e.pop(0)
def rec(se,ind):
if len(se)==3:
global ans
ans=min(ans,sum(se))
else:
for el,i in e[len(se)]:
if i not in ind:
rec(se+[el],ind|{i})
print(e)
rec([],set())
print(ans) | Title: Dasha and Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length *n* which satisfies the following requirements:
- There is at least one digit in the string, - There is at least one lowercase (small) letter of the Latin alphabet in the string, - There is at least one of three listed symbols in the string: '#', '*', '&'.
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length *m*, on each of these *n* strings there is a pointer on some character. The *i*-th character displayed on the screen is the pointed character in the *i*-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index *m*, and when we move it to the right from the position *m* it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.
Input Specification:
The first line contains two integers *n*, *m* (3<=≤<=*n*<=≤<=50,<=1<=≤<=*m*<=≤<=50) — the length of the password and the length of strings which are assigned to password symbols.
Each of the next *n* lines contains the string which is assigned to the *i*-th symbol of the password string. Its length is *m*, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password.
Output Specification:
Print one integer — the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.
Demo Input:
['3 4\n1**2\na3*0\nc4**\n', '5 5\n#*&#*\n*a1c&\n&q2w*\n#a3c#\n*&#*&\n']
Demo Output:
['1\n', '3\n']
Note:
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
In the second test one of possible algorithms will be:
- to move the pointer of the second symbol once to the right. - to move the pointer of the third symbol twice to the right. | ```python
n,leng=map(int,input().split())
a=leng
d=leng
c=leng
from re import search
e=[[],[],[],[]]
for i in range(n):
s=input()
sr=s[::-1]
let=search('[a-z]',s)
dig=search('\d', s)
sig=search('\*|#|&', s)
if let:
let=min(let.start(),search('[a-z]',sr).start()+1)
e[1].append([let, i])
if dig:
dig = min(dig.start(), search('\d', sr).start() + 1)
e[2].append([dig, i])
if sig:
sig = min(sig.start(), search('\*|#|&', sr).start() + 1)
e[3].append([sig, i])
if len(e[1])>3:
e[1].sort()
e[1].pop()
if len(e[2])>3:
e[2].sort()
e[2].pop()
if len(e[3])>3:
e[3].sort()
e[3].pop()
ans=leng*3
e.pop(0)
def rec(se,ind):
if len(se)==3:
global ans
ans=min(ans,sum(se))
else:
for el,i in e[len(se)]:
if i not in ind:
rec(se+[el],ind|{i})
print(e)
rec([],set())
print(ans)
``` | 0 | |
1,006 | B | Polycarp's Practice | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"sortings"
] | null | null | Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$. | The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them). | In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them. | [
"8 3\n5 4 2 6 5 1 9 2\n",
"5 1\n1 1 1 1 1\n",
"4 2\n1 2000 2000 2\n"
] | [
"20\n3 2 3",
"1\n5\n",
"4000\n2 2\n"
] | The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$. | 0 | [
{
"input": "8 3\n5 4 2 6 5 1 9 2",
"output": "20\n4 1 3"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "1\n5"
},
{
"input": "4 2\n1 2000 2000 2",
"output": "4000\n2 2"
},
{
"input": "1 1\n2000",
"output": "2000\n1"
},
{
"input": "1 1\n1234",
"output": "1234\n1"
... | 1,667,977,251 | 2,751 | PyPy 3-64 | OK | TESTS | 37 | 77 | 3,891,200 | def main():
n,k = map(int, input().split())
nums = list(map(int, input().split()))
a = [False] * len(nums)
order = sorted([(nums[i], i) for i in range(len(nums))])
ans = 0
count = k
i = len(nums)-1
while count:
ans += order[i][0]
i-=1
count-=1
count = k
i = len(nums)-1
while count:
a[order[i][1]] = True
count -= 1
i-=1
answer = []
count = 0
cur = []
for i in range(len(nums)):
if a[i] == True:
if count == 1:
answer.append(len(cur))
cur = [nums[i]]
count = 1
else:
cur.append(nums[i])
count += 1
else:
cur.append(nums[i])
if cur:
answer.append(len(cur))
print(ans)
print(*answer)
main()
# import sys, threading
# if __name__ == '__main__':
# sys.setrecursionlimit(1 << 30)
# threading.stack_size(1 << 27)
# main_thread = threading.Thread(target=main)
# main_thread.start()
# main_thread.join() | Title: Polycarp's Practice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$.
Input Specification:
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).
Output Specification:
In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them.
Demo Input:
['8 3\n5 4 2 6 5 1 9 2\n', '5 1\n1 1 1 1 1\n', '4 2\n1 2000 2000 2\n']
Demo Output:
['20\n3 2 3', '1\n5\n', '4000\n2 2\n']
Note:
The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$. | ```python
def main():
n,k = map(int, input().split())
nums = list(map(int, input().split()))
a = [False] * len(nums)
order = sorted([(nums[i], i) for i in range(len(nums))])
ans = 0
count = k
i = len(nums)-1
while count:
ans += order[i][0]
i-=1
count-=1
count = k
i = len(nums)-1
while count:
a[order[i][1]] = True
count -= 1
i-=1
answer = []
count = 0
cur = []
for i in range(len(nums)):
if a[i] == True:
if count == 1:
answer.append(len(cur))
cur = [nums[i]]
count = 1
else:
cur.append(nums[i])
count += 1
else:
cur.append(nums[i])
if cur:
answer.append(len(cur))
print(ans)
print(*answer)
main()
# import sys, threading
# if __name__ == '__main__':
# sys.setrecursionlimit(1 << 30)
# threading.stack_size(1 << 27)
# main_thread = threading.Thread(target=main)
# main_thread.start()
# main_thread.join()
``` | 3 | |
448 | A | Rewards | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bizon the Champion is called the Champion for a reason.
Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals.
Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:
- any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals.
Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled. | The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100).
The numbers in the lines are separated by single spaces. | Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes). | [
"1 1 1\n1 1 1\n4\n",
"1 1 3\n2 3 4\n2\n",
"1 0 0\n1 0 0\n1\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "1 1 1\n1 1 1\n4",
"output": "YES"
},
{
"input": "1 1 3\n2 3 4\n2",
"output": "YES"
},
{
"input": "1 0 0\n1 0 0\n1",
"output": "NO"
},
{
"input": "0 0 0\n0 0 0\n1",
"output": "YES"
},
{
"input": "100 100 100\n100 100 100\n100",
"output": "YES"
},
... | 1,643,270,898 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 31 | 0 | a1,a2,a3 = map(int, input().split())
b1,b2,b3 = map(int, input().split())
n = int(input())
c = a1 + a2 + a3
c2 = b1 + b2 + b3
if (c2 >= 1 and n == 1) and not ((c <= 10 and c2 <= 10 and n == 2) or (n > 2 and c >=1 and c2 >= 1)):
print("NO")
else:
print("YES") | Title: Rewards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bizon the Champion is called the Champion for a reason.
Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals.
Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:
- any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals.
Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.
Input Specification:
The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100).
The numbers in the lines are separated by single spaces.
Output Specification:
Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).
Demo Input:
['1 1 1\n1 1 1\n4\n', '1 1 3\n2 3 4\n2\n', '1 0 0\n1 0 0\n1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none | ```python
a1,a2,a3 = map(int, input().split())
b1,b2,b3 = map(int, input().split())
n = int(input())
c = a1 + a2 + a3
c2 = b1 + b2 + b3
if (c2 >= 1 and n == 1) and not ((c <= 10 and c2 <= 10 and n == 2) or (n > 2 and c >=1 and c2 >= 1)):
print("NO")
else:
print("YES")
``` | 0 | |
639 | C | Bear and Polynomials | PROGRAMMING | 2,200 | [
"hashing",
"implementation",
"math"
] | null | null | Limak is a little polar bear. He doesn't have many toys and thus he often plays with polynomials.
He considers a polynomial valid if its degree is *n* and its coefficients are integers not exceeding *k* by the absolute value. More formally:
Let *a*0,<=*a*1,<=...,<=*a**n* denote the coefficients, so . Then, a polynomial *P*(*x*) is valid if all the following conditions are satisfied:
- *a**i* is integer for every *i*; - |*a**i*|<=≤<=*k* for every *i*; - *a**n*<=≠<=0.
Limak has recently got a valid polynomial *P* with coefficients *a*0,<=*a*1,<=*a*2,<=...,<=*a**n*. He noticed that *P*(2)<=≠<=0 and he wants to change it. He is going to change one coefficient to get a valid polynomial *Q* of degree *n* that *Q*(2)<==<=0. Count the number of ways to do so. You should count two ways as a distinct if coefficients of target polynoms differ. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=200<=000,<=1<=≤<=*k*<=≤<=109) — the degree of the polynomial and the limit for absolute values of coefficients.
The second line contains *n*<=+<=1 integers *a*0,<=*a*1,<=...,<=*a**n* (|*a**i*|<=≤<=*k*,<=*a**n*<=≠<=0) — describing a valid polynomial . It's guaranteed that *P*(2)<=≠<=0. | Print the number of ways to change one coefficient to get a valid polynomial *Q* that *Q*(2)<==<=0. | [
"3 1000000000\n10 -9 -3 5\n",
"3 12\n10 -9 -3 5\n",
"2 20\n14 -7 19\n"
] | [
"3\n",
"2\n",
"0\n"
] | In the first sample, we are given a polynomial *P*(*x*) = 10 - 9*x* - 3*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup>.
Limak can change one coefficient in three ways:
1. He can set *a*<sub class="lower-index">0</sub> = - 10. Then he would get *Q*(*x*) = - 10 - 9*x* - 3*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup> and indeed *Q*(2) = - 10 - 18 - 12 + 40 = 0. 1. Or he can set *a*<sub class="lower-index">2</sub> = - 8. Then *Q*(*x*) = 10 - 9*x* - 8*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup> and indeed *Q*(2) = 10 - 18 - 32 + 40 = 0. 1. Or he can set *a*<sub class="lower-index">1</sub> = - 19. Then *Q*(*x*) = 10 - 19*x* - 3*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup> and indeed *Q*(2) = 10 - 38 - 12 + 40 = 0.
In the second sample, we are given the same polynomial. This time though, *k* is equal to 12 instead of 10<sup class="upper-index">9</sup>. Two first of ways listed above are still valid but in the third way we would get |*a*<sub class="lower-index">1</sub>| > *k* what is not allowed. Thus, the answer is 2 this time. | 1,000 | [
{
"input": "3 1000000000\n10 -9 -3 5",
"output": "3"
},
{
"input": "3 12\n10 -9 -3 5",
"output": "2"
},
{
"input": "2 20\n14 -7 19",
"output": "0"
},
{
"input": "5 5\n0 -4 -2 -2 0 5",
"output": "1"
},
{
"input": "6 10\n-2 -1 7 -3 2 7 -6",
"output": "2"
},
... | 1,459,189,830 | 6,930 | PyPy 3 | WRONG_ANSWER | PRETESTS | 3 | 93 | 23,040,000 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
n, k = map(int, input().split())
a = [i for i in map(int, input().split())]
P = 0
power = 1
for i in a:
P += i * power
power *= 2
m = 0
total = 0
while P % 2 == 0 and m < len(a) - 1:
a_m1 = a[m] - P
if abs(a_m1) <= k:
total += 1
P //= 2
m += 1
if m < len(a) - 1 and abs(a[m] - P) <= k:
total += 1
print(total)
| Title: Bear and Polynomials
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little polar bear. He doesn't have many toys and thus he often plays with polynomials.
He considers a polynomial valid if its degree is *n* and its coefficients are integers not exceeding *k* by the absolute value. More formally:
Let *a*0,<=*a*1,<=...,<=*a**n* denote the coefficients, so . Then, a polynomial *P*(*x*) is valid if all the following conditions are satisfied:
- *a**i* is integer for every *i*; - |*a**i*|<=≤<=*k* for every *i*; - *a**n*<=≠<=0.
Limak has recently got a valid polynomial *P* with coefficients *a*0,<=*a*1,<=*a*2,<=...,<=*a**n*. He noticed that *P*(2)<=≠<=0 and he wants to change it. He is going to change one coefficient to get a valid polynomial *Q* of degree *n* that *Q*(2)<==<=0. Count the number of ways to do so. You should count two ways as a distinct if coefficients of target polynoms differ.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=200<=000,<=1<=≤<=*k*<=≤<=109) — the degree of the polynomial and the limit for absolute values of coefficients.
The second line contains *n*<=+<=1 integers *a*0,<=*a*1,<=...,<=*a**n* (|*a**i*|<=≤<=*k*,<=*a**n*<=≠<=0) — describing a valid polynomial . It's guaranteed that *P*(2)<=≠<=0.
Output Specification:
Print the number of ways to change one coefficient to get a valid polynomial *Q* that *Q*(2)<==<=0.
Demo Input:
['3 1000000000\n10 -9 -3 5\n', '3 12\n10 -9 -3 5\n', '2 20\n14 -7 19\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
In the first sample, we are given a polynomial *P*(*x*) = 10 - 9*x* - 3*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup>.
Limak can change one coefficient in three ways:
1. He can set *a*<sub class="lower-index">0</sub> = - 10. Then he would get *Q*(*x*) = - 10 - 9*x* - 3*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup> and indeed *Q*(2) = - 10 - 18 - 12 + 40 = 0. 1. Or he can set *a*<sub class="lower-index">2</sub> = - 8. Then *Q*(*x*) = 10 - 9*x* - 8*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup> and indeed *Q*(2) = 10 - 18 - 32 + 40 = 0. 1. Or he can set *a*<sub class="lower-index">1</sub> = - 19. Then *Q*(*x*) = 10 - 19*x* - 3*x*<sup class="upper-index">2</sup> + 5*x*<sup class="upper-index">3</sup> and indeed *Q*(2) = 10 - 38 - 12 + 40 = 0.
In the second sample, we are given the same polynomial. This time though, *k* is equal to 12 instead of 10<sup class="upper-index">9</sup>. Two first of ways listed above are still valid but in the third way we would get |*a*<sub class="lower-index">1</sub>| > *k* what is not allowed. Thus, the answer is 2 this time. | ```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n, k = map(int, input().split())
a = [i for i in map(int, input().split())]
P = 0
power = 1
for i in a:
P += i * power
power *= 2
m = 0
total = 0
while P % 2 == 0 and m < len(a) - 1:
a_m1 = a[m] - P
if abs(a_m1) <= k:
total += 1
P //= 2
m += 1
if m < len(a) - 1 and abs(a[m] - P) <= k:
total += 1
print(total)
``` | 0 | |
847 | M | Weather Tomorrow | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral.
Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day.
Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days for which the average air temperature is known.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=≤<=*t**i*<=≤<=1000) — where *t**i* is the average temperature in the *i*-th day. | Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000. | [
"5\n10 5 0 -5 -10\n",
"4\n1 1 1 1\n",
"3\n5 1 -5\n",
"2\n900 1000\n"
] | [
"-15\n",
"1\n",
"-5\n",
"1100\n"
] | In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is - 10 - 5 = - 15.
In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1.
In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to - 5.
In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000 + 100 = 1100. | 0 | [
{
"input": "5\n10 5 0 -5 -10",
"output": "-15"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "3\n5 1 -5",
"output": "-5"
},
{
"input": "2\n900 1000",
"output": "1100"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "3\n2 5 8",
"output... | 1,585,251,283 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | def func(n,a):
d=a[1]-a[0]
for i in range(2,n):
if a[i]-a[i-1]!=d:
return(a[-1])
return a[-1]+d
n=int(input())
a=list(map(int,input().split()))
func(n,a)
| Title: Weather Tomorrow
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral.
Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day.
Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days for which the average air temperature is known.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=≤<=*t**i*<=≤<=1000) — where *t**i* is the average temperature in the *i*-th day.
Output Specification:
Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000.
Demo Input:
['5\n10 5 0 -5 -10\n', '4\n1 1 1 1\n', '3\n5 1 -5\n', '2\n900 1000\n']
Demo Output:
['-15\n', '1\n', '-5\n', '1100\n']
Note:
In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is - 10 - 5 = - 15.
In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1.
In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to - 5.
In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000 + 100 = 1100. | ```python
def func(n,a):
d=a[1]-a[0]
for i in range(2,n):
if a[i]-a[i-1]!=d:
return(a[-1])
return a[-1]+d
n=int(input())
a=list(map(int,input().split()))
func(n,a)
``` | 0 | |
146 | B | Lucky Mask | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer *n* the number that is obtained after successive writing of all lucky digits of number *n* from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer *a* and a lucky number *b*. Help him find the minimum number *c* (*c*<=><=*a*) such that the mask of number *c* equals *b*. | The only line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=105). It is guaranteed that number *b* is lucky. | In the only line print a single number — the number *c* that is sought by Petya. | [
"1 7\n",
"100 47\n"
] | [
"7\n",
"147\n"
] | none | 1,000 | [
{
"input": "1 7",
"output": "7"
},
{
"input": "100 47",
"output": "147"
},
{
"input": "458 47",
"output": "467"
},
{
"input": "7 7",
"output": "17"
},
{
"input": "547 47",
"output": "647"
},
{
"input": "77 77",
"output": "177"
},
{
"input":... | 1,620,751,042 | 4,042 | Python 3 | OK | TESTS | 43 | 498 | 6,758,400 | def mask(a):
x1 = str(a)
new_str = ""
len1 = 0
for i in range(len(x1)):
if x1[i] =='4' or x1[i] =='7':
new_str += x1[i]
#print(new_str)
if len(new_str)>0:
len1 = int(new_str)
#print(n)
return len1
a, b = map(int,input().split())
while 1:
a += 1
if mask(a)==b:
print(a)
break | Title: Lucky Mask
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer *n* the number that is obtained after successive writing of all lucky digits of number *n* from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer *a* and a lucky number *b*. Help him find the minimum number *c* (*c*<=><=*a*) such that the mask of number *c* equals *b*.
Input Specification:
The only line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=105). It is guaranteed that number *b* is lucky.
Output Specification:
In the only line print a single number — the number *c* that is sought by Petya.
Demo Input:
['1 7\n', '100 47\n']
Demo Output:
['7\n', '147\n']
Note:
none | ```python
def mask(a):
x1 = str(a)
new_str = ""
len1 = 0
for i in range(len(x1)):
if x1[i] =='4' or x1[i] =='7':
new_str += x1[i]
#print(new_str)
if len(new_str)>0:
len1 = int(new_str)
#print(n)
return len1
a, b = map(int,input().split())
while 1:
a += 1
if mask(a)==b:
print(a)
break
``` | 3 | |
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "... | 1,602,505,134 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 278 | 0 | arr=[0]*2
arr=list(map(int, input().split()))
n=arr[0]
m=arr[1]
m %= int((n * (n + 1) / 2))
for i in range(1,n+1):
if m < i:
break
m -= i
print(m)
| Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
arr=[0]*2
arr=list(map(int, input().split()))
n=arr[0]
m=arr[1]
m %= int((n * (n + 1) / 2))
for i in range(1,n+1):
if m < i:
break
m -= i
print(m)
``` | 3.9305 |
219 | B | Special Offer! Super Price 999 Bourles! | PROGRAMMING | 1,400 | [
"implementation"
] | null | null | Polycarpus is an amateur businessman. Recently he was surprised to find out that the market for paper scissors is completely free! Without further ado, Polycarpus decided to start producing and selling such scissors.
Polycaprus calculated that the optimal celling price for such scissors would be *p* bourles. However, he read somewhere that customers are attracted by prices that say something like "Special Offer! Super price 999 bourles!". So Polycarpus decided to lower the price a little if it leads to the desired effect.
Polycarpus agrees to lower the price by no more than *d* bourles so that the number of nines at the end of the resulting price is maximum. If there are several ways to do it, he chooses the maximum possible price.
Note, Polycarpus counts only the trailing nines in a price. | The first line contains two integers *p* and *d* (1<=≤<=*p*<=≤<=1018; 0<=≤<=*d*<=<<=*p*) — the initial price of scissors and the maximum possible price reduction.
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. | Print the required price — the maximum price that ends with the largest number of nines and that is less than *p* by no more than *d*.
The required number shouldn't have leading zeroes. | [
"1029 102\n",
"27191 17\n"
] | [
"999\n",
"27189\n"
] | none | 1,000 | [
{
"input": "1029 102",
"output": "999"
},
{
"input": "27191 17",
"output": "27189"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "9 0",
"output": "9"
},
{
"input": "20 1",
"output": "19"
},
{
"input": "100 23",
"output": "99"
},
{
"input... | 1,630,743,026 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 1,000 | 6,963,200 | # -*- coding: utf-8 -*-
"""Untitled112.ipynb
Automatically generated by Colaboratory.
Original file is located at
https://colab.research.google.com/drive/1BjNvsi4iTWnv6mWwHBPOZW6Yi-sCMdO0
"""
l1=list(map(int,input().split()))
a=l1[0]
b=l1[1]
d={}
for x in range(a-b,a+1):
c=str(x)
n=0
for i in range(len(c)-1,-1,-1):
if c[i]=="9":
n=n+1
else:
break
d[n]=x
print(d[max(d)])
d | Title: Special Offer! Super Price 999 Bourles!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is an amateur businessman. Recently he was surprised to find out that the market for paper scissors is completely free! Without further ado, Polycarpus decided to start producing and selling such scissors.
Polycaprus calculated that the optimal celling price for such scissors would be *p* bourles. However, he read somewhere that customers are attracted by prices that say something like "Special Offer! Super price 999 bourles!". So Polycarpus decided to lower the price a little if it leads to the desired effect.
Polycarpus agrees to lower the price by no more than *d* bourles so that the number of nines at the end of the resulting price is maximum. If there are several ways to do it, he chooses the maximum possible price.
Note, Polycarpus counts only the trailing nines in a price.
Input Specification:
The first line contains two integers *p* and *d* (1<=≤<=*p*<=≤<=1018; 0<=≤<=*d*<=<<=*p*) — the initial price of scissors and the maximum possible price reduction.
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.
Output Specification:
Print the required price — the maximum price that ends with the largest number of nines and that is less than *p* by no more than *d*.
The required number shouldn't have leading zeroes.
Demo Input:
['1029 102\n', '27191 17\n']
Demo Output:
['999\n', '27189\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""Untitled112.ipynb
Automatically generated by Colaboratory.
Original file is located at
https://colab.research.google.com/drive/1BjNvsi4iTWnv6mWwHBPOZW6Yi-sCMdO0
"""
l1=list(map(int,input().split()))
a=l1[0]
b=l1[1]
d={}
for x in range(a-b,a+1):
c=str(x)
n=0
for i in range(len(c)-1,-1,-1):
if c[i]=="9":
n=n+1
else:
break
d[n]=x
print(d[max(d)])
d
``` | 0 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,588,935,184 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 109 | 6,656,000 | n = input()
m = input()
num = ""
for x in range(len(n)):
if n[x]==m[x]:
num+="0"
else:
num+="1"
print(num) | 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
n = input()
m = input()
num = ""
for x in range(len(n)):
if n[x]==m[x]:
num+="0"
else:
num+="1"
print(num)
``` | 3.960352 |
834 | A | The Useless Toy | PROGRAMMING | 900 | [
"implementation"
] | null | null | Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this. | There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position. | Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise. | [
"^ >\n1\n",
"< ^\n3\n",
"^ v\n6\n"
] | [
"cw\n",
"ccw\n",
"undefined\n"
] | none | 500 | [
{
"input": "^ >\n1",
"output": "cw"
},
{
"input": "< ^\n3",
"output": "ccw"
},
{
"input": "^ v\n6",
"output": "undefined"
},
{
"input": "^ >\n999999999",
"output": "ccw"
},
{
"input": "> v\n1",
"output": "cw"
},
{
"input": "v <\n1",
"output": "cw"
... | 1,603,000,331 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 108 | 307,200 | a=list(input().split())
n=int(input())
if (a[0]=='v' and a[1]=='^') or (a[0]=='^' and a[1]=='v') or (a[0]=='<' and a[1]=='>') or (a[0]=='>' and a[1]=='<'):
print("undefined")
else:
if (a[0]=='v' and a[1]=='<') or (a[0]=='<' and a[1]=='^') or (a[0]=='^' and a[1]=='>') or (a[0]=='>' and a[1]=='v'):
if n < 3 or n % 2 == 0:
print("cw")
else:
print("ccw")
else:
if n > 3 or n % 2 == 0:
print("ccw")
else:
print("cw") | Title: The Useless Toy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this.
Input Specification:
There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position.
Output Specification:
Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise.
Demo Input:
['^ >\n1\n', '< ^\n3\n', '^ v\n6\n']
Demo Output:
['cw\n', 'ccw\n', 'undefined\n']
Note:
none | ```python
a=list(input().split())
n=int(input())
if (a[0]=='v' and a[1]=='^') or (a[0]=='^' and a[1]=='v') or (a[0]=='<' and a[1]=='>') or (a[0]=='>' and a[1]=='<'):
print("undefined")
else:
if (a[0]=='v' and a[1]=='<') or (a[0]=='<' and a[1]=='^') or (a[0]=='^' and a[1]=='>') or (a[0]=='>' and a[1]=='v'):
if n < 3 or n % 2 == 0:
print("cw")
else:
print("ccw")
else:
if n > 3 or n % 2 == 0:
print("ccw")
else:
print("cw")
``` | 0 | |
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... | 1,649,068,628 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 61 | 0 | n,m = map(int,input().split())
l = list(map(int,input().split()))
ma = max(l)
for i in range(n-1,-1,-1):
if l[i]==ma:
print(i+1)
break | 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
n,m = map(int,input().split())
l = list(map(int,input().split()))
ma = max(l)
for i in range(n-1,-1,-1):
if l[i]==ma:
print(i+1)
break
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct. | Print a single integer — the required sum modulo 109<=+<=7. | [
"2\n4 7\n",
"3\n4 3 1\n"
] | [
"3\n",
"9\n"
] | There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9. | 0 | [
{
"input": "2\n4 7",
"output": "3"
},
{
"input": "3\n4 3 1",
"output": "9"
},
{
"input": "20\n8 11 13 19 21 34 36 44 57 58 61 63 76 78 79 81 85 86 90 95",
"output": "83396599"
},
{
"input": "20\n1 8 9 12 15 17 18 24 30 33 36 41 53 54 59 62 64 66 72 73",
"output": "6805914... | 1,495,310,287 | 6,787 | Python 3 | WRONG_ANSWER | PRETESTS | 2 | 46 | 0 | mod = 1000000007
n = int(input())
comps = list(map(int,input().split()))
comps.sort()
sum = 0
for i in range(n-1):
count = 1
for j in range(i+1,n):
sum = (sum + ((comps[j] - comps[i]) * count)) % mod
count += 1
print(sum % mod)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct.
Output Specification:
Print a single integer — the required sum modulo 109<=+<=7.
Demo Input:
['2\n4 7\n', '3\n4 3 1\n']
Demo Output:
['3\n', '9\n']
Note:
There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9. | ```python
mod = 1000000007
n = int(input())
comps = list(map(int,input().split()))
comps.sort()
sum = 0
for i in range(n-1):
count = 1
for j in range(i+1,n):
sum = (sum + ((comps[j] - comps[i]) * count)) % mod
count += 1
print(sum % mod)
``` | 0 | |
182 | D | Common Divisors | PROGRAMMING | 1,400 | [
"brute force",
"hashing",
"implementation",
"math",
"strings"
] | null | null | Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.
String *a* is the divisor of string *b* if and only if there exists a positive integer *x* such that if we write out string *a* consecutively *x* times, we get string *b*. For example, string "abab" has two divisors — "ab" and "abab".
Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him. | The first input line contains a non-empty string *s*1.
The second input line contains a non-empty string *s*2.
Lengths of strings *s*1 and *s*2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters. | Print the number of common divisors of strings *s*1 and *s*2. | [
"abcdabcd\nabcdabcdabcdabcd\n",
"aaa\naa\n"
] | [
"2\n",
"1\n"
] | In first sample the common divisors are strings "abcd" and "abcdabcd".
In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa". | 1,000 | [
{
"input": "abcdabcd\nabcdabcdabcdabcd",
"output": "2"
},
{
"input": "aaa\naa",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "3"
},
{
"input": "aaaaaaaaaaaaaa\naaaaaaaaaaaaaa",
"output": "4"
},
{
"input": "a\nb",
... | 1,596,743,886 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 2,000 | 30,720,000 | s1=input()
s2=input()
n=len(s1)
m=len(s2)
q=min(n,m)
num=0
for i in range(1,q+1):
p=s1[:i]
if n%i==0 and p*(n//i)==s1 and m%i==0 and p*(m//i)==s2 :
num+=1
print(num) | Title: Common Divisors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.
String *a* is the divisor of string *b* if and only if there exists a positive integer *x* such that if we write out string *a* consecutively *x* times, we get string *b*. For example, string "abab" has two divisors — "ab" and "abab".
Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him.
Input Specification:
The first input line contains a non-empty string *s*1.
The second input line contains a non-empty string *s*2.
Lengths of strings *s*1 and *s*2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters.
Output Specification:
Print the number of common divisors of strings *s*1 and *s*2.
Demo Input:
['abcdabcd\nabcdabcdabcdabcd\n', 'aaa\naa\n']
Demo Output:
['2\n', '1\n']
Note:
In first sample the common divisors are strings "abcd" and "abcdabcd".
In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa". | ```python
s1=input()
s2=input()
n=len(s1)
m=len(s2)
q=min(n,m)
num=0
for i in range(1,q+1):
p=s1[:i]
if n%i==0 and p*(n//i)==s1 and m%i==0 and p*(m//i)==s2 :
num+=1
print(num)
``` | 0 | |
574 | A | Bear and Elections | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe? | The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000. | Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate. | [
"5\n5 1 11 2 8\n",
"4\n1 8 8 8\n",
"2\n7 6\n"
] | [
"4\n",
"6\n",
"0\n"
] | In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.
In the third sample Limak is a winner without bribing any citizen. | 500 | [
{
"input": "5\n5 1 11 2 8",
"output": "4"
},
{
"input": "4\n1 8 8 8",
"output": "6"
},
{
"input": "2\n7 6",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "10\n100 200 57 99 1 1000 200 200 200 500",
"output": "451"
},
{
"input": "16\... | 1,440,917,719 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 62 | 0 | #!/usr/bin/python3.4
import time
SIZE = 26
matrix = [[0] * SIZE for i in range(SIZE)]
arr = [0] * SIZE
def timer(func, *pargs, **kargs):
start = time.time()
func(*pargs, **kargs)
return time.time() - start
if __name__ == "__main__":
n = int(input())
a = list(map(int, input().split()))
value, c = a[0], 0
a = a[1:]
while True:
a.sort(reverse=True)
if value > a[0]:
break
value += 1
c += 1
a[0] -= 1
print(c)
| Title: Bear and Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.
Output Specification:
Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.
Demo Input:
['5\n5 1 11 2 8\n', '4\n1 8 8 8\n', '2\n7 6\n']
Demo Output:
['4\n', '6\n', '0\n']
Note:
In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.
In the third sample Limak is a winner without bribing any citizen. | ```python
#!/usr/bin/python3.4
import time
SIZE = 26
matrix = [[0] * SIZE for i in range(SIZE)]
arr = [0] * SIZE
def timer(func, *pargs, **kargs):
start = time.time()
func(*pargs, **kargs)
return time.time() - start
if __name__ == "__main__":
n = int(input())
a = list(map(int, input().split()))
value, c = a[0], 0
a = a[1:]
while True:
a.sort(reverse=True)
if value > a[0]:
break
value += 1
c += 1
a[0] -= 1
print(c)
``` | 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",
... | 1,683,526,458 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | n = int(input())
teams = set()
score1 = score2 = 0
for i in range(n):
goal = input().strip()
teams.add(goal)
if goal == team1:
score1 += 1
else:
score2 += 1
if score1 > score2:
print(team1)
else:
print(team2)
| 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())
teams = set()
score1 = score2 = 0
for i in range(n):
goal = input().strip()
teams.add(goal)
if goal == team1:
score1 += 1
else:
score2 += 1
if score1 > score2:
print(team1)
else:
print(team2)
``` | -1 |
39 | C | Moon Craters | PROGRAMMING | 2,100 | [
"dp",
"sortings"
] | C. Moon Craters | 1 | 256 | There are lots of theories concerning the origin of moon craters. Most scientists stick to the meteorite theory, which says that the craters were formed as a result of celestial bodies colliding with the Moon. The other version is that the craters were parts of volcanoes.
An extraterrestrial intelligence research specialist professor Okulov (the namesake of the Okulov, the author of famous textbooks on programming) put forward an alternate hypothesis. Guess what kind of a hypothesis it was –– sure, the one including extraterrestrial mind involvement. Now the professor is looking for proofs of his hypothesis.
Professor has data from the moon robot that moves linearly in one direction along the Moon surface. The moon craters are circular in form with integer-valued radii. The moon robot records only the craters whose centers lay on his path and sends to the Earth the information on the distance from the centers of the craters to the initial point of its path and on the radii of the craters.
According to the theory of professor Okulov two craters made by an extraterrestrial intelligence for the aims yet unknown either are fully enclosed one in the other or do not intersect at all. Internal or external tangency is acceptable. However the experimental data from the moon robot do not confirm this theory! Nevertheless, professor Okulov is hopeful. He perfectly understands that to create any logical theory one has to ignore some data that are wrong due to faulty measuring (or skillful disguise by the extraterrestrial intelligence that will be sooner or later found by professor Okulov!) That’s why Okulov wants to choose among the available crater descriptions the largest set that would satisfy his theory. | The first line has an integer *n* (1<=≤<=*n*<=≤<=2000) — the number of discovered craters. The next *n* lines contain crater descriptions in the "*c**i* *r**i*" format, where *c**i* is the coordinate of the center of the crater on the moon robot’s path, *r**i* is the radius of the crater. All the numbers *c**i* and *r**i* are positive integers not exceeding 109. No two craters coincide. | In the first line output the number of craters in the required largest set. In the next line output space-separated numbers of craters that this set consists of. The craters are numbered from 1 to *n* in the order in which they were given in the input data. The numbers may be output in any order. If the result is not unique, output any. | [
"4\n1 1\n2 2\n4 1\n5 1\n"
] | [
"3\n1 2 4\n"
] | none | 0 | [] | 1,681,991,629 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using i64 = long long;
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int n;
std::cin >> n;
std::vector<int> c(n), r(n);
for (int i = 0; i < n; i++) {
std::cin >> c[i] >> r[i];
}
std::vector<int> ord(n);
std::iota(ord.begin(), ord.end(), 0);
std::sort(ord.begin(), ord.end(), [&](int i, int j) {
return r[i] < r[j];
});
std::vector<int> dp(n);
std::vector<std::vector<int>> sub(n);
std::vector<int> g(n), lst(n);
for (auto i : ord) {
std::vector<std::tuple<int, int, int>> e;
for (auto j : ord) {
if (c[j] - r[j] >= c[i] - r[i] && c[j] + r[j] <= c[i] + r[i]) {
e.emplace_back(c[j] - r[j], 1, j);
e.emplace_back(c[j] + r[j], -1, j);
}
}
std::sort(e.begin(), e.end());
int v = 0;
int u = -1;
for (auto [x, t, j] : e) {
if (t == 1) {
g[j] = v + dp[j];
lst[j] = u;
} else if (g[j] > v) {
v = g[j];
u = j;
}
}
dp[i] = v + 1;
while (u != -1) {
sub[i].push_back(u);
u = lst[u];
}
}
std::vector<std::tuple<int, int, int>> e;
for (auto j : ord) {
e.emplace_back(c[j] - r[j], 1, j);
e.emplace_back(c[j] + r[j], -1, j);
}
std::sort(e.begin(), e.end());
int v = 0;
int u = -1;
for (auto [x, t, j] : e) {
if (t == 1) {
g[j] = v + dp[j];
lst[j] = u;
} else if (g[j] > v) {
v = g[j];
u = j;
}
}
int ans = v;
std::cout << ans << "\n";
std::queue<int> q;
while (u != -1) {
q.push(u);
u = lst[u];
}
while (!q.empty()) {
int x = q.front();
q.pop();
std::cout << x + 1 << " ";
for (auto y : sub[x]) {
q.push(y);
}
}
std::cout << "\n";
return 0;
} | Title: Moon Craters
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
There are lots of theories concerning the origin of moon craters. Most scientists stick to the meteorite theory, which says that the craters were formed as a result of celestial bodies colliding with the Moon. The other version is that the craters were parts of volcanoes.
An extraterrestrial intelligence research specialist professor Okulov (the namesake of the Okulov, the author of famous textbooks on programming) put forward an alternate hypothesis. Guess what kind of a hypothesis it was –– sure, the one including extraterrestrial mind involvement. Now the professor is looking for proofs of his hypothesis.
Professor has data from the moon robot that moves linearly in one direction along the Moon surface. The moon craters are circular in form with integer-valued radii. The moon robot records only the craters whose centers lay on his path and sends to the Earth the information on the distance from the centers of the craters to the initial point of its path and on the radii of the craters.
According to the theory of professor Okulov two craters made by an extraterrestrial intelligence for the aims yet unknown either are fully enclosed one in the other or do not intersect at all. Internal or external tangency is acceptable. However the experimental data from the moon robot do not confirm this theory! Nevertheless, professor Okulov is hopeful. He perfectly understands that to create any logical theory one has to ignore some data that are wrong due to faulty measuring (or skillful disguise by the extraterrestrial intelligence that will be sooner or later found by professor Okulov!) That’s why Okulov wants to choose among the available crater descriptions the largest set that would satisfy his theory.
Input Specification:
The first line has an integer *n* (1<=≤<=*n*<=≤<=2000) — the number of discovered craters. The next *n* lines contain crater descriptions in the "*c**i* *r**i*" format, where *c**i* is the coordinate of the center of the crater on the moon robot’s path, *r**i* is the radius of the crater. All the numbers *c**i* and *r**i* are positive integers not exceeding 109. No two craters coincide.
Output Specification:
In the first line output the number of craters in the required largest set. In the next line output space-separated numbers of craters that this set consists of. The craters are numbered from 1 to *n* in the order in which they were given in the input data. The numbers may be output in any order. If the result is not unique, output any.
Demo Input:
['4\n1 1\n2 2\n4 1\n5 1\n']
Demo Output:
['3\n1 2 4\n']
Note:
none | ```python
#include <bits/stdc++.h>
using i64 = long long;
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int n;
std::cin >> n;
std::vector<int> c(n), r(n);
for (int i = 0; i < n; i++) {
std::cin >> c[i] >> r[i];
}
std::vector<int> ord(n);
std::iota(ord.begin(), ord.end(), 0);
std::sort(ord.begin(), ord.end(), [&](int i, int j) {
return r[i] < r[j];
});
std::vector<int> dp(n);
std::vector<std::vector<int>> sub(n);
std::vector<int> g(n), lst(n);
for (auto i : ord) {
std::vector<std::tuple<int, int, int>> e;
for (auto j : ord) {
if (c[j] - r[j] >= c[i] - r[i] && c[j] + r[j] <= c[i] + r[i]) {
e.emplace_back(c[j] - r[j], 1, j);
e.emplace_back(c[j] + r[j], -1, j);
}
}
std::sort(e.begin(), e.end());
int v = 0;
int u = -1;
for (auto [x, t, j] : e) {
if (t == 1) {
g[j] = v + dp[j];
lst[j] = u;
} else if (g[j] > v) {
v = g[j];
u = j;
}
}
dp[i] = v + 1;
while (u != -1) {
sub[i].push_back(u);
u = lst[u];
}
}
std::vector<std::tuple<int, int, int>> e;
for (auto j : ord) {
e.emplace_back(c[j] - r[j], 1, j);
e.emplace_back(c[j] + r[j], -1, j);
}
std::sort(e.begin(), e.end());
int v = 0;
int u = -1;
for (auto [x, t, j] : e) {
if (t == 1) {
g[j] = v + dp[j];
lst[j] = u;
} else if (g[j] > v) {
v = g[j];
u = j;
}
}
int ans = v;
std::cout << ans << "\n";
std::queue<int> q;
while (u != -1) {
q.push(u);
u = lst[u];
}
while (!q.empty()) {
int x = q.front();
q.pop();
std::cout << x + 1 << " ";
for (auto y : sub[x]) {
q.push(y);
}
}
std::cout << "\n";
return 0;
}
``` | -1 |
887 | A | Div. 64 | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system. | In the only line given a non-empty binary string *s* with length up to 100. | Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise. | [
"100010001\n",
"100\n"
] | [
"yes",
"no"
] | In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system) | 500 | [
{
"input": "100010001",
"output": "yes"
},
{
"input": "100",
"output": "no"
},
{
"input": "0000001000000",
"output": "yes"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "no"
},
{
"in... | 1,509,869,946 | 2,147,483,647 | Python 3 | OK | TESTS | 98 | 62 | 0 | num = int(input())
cnt = 0
for i in str(num):
if i == '0':
cnt += 1
if num > 0 and cnt >= 6:
print("yes")
else:
print("no") | Title: Div. 64
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.
Input Specification:
In the only line given a non-empty binary string *s* with length up to 100.
Output Specification:
Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.
Demo Input:
['100010001\n', '100\n']
Demo Output:
['yes', 'no']
Note:
In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system) | ```python
num = int(input())
cnt = 0
for i in str(num):
if i == '0':
cnt += 1
if num > 0 and cnt >= 6:
print("yes")
else:
print("no")
``` | 3 | |
957 | A | Tritonic Iridescence | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them. | The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one). | If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower). | [
"5\nCY??Y\n",
"5\nC?C?Y\n",
"5\n?CYC?\n",
"5\nC??MM\n",
"3\nMMY\n"
] | [
"Yes\n",
"Yes\n",
"Yes\n",
"No\n",
"No\n"
] | For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | 500 | [
{
"input": "5\nCY??Y",
"output": "Yes"
},
{
"input": "5\nC?C?Y",
"output": "Yes"
},
{
"input": "5\n?CYC?",
"output": "Yes"
},
{
"input": "5\nC??MM",
"output": "No"
},
{
"input": "3\nMMY",
"output": "No"
},
{
"input": "15\n??YYYYYY??YYYY?",
"output"... | 1,521,908,425 | 2,725 | Python 3 | WRONG_ANSWER | PRETESTS | 6 | 124 | 7,065,600 | n = int(input())
query = list(input())
arr = [0 for i in range(n)]
ans=[]
low = 0
high = len(query)
while low!=high:
flag = 0
if query[low]=="?":
itr = low+1
arr[low] = 1
if low != 0 and low != len(query) - 1:
while query[itr] == "?":
arr[low] += 1
itr += 1
if query[low-1] == query[low-1+arr[low]]:
if arr[low]%2 != 0 and arr[low] > 3:
ans.append("no")
break
low = itr
flag = 1
else:
if low!=len(query)-1:
if query[low]==query[low+1]:
ans.append("no")
break
if flag == 0:
low += 1
if "no" in ans:
print("No")
else:
print("Yes") | Title: Tritonic Iridescence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output Specification:
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Demo Input:
['5\nCY??Y\n', '5\nC?C?Y\n', '5\n?CYC?\n', '5\nC??MM\n', '3\nMMY\n']
Demo Output:
['Yes\n', 'Yes\n', 'Yes\n', 'No\n', 'No\n']
Note:
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | ```python
n = int(input())
query = list(input())
arr = [0 for i in range(n)]
ans=[]
low = 0
high = len(query)
while low!=high:
flag = 0
if query[low]=="?":
itr = low+1
arr[low] = 1
if low != 0 and low != len(query) - 1:
while query[itr] == "?":
arr[low] += 1
itr += 1
if query[low-1] == query[low-1+arr[low]]:
if arr[low]%2 != 0 and arr[low] > 3:
ans.append("no")
break
low = itr
flag = 1
else:
if low!=len(query)-1:
if query[low]==query[low+1]:
ans.append("no")
break
if flag == 0:
low += 1
if "no" in ans:
print("No")
else:
print("Yes")
``` | 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... | 1,621,337,723 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 124 | 0 | n=int(input())
l=[]
for i in range(n):
m=list(map(int,input().split()))
l.append(m)
l1=[]
for i in range(len(l)):
s=0
for j in range(len(l)):
s=s+l[j][i]
l1.append(s*s)
su=sum(l1)
if su == 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())
l=[]
for i in range(n):
m=list(map(int,input().split()))
l.append(m)
l1=[]
for i in range(len(l)):
s=0
for j in range(len(l)):
s=s+l[j][i]
l1.append(s*s)
su=sum(l1)
if su == 0:
print("yes")
else:
print("no")
``` | 0 |
997 | A | Convert to Ones | PROGRAMMING | 1,500 | [
"brute force",
"greedy",
"implementation",
"math"
] | null | null | You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones.
Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$.
You can apply the following operations any number of times:
- Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones? | The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string $a$ of length $n$, consisting of zeros and ones. | Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations. | [
"5 1 10\n01000\n",
"5 10 1\n01000\n",
"7 2 3\n1111111\n"
] | [
"11\n",
"2\n",
"0\n"
] | In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «10000» $\to$ «11111».
The total cost of operations is $1 + 10 = 11$.
In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «11000» $\to$ «11111».
The overall cost is $1 + 1 = 2$.
In the third example, string already consists only of ones, so the answer is $0$. | 500 | [
{
"input": "5 1 10\n01000",
"output": "11"
},
{
"input": "5 10 1\n01000",
"output": "2"
},
{
"input": "7 2 3\n1111111",
"output": "0"
},
{
"input": "1 60754033 959739508\n0",
"output": "959739508"
},
{
"input": "1 431963980 493041212\n1",
"output": "0"
},
... | 1,683,607,693 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 115 | 93 | 6,860,800 | # Problem: A. Convert to Ones
# Contest: Codeforces - Codeforces Round 493 (Div. 1)
# URL: https://codeforces.com/problemset/problem/997/A
# Memory Limit: 256 MB
# Time Limit: 1000 ms
import sys
import bisect
import random
import io, os
from bisect import *
from collections import *
from contextlib import redirect_stdout
from itertools import *
from array import *
from functools import lru_cache, reduce
from types import GeneratorType
from heapq import *
from math import sqrt, gcd, inf
if sys.version >= '3.8': # ACW没有comb
from math import comb
RI = lambda: map(int, sys.stdin.buffer.readline().split())
RS = lambda: map(bytes.decode, sys.stdin.buffer.readline().strip().split())
RILST = lambda: list(RI())
DEBUG = lambda *x: sys.stderr.write(f'{str(x)}\n')
# print = lambda d: sys.stdout.write(str(d) + "\n") # 打开可以快写，但是无法使用print(*ans,sep=' ')这种语法
MOD = 10**9 + 7
PROBLEM = """https://codeforces.com/problemset/problem/997/A
输入 n(1≤n≤3e5) x(0≤x≤1e9) y(0≤y≤1e9) 和长为 n 的 01 字符串 s。
你可以执行任意次操作，每次选择其中一种操作执行。
1. 花费 x，reverse s 的一个子串，例如 1110 -> 0111。
2. 花费 y，flip s 的一个子串，例如 1110 -> 0001。
目标：使 s 中只有 1。
输出最少花费。
输入
5 1 10
01000
输出
11
输入
5 10 1
01000
输出
2
输入
7 2 3
1111111
输出
0
"""
"""https://codeforces.com/contest/997/submission/205172377
如果没有 0，输出 0。
如果 x < y，那么可以把所有 0 通过多次 reverse 操作聚在一起，然后再操作一次 flip。
reverse 的操作次数就是 0 组的个数减一。例如 001101000 中有三个 0 组，需要两次 reverse。
如果 x >= y，那么把每个 0 组 flip 掉即可。"""
# ms
def solve():
n,x,y = RI()
s, = RS()
if '0' not in s :
return print(0)
f = [0]*n
if s[0] == '0':
f[0] = 1
for i in range(1,n):
if s[i] == '0':
f[i] = f[i-1] + 1
zero = f.count(1)
ans = 0
if x < y :
ans = (zero - 1)*x + y
else:
ans = zero * y
print(ans)
if __name__ == '__main__':
t = 0
if t:
t, = RI()
for _ in range(t):
solve()
else:
solve()
| Title: Convert to Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones.
Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$.
You can apply the following operations any number of times:
- Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input Specification:
The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string $a$ of length $n$, consisting of zeros and ones.
Output Specification:
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations.
Demo Input:
['5 1 10\n01000\n', '5 10 1\n01000\n', '7 2 3\n1111111\n']
Demo Output:
['11\n', '2\n', '0\n']
Note:
In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «10000» $\to$ «11111».
The total cost of operations is $1 + 10 = 11$.
In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «11000» $\to$ «11111».
The overall cost is $1 + 1 = 2$.
In the third example, string already consists only of ones, so the answer is $0$. | ```python
# Problem: A. Convert to Ones
# Contest: Codeforces - Codeforces Round 493 (Div. 1)
# URL: https://codeforces.com/problemset/problem/997/A
# Memory Limit: 256 MB
# Time Limit: 1000 ms
import sys
import bisect
import random
import io, os
from bisect import *
from collections import *
from contextlib import redirect_stdout
from itertools import *
from array import *
from functools import lru_cache, reduce
from types import GeneratorType
from heapq import *
from math import sqrt, gcd, inf
if sys.version >= '3.8': # ACW没有comb
from math import comb
RI = lambda: map(int, sys.stdin.buffer.readline().split())
RS = lambda: map(bytes.decode, sys.stdin.buffer.readline().strip().split())
RILST = lambda: list(RI())
DEBUG = lambda *x: sys.stderr.write(f'{str(x)}\n')
# print = lambda d: sys.stdout.write(str(d) + "\n") # 打开可以快写，但是无法使用print(*ans,sep=' ')这种语法
MOD = 10**9 + 7
PROBLEM = """https://codeforces.com/problemset/problem/997/A
输入 n(1≤n≤3e5) x(0≤x≤1e9) y(0≤y≤1e9) 和长为 n 的 01 字符串 s。
你可以执行任意次操作，每次选择其中一种操作执行。
1. 花费 x，reverse s 的一个子串，例如 1110 -> 0111。
2. 花费 y，flip s 的一个子串，例如 1110 -> 0001。
目标：使 s 中只有 1。
输出最少花费。
输入
5 1 10
01000
输出
11
输入
5 10 1
01000
输出
2
输入
7 2 3
1111111
输出
0
"""
"""https://codeforces.com/contest/997/submission/205172377
如果没有 0，输出 0。
如果 x < y，那么可以把所有 0 通过多次 reverse 操作聚在一起，然后再操作一次 flip。
reverse 的操作次数就是 0 组的个数减一。例如 001101000 中有三个 0 组，需要两次 reverse。
如果 x >= y，那么把每个 0 组 flip 掉即可。"""
# ms
def solve():
n,x,y = RI()
s, = RS()
if '0' not in s :
return print(0)
f = [0]*n
if s[0] == '0':
f[0] = 1
for i in range(1,n):
if s[i] == '0':
f[i] = f[i-1] + 1
zero = f.count(1)
ans = 0
if x < y :
ans = (zero - 1)*x + y
else:
ans = zero * y
print(ans)
if __name__ == '__main__':
t = 0
if t:
t, = RI()
for _ in range(t):
solve()
else:
solve()
``` | 3 | |
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 ... | 1,686,963,073 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 92 | 0 | n, a, b = list(map(int, input().split()))
ans = min(n - a, b + 1)
print(ans)
| 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 = list(map(int, input().split()))
ans = min(n - a, b + 1)
print(ans)
``` | 3 | |
631 | B | Print Check | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | null | null | Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.
Printer works with a rectangular sheet of paper of size *n*<=×<=*m*. Consider the list as a table consisting of *n* rows and *m* columns. Rows are numbered from top to bottom with integers from 1 to *n*, while columns are numbered from left to right with integers from 1 to *m*. Initially, all cells are painted in color 0.
Your program has to support two operations:
1. Paint all cells in row *r**i* in color *a**i*; 1. Paint all cells in column *c**i* in color *a**i*.
If during some operation *i* there is a cell that have already been painted, the color of this cell also changes to *a**i*.
Your program has to print the resulting table after *k* operation. | The first line of the input contains three integers *n*, *m* and *k* (1<=<=≤<=<=*n*,<=<=*m*<=<=≤<=5000, *n*·*m*<=≤<=100<=000, 1<=≤<=*k*<=≤<=100<=000) — the dimensions of the sheet and the number of operations, respectively.
Each of the next *k* lines contains the description of exactly one query:
- 1 *r**i* *a**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=109), means that row *r**i* is painted in color *a**i*; - 2 *c**i* *a**i* (1<=≤<=*c**i*<=≤<=*m*, 1<=≤<=*a**i*<=≤<=109), means that column *c**i* is painted in color *a**i*. | Print *n* lines containing *m* integers each — the resulting table after all operations are applied. | [
"3 3 3\n1 1 3\n2 2 1\n1 2 2\n",
"5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1\n"
] | [
"3 1 3 \n2 2 2 \n0 1 0 \n",
"1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 \n"
] | The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray. | 1,000 | [
{
"input": "3 3 3\n1 1 3\n2 2 1\n1 2 2",
"output": "3 1 3 \n2 2 2 \n0 1 0 "
},
{
"input": "5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1",
"output": "1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 "
},
{
"input": "5 5 4\n1 2 1\n1 4 1\n2 2 1\n2 4 1",
"output": "0 1 0 1 0 \n1 1 1 1 1 \n0 1 0 1 0 \n1... | 1,619,954,818 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 11,059,200 | n, m, k = map(int, input().split())
matrix = []
for i in range(n):
matrix.append([0 for x in range(m)])
for q in range(k):
type, r, color = map(int, input().split())
if type == 1:
matrix[r-1] = [color for c in range(m)]
if type == 2:
for i in range(n):
matrix[i][r-1] = color
for i in matrix:
print(' '.join([str(b) for b in i]))
| Title: Print Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.
Printer works with a rectangular sheet of paper of size *n*<=×<=*m*. Consider the list as a table consisting of *n* rows and *m* columns. Rows are numbered from top to bottom with integers from 1 to *n*, while columns are numbered from left to right with integers from 1 to *m*. Initially, all cells are painted in color 0.
Your program has to support two operations:
1. Paint all cells in row *r**i* in color *a**i*; 1. Paint all cells in column *c**i* in color *a**i*.
If during some operation *i* there is a cell that have already been painted, the color of this cell also changes to *a**i*.
Your program has to print the resulting table after *k* operation.
Input Specification:
The first line of the input contains three integers *n*, *m* and *k* (1<=<=≤<=<=*n*,<=<=*m*<=<=≤<=5000, *n*·*m*<=≤<=100<=000, 1<=≤<=*k*<=≤<=100<=000) — the dimensions of the sheet and the number of operations, respectively.
Each of the next *k* lines contains the description of exactly one query:
- 1 *r**i* *a**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=109), means that row *r**i* is painted in color *a**i*; - 2 *c**i* *a**i* (1<=≤<=*c**i*<=≤<=*m*, 1<=≤<=*a**i*<=≤<=109), means that column *c**i* is painted in color *a**i*.
Output Specification:
Print *n* lines containing *m* integers each — the resulting table after all operations are applied.
Demo Input:
['3 3 3\n1 1 3\n2 2 1\n1 2 2\n', '5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1\n']
Demo Output:
['3 1 3 \n2 2 2 \n0 1 0 \n', '1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 \n']
Note:
The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray. | ```python
n, m, k = map(int, input().split())
matrix = []
for i in range(n):
matrix.append([0 for x in range(m)])
for q in range(k):
type, r, color = map(int, input().split())
if type == 1:
matrix[r-1] = [color for c in range(m)]
if type == 2:
for i in range(n):
matrix[i][r-1] = color
for i in matrix:
print(' '.join([str(b) for b in i]))
``` | 0 | |
114 | A | Cifera | PROGRAMMING | 1,000 | [
"math"
] | null | null | When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it. | The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1). | You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*. | [
"5\n25\n",
"3\n8\n"
] | [
"YES\n1\n",
"NO\n"
] | none | 500 | [
{
"input": "5\n25",
"output": "YES\n1"
},
{
"input": "3\n8",
"output": "NO"
},
{
"input": "123\n123",
"output": "YES\n0"
},
{
"input": "99\n970300",
"output": "NO"
},
{
"input": "1000\n6666666",
"output": "NO"
},
{
"input": "59\n3571",
"output": "N... | 1,566,295,310 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 19 | 280 | 0 | from math import log
Base = int(input())
Symbol = int(input())
if log(Symbol,Base)==int(log(Symbol,Base)):
print("YES",int(log(Symbol,Base))-1,sep='\n')
else:
print("NO")
# UB_CodeForces
# Advice: Be on your own
# Location: Based on popular demand "" "" "" ""
# Caption: Last breath
# CodeNumber: 574
| Title: Cifera
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
Input Specification:
The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1).
Output Specification:
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*.
Demo Input:
['5\n25\n', '3\n8\n']
Demo Output:
['YES\n1\n', 'NO\n']
Note:
none | ```python
from math import log
Base = int(input())
Symbol = int(input())
if log(Symbol,Base)==int(log(Symbol,Base)):
print("YES",int(log(Symbol,Base))-1,sep='\n')
else:
print("NO")
# UB_CodeForces
# Advice: Be on your own
# Location: Based on popular demand "" "" "" ""
# Caption: Last breath
# CodeNumber: 574
``` | 0 | |
873 | C | Strange Game On Matrix | PROGRAMMING | 1,600 | [
"greedy",
"two pointers"
] | null | null | Ivan is playing a strange game.
He has a matrix *a* with *n* rows and *m* columns. Each element of the matrix is equal to either 0 or 1. Rows and columns are 1-indexed. Ivan can replace any number of ones in this matrix with zeroes. After that, his score in the game will be calculated as follows:
1. Initially Ivan's score is 0; 1. In each column, Ivan will find the topmost 1 (that is, if the current column is *j*, then he will find minimum *i* such that *a**i*,<=*j*<==<=1). If there are no 1's in the column, this column is skipped; 1. Ivan will look at the next *min*(*k*,<=*n*<=-<=*i*<=+<=1) elements in this column (starting from the element he found) and count the number of 1's among these elements. This number will be added to his score.
Of course, Ivan wants to maximize his score in this strange game. Also he doesn't want to change many elements, so he will replace the minimum possible number of ones with zeroes. Help him to determine the maximum possible score he can get and the minimum possible number of replacements required to achieve that score. | The first line contains three integer numbers *n*, *m* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100).
Then *n* lines follow, *i*-th of them contains *m* integer numbers — the elements of *i*-th row of matrix *a*. Each number is either 0 or 1. | Print two numbers: the maximum possible score Ivan can get and the minimum number of replacements required to get this score. | [
"4 3 2\n0 1 0\n1 0 1\n0 1 0\n1 1 1\n",
"3 2 1\n1 0\n0 1\n0 0\n"
] | [
"4 1\n",
"2 0\n"
] | In the first example Ivan will replace the element *a*<sub class="lower-index">1, 2</sub>. | 0 | [
{
"input": "4 3 2\n0 1 0\n1 0 1\n0 1 0\n1 1 1",
"output": "4 1"
},
{
"input": "3 2 1\n1 0\n0 1\n0 0",
"output": "2 0"
},
{
"input": "3 4 2\n0 1 1 1\n1 0 1 1\n1 0 0 1",
"output": "7 0"
},
{
"input": "3 57 3\n1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1... | 1,600,685,734 | 2,134 | PyPy 3 | OK | TESTS | 20 | 155 | 1,945,600 | import sys
n, m, k = map(int, input().split())
matrix = [[0]*m] + [list(map(int, input().split())) for _ in range(n)]
for j in range(m):
for i in range(1, n+1):
matrix[i][j] += matrix[i-1][j]
ans = [0, 0]
for j in range(m):
max_one, rem = 0, 0
for i in range(k, n+1):
if max_one < matrix[i][j] - matrix[i-k][j]:
max_one = matrix[i][j] - matrix[i-k][j]
rem = matrix[i-k][j]
ans[0] += max_one
ans[1] += rem
print(*ans)
| Title: Strange Game On Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan is playing a strange game.
He has a matrix *a* with *n* rows and *m* columns. Each element of the matrix is equal to either 0 or 1. Rows and columns are 1-indexed. Ivan can replace any number of ones in this matrix with zeroes. After that, his score in the game will be calculated as follows:
1. Initially Ivan's score is 0; 1. In each column, Ivan will find the topmost 1 (that is, if the current column is *j*, then he will find minimum *i* such that *a**i*,<=*j*<==<=1). If there are no 1's in the column, this column is skipped; 1. Ivan will look at the next *min*(*k*,<=*n*<=-<=*i*<=+<=1) elements in this column (starting from the element he found) and count the number of 1's among these elements. This number will be added to his score.
Of course, Ivan wants to maximize his score in this strange game. Also he doesn't want to change many elements, so he will replace the minimum possible number of ones with zeroes. Help him to determine the maximum possible score he can get and the minimum possible number of replacements required to achieve that score.
Input Specification:
The first line contains three integer numbers *n*, *m* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100).
Then *n* lines follow, *i*-th of them contains *m* integer numbers — the elements of *i*-th row of matrix *a*. Each number is either 0 or 1.
Output Specification:
Print two numbers: the maximum possible score Ivan can get and the minimum number of replacements required to get this score.
Demo Input:
['4 3 2\n0 1 0\n1 0 1\n0 1 0\n1 1 1\n', '3 2 1\n1 0\n0 1\n0 0\n']
Demo Output:
['4 1\n', '2 0\n']
Note:
In the first example Ivan will replace the element *a*<sub class="lower-index">1, 2</sub>. | ```python
import sys
n, m, k = map(int, input().split())
matrix = [[0]*m] + [list(map(int, input().split())) for _ in range(n)]
for j in range(m):
for i in range(1, n+1):
matrix[i][j] += matrix[i-1][j]
ans = [0, 0]
for j in range(m):
max_one, rem = 0, 0
for i in range(k, n+1):
if max_one < matrix[i][j] - matrix[i-k][j]:
max_one = matrix[i][j] - matrix[i-k][j]
rem = matrix[i-k][j]
ans[0] += max_one
ans[1] += rem
print(*ans)
``` | 3 | |
143 | A | Help Vasilisa the Wise 2 | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task. | The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement. | Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any. | [
"3 7\n4 6\n5 5\n",
"11 10\n13 8\n5 16\n",
"1 2\n3 4\n5 6\n",
"10 10\n10 10\n10 10\n"
] | [
"1 2\n3 4\n",
"4 7\n9 1\n",
"-1\n",
"-1\n"
] | Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9. | 500 | [
{
"input": "3 7\n4 6\n5 5",
"output": "1 2\n3 4"
},
{
"input": "11 10\n13 8\n5 16",
"output": "4 7\n9 1"
},
{
"input": "1 2\n3 4\n5 6",
"output": "-1"
},
{
"input": "10 10\n10 10\n10 10",
"output": "-1"
},
{
"input": "5 13\n8 10\n11 7",
"output": "3 2\n5 8"
... | 1,675,929,940 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 50 | 186 | 0 | r1, r2 = map(int, input().split())
c1, c2 = map(int, input().split())
d1, d2 = map(int, input().split())
x = (r1 + c1 - d2) // 2
y = (r1 + c2 - d1) // 2
z = (r2 + c1 - d1) // 2
w = (r2 + c2 - d2) // 2
ans = {x, y, z, w}
if len(ans) != 4 or min(ans) < 1 or max(ans) > 9:
print(-1)
else:
print(x, y)
print(z, w) | Title: Help Vasilisa the Wise 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
Input Specification:
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
Output Specification:
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
Demo Input:
['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n']
Demo Output:
['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n']
Note:
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9. | ```python
r1, r2 = map(int, input().split())
c1, c2 = map(int, input().split())
d1, d2 = map(int, input().split())
x = (r1 + c1 - d2) // 2
y = (r1 + c2 - d1) // 2
z = (r2 + c1 - d1) // 2
w = (r2 + c2 - d2) // 2
ans = {x, y, z, w}
if len(ans) != 4 or min(ans) < 1 or max(ans) > 9:
print(-1)
else:
print(x, y)
print(z, w)
``` | 0 | |
584 | A | Olesya and Rodion | PROGRAMMING | 1,000 | [
"math"
] | null | null | Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. | The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. | Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. | [
"3 2\n"
] | [
"712"
] | none | 500 | [
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input"... | 1,678,476,550 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 11 | 46 | 0 | n,m = list(map(int,input().split()))
x = str(m%10) * n
print(-1) if x == '0' else print(x) | Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none | ```python
n,m = list(map(int,input().split()))
x = str(m%10) * n
print(-1) if x == '0' else print(x)
``` | 0 | |
139 | A | Petr and Book | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book. | The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero. | Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. | [
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] | [
"6\n",
"1\n"
] | Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book. | 500 | [
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1... | 1,594,059,251 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 216 | 6,656,000 | n = int(input())
a = [int(x) for x in input().split()]
ans = 0
while(n > 0):
if ans == 7:
ans = 0
n -= a[ans]
ans += 1
print(ans)
| Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book. | ```python
n = int(input())
a = [int(x) for x in input().split()]
ans = 0
while(n > 0):
if ans == 7:
ans = 0
n -= a[ans]
ans += 1
print(ans)
``` | 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"
},
{
... | 1,598,275,630 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 109 | 6,656,000 | from sys import stdin
n, x = list(map(int, stdin.readline().split()))
res = 0
for _ in range(n):
a, d = stdin.readline().split()
d = int(d)
if a == '+':
x += d
else:
if x >= d:
x -= d
else:
res += 1
print(x, res) | 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
from sys import stdin
n, x = list(map(int, stdin.readline().split()))
res = 0
for _ in range(n):
a, d = stdin.readline().split()
d = int(d)
if a == '+':
x += d
else:
if x >= d:
x -= d
else:
res += 1
print(x, res)
``` | 3 | |
405 | A | Gravity Flip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch! | The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column. | Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch. | [
"4\n3 2 1 2\n",
"3\n2 3 8\n"
] | [
"1 2 2 3 \n",
"2 3 8 \n"
] | The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | 500 | [
{
"input": "4\n3 2 1 2",
"output": "1 2 2 3 "
},
{
"input": "3\n2 3 8",
"output": "2 3 8 "
},
{
"input": "5\n2 1 2 1 2",
"output": "1 1 2 2 2 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n4 3",
"output": "3 4 "
},
{
"input": "6\n100 40 60 20... | 1,694,860,901 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | n = int(input())
b = [int(n) for(n) in input().split()]
b.sort()
for i in range(n):
print(b[i], end=" ") | Title: Gravity Flip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch!
Input Specification:
The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column.
Output Specification:
Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch.
Demo Input:
['4\n3 2 1 2\n', '3\n2 3 8\n']
Demo Output:
['1 2 2 3 \n', '2 3 8 \n']
Note:
The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | ```python
n = int(input())
b = [int(n) for(n) in input().split()]
b.sort()
for i in range(n):
print(b[i], end=" ")
``` | 3 | |
514 | A | Chewbaсca and Number | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero. | The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca. | Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes. | [
"27\n",
"4545\n"
] | [
"22\n",
"4444\n"
] | none | 500 | [
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
... | 1,696,077,350 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | a = list(input())
b = []
for i in a:
if int(i) > 9-int(i):
b.append(str(9-int(i)))
else:
b.append(i)
print(b)
ans = ''
for j in range(len(b)):
ans += b[j]
print(ans) | Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none | ```python
a = list(input())
b = []
for i in a:
if int(i) > 9-int(i):
b.append(str(9-int(i)))
else:
b.append(i)
print(b)
ans = ''
for j in range(len(b)):
ans += b[j]
print(ans)
``` | 0 | |
581 | A | Vasya the Hipster | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him? | The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got. | Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day. | [
"3 1\n",
"2 3\n",
"7 3\n"
] | [
"1 1\n",
"2 0\n",
"3 2\n"
] | In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | 500 | [
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input":... | 1,666,912,172 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 46 | 0 | b,r=map(int, input().split())
c=0
while b>0 and r>0:
b-=1
r-=1
c+=1
print(c,max(r,b)//2)
| Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | ```python
b,r=map(int, input().split())
c=0
while b>0 and r>0:
b-=1
r-=1
c+=1
print(c,max(r,b)//2)
``` | 3 | |
940 | B | Our Tanya is Crying Out Loud | PROGRAMMING | 1,400 | [
"dp",
"greedy"
] | null | null | Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109). | Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1. | [
"9\n2\n3\n1\n",
"5\n5\n2\n20\n",
"19\n3\n4\n2\n"
] | [
"6\n",
"8\n",
"12\n"
] | In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | 1,250 | [
{
"input": "9\n2\n3\n1",
"output": "6"
},
{
"input": "5\n5\n2\n20",
"output": "8"
},
{
"input": "19\n3\n4\n2",
"output": "12"
},
{
"input": "1845999546\n999435865\n1234234\n2323423",
"output": "1044857680578777"
},
{
"input": "1604353664\n1604353665\n9993432\n1",
... | 1,521,193,950 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 7,065,600 | # -*-coding:utf-8 -*-
'''
Created on 2018年3月16日
@author: 溟檠
'''
x=int(input())
k=int(input())
A=int(input())
B=int(input())
count=0
# x5 k5 A2 B20
counter=0
while x>1 :
if not (x%k==0) :
count+=(x%k)*A
x-=x%k
print(count,end="")
else :
counta=(x-x//k)*A
countb=B
if counta>countb:
count+=countb
else :
count+=counta
x//=k
print(count,end="") | Title: Our Tanya is Crying Out Loud
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109).
Output Specification:
Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1.
Demo Input:
['9\n2\n3\n1\n', '5\n5\n2\n20\n', '19\n3\n4\n2\n']
Demo Output:
['6\n', '8\n', '12\n']
Note:
In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | ```python
# -*-coding:utf-8 -*-
'''
Created on 2018年3月16日
@author: 溟檠
'''
x=int(input())
k=int(input())
A=int(input())
B=int(input())
count=0
# x5 k5 A2 B20
counter=0
while x>1 :
if not (x%k==0) :
count+=(x%k)*A
x-=x%k
print(count,end="")
else :
counta=(x-x//k)*A
countb=B
if counta>countb:
count+=countb
else :
count+=counta
x//=k
print(count,end="")
``` | 0 | |
584 | A | Olesya and Rodion | PROGRAMMING | 1,000 | [
"math"
] | null | null | Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. | The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. | Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. | [
"3 2\n"
] | [
"712"
] | none | 500 | [
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input"... | 1,689,428,854 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | print("_RANDOM_GUESS_1689428854.3151317")# 1689428854.3151517 | Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none | ```python
print("_RANDOM_GUESS_1689428854.3151317")# 1689428854.3151517
``` | 0 | |
771 | A | Bear and Friendship Condition | PROGRAMMING | 1,500 | [
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes. | The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input. | If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes). | [
"4 3\n1 3\n3 4\n1 4\n",
"4 4\n3 1\n2 3\n3 4\n1 2\n",
"10 4\n4 3\n5 10\n8 9\n1 2\n",
"3 2\n1 2\n2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] | The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | 250 | [
{
"input": "4 3\n1 3\n3 4\n1 4",
"output": "YES"
},
{
"input": "4 4\n3 1\n2 3\n3 4\n1 2",
"output": "NO"
},
{
"input": "10 4\n4 3\n5 10\n8 9\n1 2",
"output": "YES"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 0",
"output": "YES"
},
{
... | 1,577,249,866 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 108 | 409,600 | #Link to the problem : https://codeforces.com/problemset/problem/771/A
# We just need to check if each connected component is a clique
# Run dfs and count the number of vertices and edges
# it's a clique if and only if numberOfEdges = NumberOfVertices(NumberOfvertices - 1)/2
from collections import defaultdict, deque
def is_reasonable(graph):
to_visit = set(graph.keys())
while to_visit:
count_nodes = count_edges = 0
queue = deque([to_visit.pop()])
print(queue)
while queue:
node = queue.pop()
count_nodes += 1
for child in graph[node]:
count_edges += 1
if child in to_visit:
queue.appendleft(child)
to_visit.remove(child)
if count_edges != count_nodes * (count_nodes - 1):
return False
return True
def main():
n, m = map(int, input().split())
graph = defaultdict(list)
for _ in range(m):
u, v = map(int, input().split())
graph[u].append(v)
graph[v].append(u)
print('YES' if is_reasonable(graph) else 'NO')
if __name__ == "__main__":
main() | Title: Bear and Friendship Condition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input Specification:
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
Output Specification:
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Demo Input:
['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | ```python
#Link to the problem : https://codeforces.com/problemset/problem/771/A
# We just need to check if each connected component is a clique
# Run dfs and count the number of vertices and edges
# it's a clique if and only if numberOfEdges = NumberOfVertices(NumberOfvertices - 1)/2
from collections import defaultdict, deque
def is_reasonable(graph):
to_visit = set(graph.keys())
while to_visit:
count_nodes = count_edges = 0
queue = deque([to_visit.pop()])
print(queue)
while queue:
node = queue.pop()
count_nodes += 1
for child in graph[node]:
count_edges += 1
if child in to_visit:
queue.appendleft(child)
to_visit.remove(child)
if count_edges != count_nodes * (count_nodes - 1):
return False
return True
def main():
n, m = map(int, input().split())
graph = defaultdict(list)
for _ in range(m):
u, v = map(int, input().split())
graph[u].append(v)
graph[v].append(u)
print('YES' if is_reasonable(graph) else 'NO')
if __name__ == "__main__":
main()
``` | 0 | |
910 | A | The Way to Home | PROGRAMMING | 800 | [
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null | A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1. | The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one. | If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1. | [
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] | [
"2\n",
"-1\n",
"3\n",
"4\n"
] | In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | 500 | [
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
... | 1,678,471,761 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 46 | 0 | n,d = map(int,input().split())
path = input()
count = 0
x = 0
while x!=(n-1):
y = path[:x+d+1].rfind('1')
if x == y:
count = -1
break
else:
count += 1
x = y
print(count) | Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | ```python
n,d = map(int,input().split())
path = input()
count = 0
x = 0
while x!=(n-1):
y = path[:x+d+1].rfind('1')
if x == y:
count = -1
break
else:
count += 1
x = y
print(count)
``` | 3 | |
276 | B | Little Girl and Game | PROGRAMMING | 1,300 | [
"games",
"greedy"
] | null | null | The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string *s*, consisting of lowercase English letters. They play a game that is described by the following rules:
- The players move in turns; In one move the player can remove an arbitrary letter from string *s*. - If the player before his turn can reorder the letters in string *s* so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second. | The input contains a single line, containing string *s* (1<=≤<=|*s*|<=<=≤<=<=103). String *s* consists of lowercase English letters. | In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes. | [
"aba\n",
"abca\n"
] | [
"First\n",
"Second\n"
] | none | 1,000 | [
{
"input": "aba",
"output": "First"
},
{
"input": "abca",
"output": "Second"
},
{
"input": "aabb",
"output": "First"
},
{
"input": "ctjxzuimsxnarlciuynqeoqmmbqtagszuo",
"output": "Second"
},
{
"input": "gevqgtaorjixsxnbcoybr",
"output": "First"
},
{
"i... | 1,638,426,679 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 10,649,600 | def minDeletions(X, i, j):
if i >= j:
return 0
if X[i] == X[j]:
return minDeletions(X, i+1, j-1)
return 1 + min(minDeletions(X, i, j - 1), minDeletions(X, i + 1, j))
s = input()
if minDeletions(s, 0, len(s)-1)&1:
print("Second")
else:
print("First") | Title: Little Girl and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string *s*, consisting of lowercase English letters. They play a game that is described by the following rules:
- The players move in turns; In one move the player can remove an arbitrary letter from string *s*. - If the player before his turn can reorder the letters in string *s* so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input Specification:
The input contains a single line, containing string *s* (1<=≤<=|*s*|<=<=≤<=<=103). String *s* consists of lowercase English letters.
Output Specification:
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Demo Input:
['aba\n', 'abca\n']
Demo Output:
['First\n', 'Second\n']
Note:
none | ```python
def minDeletions(X, i, j):
if i >= j:
return 0
if X[i] == X[j]:
return minDeletions(X, i+1, j-1)
return 1 + min(minDeletions(X, i, j - 1), minDeletions(X, i + 1, j))
s = input()
if minDeletions(s, 0, len(s)-1)&1:
print("Second")
else:
print("First")
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking! | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters. | Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences. | [
"7\naogogob\n",
"13\nogogmgogogogo\n",
"9\nogoogoogo\n"
] | [
"a***b\n",
"***gmg***\n",
"*********\n"
] | The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | 0 | [
{
"input": "7\naogogob",
"output": "a***b"
},
{
"input": "13\nogogmgogogogo",
"output": "***gmg***"
},
{
"input": "9\nogoogoogo",
"output": "*********"
},
{
"input": "32\nabcdefogoghijklmnogoopqrstuvwxyz",
"output": "abcdef***ghijklmn***opqrstuvwxyz"
},
{
"input":... | 1,479,633,378 | 678 | Python 3 | WRONG_ANSWER | TESTS | 7 | 61 | 0 | n=int(input())
A=input()
i=0
while i<len(A):
if A[i]=='o' and i<len(A)-2:
if A[i+1]=='g' and A[i+2]=='o':
print("***",end='')
i+=2
else:
print(A[i],end='')
elif A[i]=='g' and i<len(A)-1:
if A[i-1]=='o' and A[i+1]=='o':
print("",end='')
i+=1
else:
print(A[i],end='')
else:
print(A[i],end='')
i+=1
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
Output Specification:
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Demo Input:
['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n']
Demo Output:
['a***b\n', '***gmg***\n', '*********\n']
Note:
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | ```python
n=int(input())
A=input()
i=0
while i<len(A):
if A[i]=='o' and i<len(A)-2:
if A[i+1]=='g' and A[i+2]=='o':
print("***",end='')
i+=2
else:
print(A[i],end='')
elif A[i]=='g' and i<len(A)-1:
if A[i-1]=='o' and A[i+1]=='o':
print("",end='')
i+=1
else:
print(A[i],end='')
else:
print(A[i],end='')
i+=1
``` | 0 | |
764 | A | Taymyr is calling you | PROGRAMMING | 800 | [
"brute force",
"implementation",
"math"
] | null | null | Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. | The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). | Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. | [
"1 1 10\n",
"1 2 5\n",
"2 3 9\n"
] | [
"10\n",
"2\n",
"1\n"
] | Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | 500 | [
{
"input": "1 1 10",
"output": "10"
},
{
"input": "1 2 5",
"output": "2"
},
{
"input": "2 3 9",
"output": "1"
},
{
"input": "4 8 9",
"output": "1"
},
{
"input": "7 9 2",
"output": "0"
},
{
"input": "10000 10000 10000",
"output": "1"
},
{
"i... | 1,639,054,200 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | import math
n,m,z = map(int,input().split())
j = n*m/math.gcd(n,m)
if j <= z:
print(int(j)) | Title: Taymyr is calling you
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input Specification:
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
Output Specification:
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Demo Input:
['1 1 10\n', '1 2 5\n', '2 3 9\n']
Demo Output:
['10\n', '2\n', '1\n']
Note:
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | ```python
import math
n,m,z = map(int,input().split())
j = n*m/math.gcd(n,m)
if j <= z:
print(int(j))
``` | 0 | |
570 | B | Simple Game | PROGRAMMING | 1,300 | [
"constructive algorithms",
"games",
"greedy",
"implementation",
"math"
] | null | null | One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive). | The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively. | Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them. | [
"3 1\n",
"4 3\n"
] | [
"2",
"2"
] | In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less. | 1,000 | [
{
"input": "3 1",
"output": "2"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "10 5",
"output": "6"
},
{
"input": "20 13",
"output": "12"
},
{
"input": "51 1",
"output": "2"
},
{
"input": "100 50",
... | 1,569,044,790 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 16 | 155 | 0 | n, m = list(map(int, input().split()))
if m <= n // 2:
print(m + 1)
else:
print(m - 1)
| Title: Simple Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive).
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively.
Output Specification:
Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.
Demo Input:
['3 1\n', '4 3\n']
Demo Output:
['2', '2']
Note:
In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less. | ```python
n, m = list(map(int, input().split()))
if m <= n // 2:
print(m + 1)
else:
print(m - 1)
``` | 0 | |
165 | A | Supercentral Point | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | [
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] | [
"2\n",
"1\n"
] | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | 500 | [
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n... | 1,676,864,094 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 92 | 0 | n=int(input())
p=[]
for i in range(n):
punto=list(map(int,input().split()))
p.append(punto)
px=sorted(p, key=lambda i: i[0])
py=sorted(px, key=lambda i: i[1])
px=sorted(py, key=lambda i: i[0])
c=[]
for i in range(1,n-1):
if px[i][0]==px[i-1][0] and px[i][0]==px[i+1][0]:
c.append((px[i][0]+1000)*10000+(px[i][1]+1000))
if py[i][1]==py[i-1][1] and py[i][1]==py[i+1][1]:
c.append((py[i][0]+1000)*10000+(py[i][1]+1000))
print(len(c)-len(set(c))) | Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | ```python
n=int(input())
p=[]
for i in range(n):
punto=list(map(int,input().split()))
p.append(punto)
px=sorted(p, key=lambda i: i[0])
py=sorted(px, key=lambda i: i[1])
px=sorted(py, key=lambda i: i[0])
c=[]
for i in range(1,n-1):
if px[i][0]==px[i-1][0] and px[i][0]==px[i+1][0]:
c.append((px[i][0]+1000)*10000+(px[i][1]+1000))
if py[i][1]==py[i-1][1] and py[i][1]==py[i+1][1]:
c.append((py[i][0]+1000)*10000+(py[i][1]+1000))
print(len(c)-len(set(c)))
``` | 3 | |
363 | B | Fence | PROGRAMMING | 1,100 | [
"brute force",
"dp"
] | null | null | There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic). | The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence. | Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them. | [
"7 3\n1 2 6 1 1 7 1\n"
] | [
"3\n"
] | In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8. | 1,000 | [
{
"input": "7 3\n1 2 6 1 1 7 1",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 1\n10 20",
"output": "1"
},
{
"input": "10 5\n1 2 3 1 2 2 3 1 4 5",
"output": "1"
},
{
"input": "10 2\n3 1 4 1 4 6 2 1 4 6",
"output": "7"
},
{
"inp... | 1,670,771,504 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 124 | 10,547,200 | n, k = map(int, input().split())
mas = list(map(int, input().split()))
s = 0
for j in range(k):
s += mas[j]
min_s = s
min_idx = 0
for i in range(1, n-k):
s = s - mas[i-1]+mas[i+k-1]
if s < min_s:
min_s = s
min_idx = i
print(min_idx+1)
| Title: Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
Output Specification:
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
Demo Input:
['7 3\n1 2 6 1 1 7 1\n']
Demo Output:
['3\n']
Note:
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8. | ```python
n, k = map(int, input().split())
mas = list(map(int, input().split()))
s = 0
for j in range(k):
s += mas[j]
min_s = s
min_idx = 0
for i in range(1, n-k):
s = s - mas[i-1]+mas[i+k-1]
if s < min_s:
min_s = s
min_idx = i
print(min_idx+1)
``` | 0 | |
608 | A | Saitama Destroys Hotel | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to *s* and elevator initially starts on floor *s* at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0. | The first line of input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=100, 1<=≤<=*s*<=≤<=1000) — the number of passengers and the number of the top floor respectively.
The next *n* lines each contain two space-separated integers *f**i* and *t**i* (1<=≤<=*f**i*<=≤<=*s*, 1<=≤<=*t**i*<=≤<=1000) — the floor and the time of arrival in seconds for the passenger number *i*. | Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0. | [
"3 7\n2 1\n3 8\n5 2\n",
"5 10\n2 77\n3 33\n8 21\n9 12\n10 64\n"
] | [
"11\n",
"79\n"
] | In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | 500 | [
{
"input": "3 7\n2 1\n3 8\n5 2",
"output": "11"
},
{
"input": "5 10\n2 77\n3 33\n8 21\n9 12\n10 64",
"output": "79"
},
{
"input": "1 1000\n1000 1000",
"output": "2000"
},
{
"input": "1 1\n1 1",
"output": "2"
},
{
"input": "1 1000\n1 1",
"output": "1000"
},
... | 1,597,945,175 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 109 | 0 | n,s = map(int,input().split())
max = 0
idx = 0
for i in range(n):
f,t = map(int,input().split())
if t > max:
max = t
idx = f
print(max+idx) | Title: Saitama Destroys Hotel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to *s* and elevator initially starts on floor *s* at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input Specification:
The first line of input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=100, 1<=≤<=*s*<=≤<=1000) — the number of passengers and the number of the top floor respectively.
The next *n* lines each contain two space-separated integers *f**i* and *t**i* (1<=≤<=*f**i*<=≤<=*s*, 1<=≤<=*t**i*<=≤<=1000) — the floor and the time of arrival in seconds for the passenger number *i*.
Output Specification:
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Demo Input:
['3 7\n2 1\n3 8\n5 2\n', '5 10\n2 77\n3 33\n8 21\n9 12\n10 64\n']
Demo Output:
['11\n', '79\n']
Note:
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | ```python
n,s = map(int,input().split())
max = 0
idx = 0
for i in range(n):
f,t = map(int,input().split())
if t > max:
max = t
idx = f
print(max+idx)
``` | 0 | |
967 | B | Watering System | PROGRAMMING | 1,000 | [
"math",
"sortings"
] | null | null | Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole? | The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes. | Print a single integer — the number of holes Arkady should block. | [
"4 10 3\n2 2 2 2\n",
"4 80 20\n3 2 1 4\n",
"5 10 10\n1000 1 1 1 1\n"
] | [
"1\n",
"0\n",
"4\n"
] | In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole. | 1,000 | [
{
"input": "4 10 3\n2 2 2 2",
"output": "1"
},
{
"input": "4 80 20\n3 2 1 4",
"output": "0"
},
{
"input": "5 10 10\n1000 1 1 1 1",
"output": "4"
},
{
"input": "10 300 100\n20 1 3 10 8 5 3 6 4 3",
"output": "1"
},
{
"input": "10 300 100\n20 25 68 40 60 37 44 85 23 ... | 1,616,138,635 | 2,147,483,647 | PyPy 3 | OK | TESTS | 26 | 140 | 9,113,600 | n, a, b = map(int, input().split())
s = list(map(int, input().split()))
c = 0
y = s[0]
k = sum(s)
s.remove(s[0])
i = 0
s.sort(reverse=True)
while ((a * y) / k) < b:
k -= s[i]
i += 1
c += 1
print(c) | Title: Watering System
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?
Input Specification:
The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes.
Output Specification:
Print a single integer — the number of holes Arkady should block.
Demo Input:
['4 10 3\n2 2 2 2\n', '4 80 20\n3 2 1 4\n', '5 10 10\n1000 1 1 1 1\n']
Demo Output:
['1\n', '0\n', '4\n']
Note:
In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole. | ```python
n, a, b = map(int, input().split())
s = list(map(int, input().split()))
c = 0
y = s[0]
k = sum(s)
s.remove(s[0])
i = 0
s.sort(reverse=True)
while ((a * y) / k) < b:
k -= s[i]
i += 1
c += 1
print(c)
``` | 3 | |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,630,635,252 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 296 | 22,016,000 | def process_input():
nums = input().split(' ')
i = [int(i) for i in nums]
return sorted(i)
def errors(s1, s2):
for i, num1 in enumerate(s1[0:-1]):
if s2[i] != num1:
return num1
return s1[-1]
_ = input()
s1 = process_input()
s2 = process_input()
s3 = process_input()
print(errors(s1, s2))
print(errors(s2, s3)) | Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
def process_input():
nums = input().split(' ')
i = [int(i) for i in nums]
return sorted(i)
def errors(s1, s2):
for i, num1 in enumerate(s1[0:-1]):
if s2[i] != num1:
return num1
return s1[-1]
_ = input()
s1 = process_input()
s2 = process_input()
s3 = process_input()
print(errors(s1, s2))
print(errors(s2, s3))
``` | 3 | |
445 | A | DZY Loves Chessboard | PROGRAMMING | 1,200 | [
"dfs and similar",
"implementation"
] | null | null | DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad. | Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists. | [
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] | [
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] | In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | 500 | [
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n--------... | 1,644,718,440 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 30 | 0 |
def resolve(actual_pos , table , symbol):
y= actual_pos
if (y == len(table) ):
# print(actual_pos , "retornou !!!")
return
for i in range(len(table[y])):
if (table[y][i] == "."):
table[y][i] = symbol
if(symbol == "B"):
symbol = "W"
else:
symbol = "B"
resolve(y + 1 , table , symbol)
n , m= input().split()
table =[]
for i in range(int(n)):
line = input()
table.append([e for e in line])
resolve(0 , table , "B")
saida = ""
for l in table:
saida += "".join(l) + "\n"
print(saida)
| Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | ```python
def resolve(actual_pos , table , symbol):
y= actual_pos
if (y == len(table) ):
# print(actual_pos , "retornou !!!")
return
for i in range(len(table[y])):
if (table[y][i] == "."):
table[y][i] = symbol
if(symbol == "B"):
symbol = "W"
else:
symbol = "B"
resolve(y + 1 , table , symbol)
n , m= input().split()
table =[]
for i in range(int(n)):
line = input()
table.append([e for e in line])
resolve(0 , table , "B")
saida = ""
for l in table:
saida += "".join(l) + "\n"
print(saida)
``` | 0 | |
6 | A | Triangle | PROGRAMMING | 900 | [
"brute force",
"geometry"
] | A. Triangle | 2 | 64 | Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him. | The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks. | Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length. | [
"4 2 1 3\n",
"7 2 2 4\n",
"3 5 9 1\n"
] | [
"TRIANGLE\n",
"SEGMENT\n",
"IMPOSSIBLE\n"
] | none | 0 | [
{
"input": "4 2 1 3",
"output": "TRIANGLE"
},
{
"input": "7 2 2 4",
"output": "SEGMENT"
},
{
"input": "3 5 9 1",
"output": "IMPOSSIBLE"
},
{
"input": "3 1 5 1",
"output": "IMPOSSIBLE"
},
{
"input": "10 10 10 10",
"output": "TRIANGLE"
},
{
"input": "11 ... | 1,560,890,760 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 186 | 0 | x = input()
x = int(x)
y = input()
y = int(y)
z = input()
z = int(z)
t = input()
t = int(t)
def check(a,b,c):
if(a+b<c or b+c<a or a+c<b):
return 0
elif(a+b==c or b+c==a or a+c==b):
return 1
else:
return 2
if(check(x,y,z)==2 or check(x,y,t)==2 or check(y,z,t)==2):
print('TRIANGLE')
elif(check(x,y,z)==1 or check(x,y,t)==1 or check(y,z,t)==1):
print('SEGMENT')
else:
print('IMPOSSIBLE') | Title: Triangle
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input Specification:
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output Specification:
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Demo Input:
['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n']
Demo Output:
['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n']
Note:
none | ```python
x = input()
x = int(x)
y = input()
y = int(y)
z = input()
z = int(z)
t = input()
t = int(t)
def check(a,b,c):
if(a+b<c or b+c<a or a+c<b):
return 0
elif(a+b==c or b+c==a or a+c==b):
return 1
else:
return 2
if(check(x,y,z)==2 or check(x,y,t)==2 or check(y,z,t)==2):
print('TRIANGLE')
elif(check(x,y,z)==1 or check(x,y,t)==1 or check(y,z,t)==1):
print('SEGMENT')
else:
print('IMPOSSIBLE')
``` | -1 |
931 | B | World Cup | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | null | null | The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet. | The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal. | In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1. | [
"4 1 2\n",
"8 2 6\n",
"8 7 5\n"
] | [
"1\n",
"Final!\n",
"2\n"
] | In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | 1,000 | [
{
"input": "4 1 2",
"output": "1"
},
{
"input": "8 2 6",
"output": "Final!"
},
{
"input": "8 7 5",
"output": "2"
},
{
"input": "128 30 98",
"output": "Final!"
},
{
"input": "256 128 256",
"output": "Final!"
},
{
"input": "256 2 127",
"output": "7"
... | 1,619,313,081 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 26 | 93 | 0 | import math
n, a, b = map(int, input().split())
teams = list(range(1, n+1))
if a > len(teams) // 2:
teams = teams[len(teams) // 2:]
else:
teams = teams[:len(teams) // 2]
#print(teams)
cnt = 0
while a in teams and b in teams:
pl = teams.index(a)
if a > len(teams) // 2:
teams = teams[len(teams)//2:]
else:
teams = teams[:len(teams) // 2]
#print(teams)
cnt += 1
if cnt == 0:
print("Final!")
else:
print(int(math.log(n, 2)) - cnt)
| Title: World Cup
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
Input Specification:
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
Output Specification:
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
Demo Input:
['4 1 2\n', '8 2 6\n', '8 7 5\n']
Demo Output:
['1\n', 'Final!\n', '2\n']
Note:
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | ```python
import math
n, a, b = map(int, input().split())
teams = list(range(1, n+1))
if a > len(teams) // 2:
teams = teams[len(teams) // 2:]
else:
teams = teams[:len(teams) // 2]
#print(teams)
cnt = 0
while a in teams and b in teams:
pl = teams.index(a)
if a > len(teams) // 2:
teams = teams[len(teams)//2:]
else:
teams = teams[:len(teams) // 2]
#print(teams)
cnt += 1
if cnt == 0:
print("Final!")
else:
print(int(math.log(n, 2)) - cnt)
``` | 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"
},
{
"... | 1,693,576,755 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 46 | 0 | num = int(input())
count = 0
for i in range(num):
x,y = input().split()
if int(y) - int(x) >= 2:
count += 1
print(count) | Title: George and Accommodation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Output Specification:
Print a single integer — the number of rooms where George and Alex can move in.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n']
Demo Output:
['0\n', '2\n']
Note:
none | ```python
num = int(input())
count = 0
for i in range(num):
x,y = input().split()
if int(y) - int(x) >= 2:
count += 1
print(count)
``` | 3 | |
932 | C | Permutation Cycle | PROGRAMMING | 1,600 | [
"brute force",
"constructive algorithms"
] | null | null | For a permutation *P*[1... *N*] of integers from 1 to *N*, function *f* is defined as follows:
Let *g*(*i*) be the minimum positive integer *j* such that *f*(*i*,<=*j*)<==<=*i*. We can show such *j* always exists.
For given *N*,<=*A*,<=*B*, find a permutation *P* of integers from 1 to *N* such that for 1<=≤<=*i*<=≤<=*N*, *g*(*i*) equals either *A* or *B*. | The only line contains three integers *N*,<=*A*,<=*B* (1<=≤<=*N*<=≤<=106,<=1<=≤<=*A*,<=*B*<=≤<=*N*). | If no such permutation exists, output -1. Otherwise, output a permutation of integers from 1 to *N*. | [
"9 2 5\n",
"3 2 1\n"
] | [
"6 5 8 3 4 1 9 2 7",
"1 2 3 "
] | In the first example, *g*(1) = *g*(6) = *g*(7) = *g*(9) = 2 and *g*(2) = *g*(3) = *g*(4) = *g*(5) = *g*(8) = 5
In the second example, *g*(1) = *g*(2) = *g*(3) = 1 | 1,500 | [
{
"input": "9 2 5",
"output": "2 1 4 3 6 7 8 9 5 "
},
{
"input": "3 2 1",
"output": "1 2 3 "
},
{
"input": "7 4 4",
"output": "-1"
},
{
"input": "1000000 999998 3",
"output": "-1"
},
{
"input": "1 1 1",
"output": "1 "
},
{
"input": "993012 997 1001",
... | 1,518,720,077 | 2,147,483,647 | Python 3 | OK | TESTS | 69 | 1,138 | 74,854,400 | #This code sucks, you know it and I know it.
#Move on and call me an idiot later.
def solve(a, b, n):
i = 0
while i * a <= n:
if (n - (i * a)) % b == 0:
x = i
y = (n - (i * a)) // b
return (x, y)
i = i + 1
return (-1, -1)
n, a, b = map(int, input().split())
aa, bb = solve(a, b, n)
l = []
if (aa, bb) == (-1, -1):
print(-1)
else:
for i in range(1,aa+1):
x = a*(i-1) + 1
y = a*i
l += [y]
l += [j for j in range(x, y)]
for i in range(1,bb+1):
x = a*aa + b*(i-1) + 1
y = a*aa + b*i
l += [y]
l += [j for j in range(x, y)]
print(" ".join(map(str, l))) | Title: Permutation Cycle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a permutation *P*[1... *N*] of integers from 1 to *N*, function *f* is defined as follows:
Let *g*(*i*) be the minimum positive integer *j* such that *f*(*i*,<=*j*)<==<=*i*. We can show such *j* always exists.
For given *N*,<=*A*,<=*B*, find a permutation *P* of integers from 1 to *N* such that for 1<=≤<=*i*<=≤<=*N*, *g*(*i*) equals either *A* or *B*.
Input Specification:
The only line contains three integers *N*,<=*A*,<=*B* (1<=≤<=*N*<=≤<=106,<=1<=≤<=*A*,<=*B*<=≤<=*N*).
Output Specification:
If no such permutation exists, output -1. Otherwise, output a permutation of integers from 1 to *N*.
Demo Input:
['9 2 5\n', '3 2 1\n']
Demo Output:
['6 5 8 3 4 1 9 2 7', '1 2 3 ']
Note:
In the first example, *g*(1) = *g*(6) = *g*(7) = *g*(9) = 2 and *g*(2) = *g*(3) = *g*(4) = *g*(5) = *g*(8) = 5
In the second example, *g*(1) = *g*(2) = *g*(3) = 1 | ```python
#This code sucks, you know it and I know it.
#Move on and call me an idiot later.
def solve(a, b, n):
i = 0
while i * a <= n:
if (n - (i * a)) % b == 0:
x = i
y = (n - (i * a)) // b
return (x, y)
i = i + 1
return (-1, -1)
n, a, b = map(int, input().split())
aa, bb = solve(a, b, n)
l = []
if (aa, bb) == (-1, -1):
print(-1)
else:
for i in range(1,aa+1):
x = a*(i-1) + 1
y = a*i
l += [y]
l += [j for j in range(x, y)]
for i in range(1,bb+1):
x = a*aa + b*(i-1) + 1
y = a*aa + b*i
l += [y]
l += [j for j in range(x, y)]
print(" ".join(map(str, l)))
``` | 3 | |
20 | C | Dijkstra? | PROGRAMMING | 1,900 | [
"graphs",
"shortest paths"
] | C. Dijkstra? | 1 | 64 | You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | [
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] | [
"1 4 3 5 ",
"1 4 3 5 "
] | none | 1,500 | [
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input... | 1,624,463,855 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 31 | 1,000 | 26,726,400 | from heapq import heappush , heappop
n,m = map(int, input().split(' '))
g = [[] for i in range(n)]
for i in range(m):
a,b,w = map(int, input().split(' '))
a -= 1
b -= 1
g[a].append((b,w))
g[b].append((a,w))
h = [(0, 0)]
d = [2**40 for i in range(n)]
parents = [None for i in range(n)]
while h:
di,v = heappop(h)
for u,w in g[v]:
if d[u] > di + w:
parents[u] = v
d[u] = di+w
heappush(h, (d[u],u))
if parents[n-1] == None:
print(-1)
exit(0)
res, v = [n],n-1
while v != 0:
v = parents[v]
res.append(v+1)
print(*res[::-1])
| Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none | ```python
from heapq import heappush , heappop
n,m = map(int, input().split(' '))
g = [[] for i in range(n)]
for i in range(m):
a,b,w = map(int, input().split(' '))
a -= 1
b -= 1
g[a].append((b,w))
g[b].append((a,w))
h = [(0, 0)]
d = [2**40 for i in range(n)]
parents = [None for i in range(n)]
while h:
di,v = heappop(h)
for u,w in g[v]:
if d[u] > di + w:
parents[u] = v
d[u] = di+w
heappush(h, (d[u],u))
if parents[n-1] == None:
print(-1)
exit(0)
res, v = [n],n-1
while v != 0:
v = parents[v]
res.append(v+1)
print(*res[::-1])
``` | 0 |
45 | C | Dancing Lessons | PROGRAMMING | 1,900 | [
"data structures"
] | C. Dancing Lessons | 2 | 256 | There are *n* people taking dancing lessons. Every person is characterized by his/her dancing skill *a**i*. At the beginning of the lesson they line up from left to right. While there is at least one couple of a boy and a girl in the line, the following process is repeated: the boy and girl who stand next to each other, having the minimal difference in dancing skills start to dance. If there are several such couples, the one first from the left starts to dance. After a couple leaves to dance, the line closes again, i.e. as a result the line is always continuous. The difference in dancing skills is understood as the absolute value of difference of *a**i* variable. Your task is to find out what pairs and in what order will start dancing. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of people. The next line contains *n* symbols B or G without spaces. B stands for a boy, G stands for a girl. The third line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=107) — the dancing skill. People are specified from left to right in the order in which they lined up. | Print the resulting number of couples *k*. Then print *k* lines containing two numerals each — the numbers of people forming the couple. The people are numbered with integers from 1 to *n* from left to right. When a couple leaves to dance you shouldn't renumber the people. The numbers in one couple should be sorted in the increasing order. Print the couples in the order in which they leave to dance. | [
"4\nBGBG\n4 2 4 3\n",
"4\nBBGG\n4 6 1 5\n",
"4\nBGBB\n1 1 2 3\n"
] | [
"2\n3 4\n1 2\n",
"2\n2 3\n1 4\n",
"1\n1 2\n"
] | none | 0 | [
{
"input": "4\nBGBG\n4 2 4 3",
"output": "2\n3 4\n1 2"
},
{
"input": "4\nBBGG\n4 6 1 5",
"output": "2\n2 3\n1 4"
},
{
"input": "4\nBGBB\n1 1 2 3",
"output": "1\n1 2"
},
{
"input": "1\nB\n490297",
"output": "0"
},
{
"input": "2\nBB\n2518190 6313112",
"output": ... | 1,462,602,023 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 1,716 | 42,803,200 | # -*- coding: utf-8 -*-
"""
Created on Sat May 7 11:48:38 2016
@author: Alex
"""
import heapq, sys
q = []
cnt = 0
n = int(input())
sex = input()
a = list(map(int, input().split()))
l = [i-1 for i in range(n)] #生成左指针列表
r = [i+1 for i in range(n)] #生成右指针列表
for i in range(n-1):
if sex[i] != sex[i+1]:
heapq.heappush(q, (abs(a[i] - a[i+1]), i, i+1))
cnt += 1
ans = []
while cnt > 0:
t, i, j = heapq.heappop(q)
cnt -= 1
if r[i] == -1 or r[j] == -1:
continue
ans.append('%d %d' % (i+1, j+1))
u, v = l[i], r[j]
r[i] = r[j] = -1
if u >= 0:
r[u] = v
if v < n:
l[v] = u
if u >= 0 and v < n and sex[u] != sex[v]:
heapq.heappush(q, (abs(a[u] - a[v]), u, v))
cnt += 1
sys.stdout.write(str(len(ans)) + '\n' + '\n'.join(ans) + '\n')\ | Title: Dancing Lessons
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* people taking dancing lessons. Every person is characterized by his/her dancing skill *a**i*. At the beginning of the lesson they line up from left to right. While there is at least one couple of a boy and a girl in the line, the following process is repeated: the boy and girl who stand next to each other, having the minimal difference in dancing skills start to dance. If there are several such couples, the one first from the left starts to dance. After a couple leaves to dance, the line closes again, i.e. as a result the line is always continuous. The difference in dancing skills is understood as the absolute value of difference of *a**i* variable. Your task is to find out what pairs and in what order will start dancing.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of people. The next line contains *n* symbols B or G without spaces. B stands for a boy, G stands for a girl. The third line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=107) — the dancing skill. People are specified from left to right in the order in which they lined up.
Output Specification:
Print the resulting number of couples *k*. Then print *k* lines containing two numerals each — the numbers of people forming the couple. The people are numbered with integers from 1 to *n* from left to right. When a couple leaves to dance you shouldn't renumber the people. The numbers in one couple should be sorted in the increasing order. Print the couples in the order in which they leave to dance.
Demo Input:
['4\nBGBG\n4 2 4 3\n', '4\nBBGG\n4 6 1 5\n', '4\nBGBB\n1 1 2 3\n']
Demo Output:
['2\n3 4\n1 2\n', '2\n2 3\n1 4\n', '1\n1 2\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Sat May 7 11:48:38 2016
@author: Alex
"""
import heapq, sys
q = []
cnt = 0
n = int(input())
sex = input()
a = list(map(int, input().split()))
l = [i-1 for i in range(n)] #生成左指针列表
r = [i+1 for i in range(n)] #生成右指针列表
for i in range(n-1):
if sex[i] != sex[i+1]:
heapq.heappush(q, (abs(a[i] - a[i+1]), i, i+1))
cnt += 1
ans = []
while cnt > 0:
t, i, j = heapq.heappop(q)
cnt -= 1
if r[i] == -1 or r[j] == -1:
continue
ans.append('%d %d' % (i+1, j+1))
u, v = l[i], r[j]
r[i] = r[j] = -1
if u >= 0:
r[u] = v
if v < n:
l[v] = u
if u >= 0 and v < n and sex[u] != sex[v]:
heapq.heappush(q, (abs(a[u] - a[v]), u, v))
cnt += 1
sys.stdout.write(str(len(ans)) + '\n' + '\n'.join(ans) + '\n')\
``` | 3.491273 |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output"... | 1,574,841,602 | 2,147,483,647 | PyPy 3 | OK | TESTS | 32 | 280 | 0 | n = int(input())
num = [int(x) for x in input().split()]
ans1 = 0
p1 = 0
ans2 = 0
p2 = 0
for i in range(n):
if (num[i] % 2 == 0):
ans1 += 1
p1 = i
else:
ans2 += 1
p2 = i
if (ans1 == 1):
print(p1 + 1)
if (ans2 == 1):
print(p2 + 1)
| Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n = int(input())
num = [int(x) for x in input().split()]
ans1 = 0
p1 = 0
ans2 = 0
p2 = 0
for i in range(n):
if (num[i] % 2 == 0):
ans1 += 1
p1 = i
else:
ans2 += 1
p2 = i
if (ans1 == 1):
print(p1 + 1)
if (ans2 == 1):
print(p2 + 1)
``` | 3.93 |
6 | A | Triangle | PROGRAMMING | 900 | [
"brute force",
"geometry"
] | A. Triangle | 2 | 64 | Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him. | The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks. | Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length. | [
"4 2 1 3\n",
"7 2 2 4\n",
"3 5 9 1\n"
] | [
"TRIANGLE\n",
"SEGMENT\n",
"IMPOSSIBLE\n"
] | none | 0 | [
{
"input": "4 2 1 3",
"output": "TRIANGLE"
},
{
"input": "7 2 2 4",
"output": "SEGMENT"
},
{
"input": "3 5 9 1",
"output": "IMPOSSIBLE"
},
{
"input": "3 1 5 1",
"output": "IMPOSSIBLE"
},
{
"input": "10 10 10 10",
"output": "TRIANGLE"
},
{
"input": "11 ... | 1,474,843,059 | 2,147,483,647 | Python 3 | OK | TESTS | 83 | 124 | 0 | s = input()
#s = '4 2 1 3'
#s = '7 2 2 4'
#s = '10 10 10 10'
s = s.split(' ')
l = [int(s[0]),int(s[1]),int(s[2]),int(s[3])]
l.sort()
tmp = l[0]
if l[0] + l[1] > l[2]:
print("TRIANGLE")
else:
l.pop(0)
if l[0] + l[1] > l[2]:
print("TRIANGLE")
else:
l.append(tmp)
l.sort()
if l[0] + l[1] == l[2] or l[1] + l[2] == l[3]:
print("SEGMENT")
else:
print("IMPOSSIBLE") | Title: Triangle
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input Specification:
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output Specification:
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Demo Input:
['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n']
Demo Output:
['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n']
Note:
none | ```python
s = input()
#s = '4 2 1 3'
#s = '7 2 2 4'
#s = '10 10 10 10'
s = s.split(' ')
l = [int(s[0]),int(s[1]),int(s[2]),int(s[3])]
l.sort()
tmp = l[0]
if l[0] + l[1] > l[2]:
print("TRIANGLE")
else:
l.pop(0)
if l[0] + l[1] > l[2]:
print("TRIANGLE")
else:
l.append(tmp)
l.sort()
if l[0] + l[1] == l[2] or l[1] + l[2] == l[3]:
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | 3.969 |
1,003 | A | Polycarp's Pockets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. | The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. | Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. | [
"6\n1 2 4 3 3 2\n",
"1\n100\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "6\n1 2 4 3 3 2",
"output": "2"
},
{
"input": "1\n100",
"output": "1"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100... | 1,603,445,584 | 2,147,483,647 | PyPy 3 | OK | TESTS | 37 | 155 | 102,400 | from collections import defaultdict
n = int(input())
arr = list(map(int,input().split()))
maxm = -10**18
freq = defaultdict(int)
for ele in arr:
freq[ele]+=1
maxm = max(freq[ele],maxm)
print(maxm) | Title: Polycarp's Pockets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
Output Specification:
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
Demo Input:
['6\n1 2 4 3 3 2\n', '1\n100\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
from collections import defaultdict
n = int(input())
arr = list(map(int,input().split()))
maxm = -10**18
freq = defaultdict(int)
for ele in arr:
freq[ele]+=1
maxm = max(freq[ele],maxm)
print(maxm)
``` | 3 | |
552 | B | Vanya and Books | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | null | null | Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library. | Print the number of digits needed to number all the books. | [
"13\n",
"4\n"
] | [
"17\n",
"4\n"
] | Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | 1,000 | [
{
"input": "13",
"output": "17"
},
{
"input": "4",
"output": "4"
},
{
"input": "100",
"output": "192"
},
{
"input": "99",
"output": "189"
},
{
"input": "1000000000",
"output": "8888888899"
},
{
"input": "1000000",
"output": "5888896"
},
{
"... | 1,694,696,765 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 46 | 0 | v1=int(input());w=len(str(v1));w1='1'+'0'*w;w2=w*(v1+1);c=0
for i in range(1,w):
c+=10**i
print(w2-(c+1)) | Title: Vanya and Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library.
Output Specification:
Print the number of digits needed to number all the books.
Demo Input:
['13\n', '4\n']
Demo Output:
['17\n', '4\n']
Note:
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | ```python
v1=int(input());w=len(str(v1));w1='1'+'0'*w;w2=w*(v1+1);c=0
for i in range(1,w):
c+=10**i
print(w2-(c+1))
``` | 3 | |
932 | A | Palindromic Supersequence | PROGRAMMING | 800 | [
"constructive algorithms"
] | null | null | You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. | First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*. | Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them. | [
"aba\n",
"ab\n"
] | [
"aba",
"aabaa"
] | In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome. | 500 | [
{
"input": "aba",
"output": "abaaba"
},
{
"input": "ab",
"output": "abba"
},
{
"input": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmk... | 1,518,771,839 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 77 | 5,734,400 | s = input()
s1 = s[::-1]
print(s, s1, sep = '') | Title: Palindromic Supersequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.
Input Specification:
First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.
Output Specification:
Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.
Demo Input:
['aba\n', 'ab\n']
Demo Output:
['aba', 'aabaa']
Note:
In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome. | ```python
s = input()
s1 = s[::-1]
print(s, s1, sep = '')
``` | 3 | |
445 | A | DZY Loves Chessboard | PROGRAMMING | 1,200 | [
"dfs and similar",
"implementation"
] | null | null | DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad. | Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists. | [
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] | [
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] | In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | 500 | [
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n--------... | 1,660,928,950 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 37 | 93 | 6,860,800 | a=list(map(int,input().split()))
n,m=a
chess=[[None for z in range(m)] for _ in range(n)]
# for _ in range(n):
# chess.append(sq)
for i in range(n):
for j in range(m):
# print(chess)
if i%2==0:
if j%2==0:
chess[i][j]='B'
else:
chess[i][j]='W'
else:
if j%2==0:
chess[i][j]='W'
else:
chess[i][j]='B'
loc=[]
for i in range(n):
a=input()
a=list(a)
for j in range(m):
if a[j]=='-':
loc.append((i,j))
for i,j in loc:
chess[i][j]='-'
for i in chess:
for j in i:
print(j,end='')
print()
# print(chess)
| Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | ```python
a=list(map(int,input().split()))
n,m=a
chess=[[None for z in range(m)] for _ in range(n)]
# for _ in range(n):
# chess.append(sq)
for i in range(n):
for j in range(m):
# print(chess)
if i%2==0:
if j%2==0:
chess[i][j]='B'
else:
chess[i][j]='W'
else:
if j%2==0:
chess[i][j]='W'
else:
chess[i][j]='B'
loc=[]
for i in range(n):
a=input()
a=list(a)
for j in range(m):
if a[j]=='-':
loc.append((i,j))
for i,j in loc:
chess[i][j]='-'
for i in chess:
for j in i:
print(j,end='')
print()
# print(chess)
``` | 3 | |
515 | C | Drazil and Factorial | PROGRAMMING | 1,400 | [
"greedy",
"math",
"sortings"
] | null | null | Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes. | Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. | [
"4\n1234\n",
"3\n555\n"
] | [
"33222\n",
"555\n"
] | In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "4\n1234",
"output": "33222"
},
{
"input": "3\n555",
"output": "555"
},
{
"input": "15\n012345781234578",
"output": "7777553333222222222222"
},
{
"input": "1\n8",
"output": "7222"
},
{
"input": "10\n1413472614",
"output": "75333332222222"
},
{
... | 1,684,825,975 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | n = int(input())
a = list(map(int,list(input())))
dict1 = {}
for i in range(n):
for j in range(2,a[i]+1):
if j in dict1:
dict1[j]+=1
else:
dict1[j] = 1
ans = []
for i in sorted(dict1,reverse = True):
if i in (2,3,5,7):
ans += [i for j in range(dict1[i])]
for j in range(2,i+1):
dict1[j]-=dict1[i]
elif i == 4:
dict1[2]+= dict1[4]*2
elif i == 6:
dict1[2]+=dict1[6]
dict1[3]+=dict1[2]
elif i == 8:
dict1[2]+=dict1[8]*3
elif i == 9:
dict1[3]+=dict1[9]*2
ans = list(map(str,ans))
print(int(''.join(ans))) | Title: Drazil and Factorial
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
Output Specification:
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Demo Input:
['4\n1234\n', '3\n555\n']
Demo Output:
['33222\n', '555\n']
Note:
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
n = int(input())
a = list(map(int,list(input())))
dict1 = {}
for i in range(n):
for j in range(2,a[i]+1):
if j in dict1:
dict1[j]+=1
else:
dict1[j] = 1
ans = []
for i in sorted(dict1,reverse = True):
if i in (2,3,5,7):
ans += [i for j in range(dict1[i])]
for j in range(2,i+1):
dict1[j]-=dict1[i]
elif i == 4:
dict1[2]+= dict1[4]*2
elif i == 6:
dict1[2]+=dict1[6]
dict1[3]+=dict1[2]
elif i == 8:
dict1[2]+=dict1[8]*3
elif i == 9:
dict1[3]+=dict1[9]*2
ans = list(map(str,ans))
print(int(''.join(ans)))
``` | 0 | |
371 | B | Fox Dividing Cheese | PROGRAMMING | 1,300 | [
"math",
"number theory"
] | null | null | Two little greedy bears have found two pieces of cheese in the forest of weight *a* and *b* grams, correspondingly. The bears are so greedy that they are ready to fight for the larger piece. That's where the fox comes in and starts the dialog: "Little bears, wait a little, I want to make your pieces equal" "Come off it fox, how are you going to do that?", the curious bears asked. "It's easy", said the fox. "If the mass of a certain piece is divisible by two, then I can eat exactly a half of the piece. If the mass of a certain piece is divisible by three, then I can eat exactly two-thirds, and if the mass is divisible by five, then I can eat four-fifths. I'll eat a little here and there and make the pieces equal".
The little bears realize that the fox's proposal contains a catch. But at the same time they realize that they can not make the two pieces equal themselves. So they agreed to her proposal, but on one condition: the fox should make the pieces equal as quickly as possible. Find the minimum number of operations the fox needs to make pieces equal. | The first line contains two space-separated integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=109). | If the fox is lying to the little bears and it is impossible to make the pieces equal, print -1. Otherwise, print the required minimum number of operations. If the pieces of the cheese are initially equal, the required number is 0. | [
"15 20\n",
"14 8\n",
"6 6\n"
] | [
"3\n",
"-1\n",
"0\n"
] | none | 1,000 | [
{
"input": "15 20",
"output": "3"
},
{
"input": "14 8",
"output": "-1"
},
{
"input": "6 6",
"output": "0"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 1024",
"output": "10"
},
{
"input": "1024 729",
"output": "16"
},
{
"input": "1024... | 1,640,892,058 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 54 | 109 | 0 | a, b = map(int, input().split())
if a==b:
print(0)
else:
s1 = 0
s2 = 0
s3 = 0
while a%2==0:
a//=2
s1+=1
while a%3==0:
a//=3
s2+=1
while a%5==0:
a//=5
s3+=1
while b%2==0:
b//=2
s1-=1
while b%3==0:
b//=3
s2-=1
while b%5==0:
b//=5
s3-=1
if a!=b:
print(-1)
else:
print(abs(s1)+abs(s2)+abs(s3)) | Title: Fox Dividing Cheese
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two little greedy bears have found two pieces of cheese in the forest of weight *a* and *b* grams, correspondingly. The bears are so greedy that they are ready to fight for the larger piece. That's where the fox comes in and starts the dialog: "Little bears, wait a little, I want to make your pieces equal" "Come off it fox, how are you going to do that?", the curious bears asked. "It's easy", said the fox. "If the mass of a certain piece is divisible by two, then I can eat exactly a half of the piece. If the mass of a certain piece is divisible by three, then I can eat exactly two-thirds, and if the mass is divisible by five, then I can eat four-fifths. I'll eat a little here and there and make the pieces equal".
The little bears realize that the fox's proposal contains a catch. But at the same time they realize that they can not make the two pieces equal themselves. So they agreed to her proposal, but on one condition: the fox should make the pieces equal as quickly as possible. Find the minimum number of operations the fox needs to make pieces equal.
Input Specification:
The first line contains two space-separated integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=109).
Output Specification:
If the fox is lying to the little bears and it is impossible to make the pieces equal, print -1. Otherwise, print the required minimum number of operations. If the pieces of the cheese are initially equal, the required number is 0.
Demo Input:
['15 20\n', '14 8\n', '6 6\n']
Demo Output:
['3\n', '-1\n', '0\n']
Note:
none | ```python
a, b = map(int, input().split())
if a==b:
print(0)
else:
s1 = 0
s2 = 0
s3 = 0
while a%2==0:
a//=2
s1+=1
while a%3==0:
a//=3
s2+=1
while a%5==0:
a//=5
s3+=1
while b%2==0:
b//=2
s1-=1
while b%3==0:
b//=3
s2-=1
while b%5==0:
b//=5
s3-=1
if a!=b:
print(-1)
else:
print(abs(s1)+abs(s2)+abs(s3))
``` | 3 | |
766 | B | Mahmoud and a Triangle | PROGRAMMING | 1,000 | [
"constructive algorithms",
"geometry",
"greedy",
"math",
"number theory",
"sortings"
] | null | null | Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area. | The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has. | In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise. | [
"5\n1 5 3 2 4\n",
"3\n4 1 2\n"
] | [
"YES\n",
"NO\n"
] | For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | 1,000 | [
{
"input": "5\n1 5 3 2 4",
"output": "YES"
},
{
"input": "3\n4 1 2",
"output": "NO"
},
{
"input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576",
"output": "NO"
},... | 1,661,911,486 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | t=int(input())
a=sorted(list(map(int,input().split())))
def tr(x):
for i in range(t-2):
if x[i]+x[i+1]>x[i+2]
print('YES')
return
print('NO')
return
tr(a) | Title: Mahmoud and a Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input Specification:
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
Output Specification:
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
Demo Input:
['5\n1 5 3 2 4\n', '3\n4 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | ```python
t=int(input())
a=sorted(list(map(int,input().split())))
def tr(x):
for i in range(t-2):
if x[i]+x[i+1]>x[i+2]
print('YES')
return
print('NO')
return
tr(a)
``` | -1 | |
279 | B | Books | PROGRAMMING | 1,400 | [
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null | When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. | Print a single integer — the maximum number of books Valera can read. | [
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] | [
"3\n",
"1\n"
] | none | 1,000 | [
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4... | 1,676,720,332 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 342 | 7,065,600 | n,free=map(int,input().split())
t=[int(i) for i in input().split()]
pointer,time,ans=0,0,0
for curr in range(n):
while pointer<n and time+t[pointer]<=free:
time+=t[pointer]
pointer+=1
if pointer - curr > ans : ans = pointer - curr
time-=t[curr]
print(ans) | Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
n,free=map(int,input().split())
t=[int(i) for i in input().split()]
pointer,time,ans=0,0,0
for curr in range(n):
while pointer<n and time+t[pointer]<=free:
time+=t[pointer]
pointer+=1
if pointer - curr > ans : ans = pointer - curr
time-=t[curr]
print(ans)
``` | 3 | |
818 | A | Diplomas and Certificates | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners. | The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas. | Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners. | [
"18 2\n",
"9 10\n",
"1000000000000 5\n",
"1000000000000 499999999999\n"
] | [
"3 6 9\n",
"0 0 9\n",
"83333333333 416666666665 500000000002\n",
"1 499999999999 500000000000\n"
] | none | 0 | [
{
"input": "18 2",
"output": "3 6 9"
},
{
"input": "9 10",
"output": "0 0 9"
},
{
"input": "1000000000000 5",
"output": "83333333333 416666666665 500000000002"
},
{
"input": "1000000000000 499999999999",
"output": "1 499999999999 500000000000"
},
{
"input": "1 1",... | 1,579,040,533 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 124 | 307,200 | n, k = map(int, input().split())
d = n//2//(k+1)
print(d, k*d, n-(k+1)*d) | Title: Diplomas and Certificates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
Input Specification:
The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
Output Specification:
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners.
Demo Input:
['18 2\n', '9 10\n', '1000000000000 5\n', '1000000000000 499999999999\n']
Demo Output:
['3 6 9\n', '0 0 9\n', '83333333333 416666666665 500000000002\n', '1 499999999999 500000000000\n']
Note:
none | ```python
n, k = map(int, input().split())
d = n//2//(k+1)
print(d, k*d, n-(k+1)*d)
``` | 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"
},
... | 1,603,887,033 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 248 | 0 | n, x, y = input().split()
n, x, y = int(n), int(x), int(y)
percent = ((y * n) / 100)
if int(percent) != percent:
percent += 1
percent = int(percent)
print(int(percent - x))
| 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 = input().split()
n, x, y = int(n), int(x), int(y)
percent = ((y * n) / 100)
if int(percent) != percent:
percent += 1
percent = int(percent)
print(int(percent - x))
``` | 0 | |
576 | A | Vasya and Petya's Game | PROGRAMMING | 1,500 | [
"math",
"number theory"
] | null | null | Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about. | A single line contains number *n* (1<=≤<=*n*<=≤<=103). | Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them. | [
"4\n",
"6\n"
] | [
"3\n2 4 3 \n",
"4\n2 4 3 5 \n"
] | The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | 500 | [
{
"input": "4",
"output": "3\n2 4 3 "
},
{
"input": "6",
"output": "4\n2 4 3 5 "
},
{
"input": "1",
"output": "0"
},
{
"input": "15",
"output": "9\n2 4 8 3 9 5 7 11 13 "
},
{
"input": "19",
"output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 "
},
{
"input": "2... | 1,591,731,335 | 2,147,483,647 | PyPy 3 | OK | TESTS | 26 | 140 | 1,228,800 | def prog():
n = int(input())
primes = [True for i in range(n+1)]
for p in range(2,int(n**0.5)+1):
if primes[p]:
for num in range(p**2,n+1,p):
primes[num] = False
sequence = []
for num in range(2,n+1):
if primes[num] == True:
num_pow = num
while num_pow <= n:
sequence.append(str(num_pow))
num_pow *= num
print(len(sequence))
print(" ".join(sequence))
prog()
| Title: Vasya and Petya's Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.
Input Specification:
A single line contains number *n* (1<=≤<=*n*<=≤<=103).
Output Specification:
Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Demo Input:
['4\n', '6\n']
Demo Output:
['3\n2 4 3 \n', '4\n2 4 3 5 \n']
Note:
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | ```python
def prog():
n = int(input())
primes = [True for i in range(n+1)]
for p in range(2,int(n**0.5)+1):
if primes[p]:
for num in range(p**2,n+1,p):
primes[num] = False
sequence = []
for num in range(2,n+1):
if primes[num] == True:
num_pow = num
while num_pow <= n:
sequence.append(str(num_pow))
num_pow *= num
print(len(sequence))
print(" ".join(sequence))
prog()
``` | 3 | |
265 | A | Colorful Stones (Simplified Edition) | PROGRAMMING | 800 | [
"implementation"
] | null | null | There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. | The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. | Print the final 1-based position of Liss in a single line. | [
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] | [
"2\n",
"3\n",
"15\n"
] | none | 500 | [
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
... | 1,664,022,702 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 92 | 0 | a = input()
b = input()
x = 0
for i in b:
if i == a[x]:
x += 1
print(x+1) | Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none | ```python
a = input()
b = input()
x = 0
for i in b:
if i == a[x]:
x += 1
print(x+1)
``` | 3 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,660,527,364 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | from math import floor
[N, M] = list(map(int, input().split()))
returned = floor(M/2 * N)
print(returned)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
from math import floor
[N, M] = list(map(int, input().split()))
returned = floor(M/2 * N)
print(returned)
``` | 3.977 |
17 | A | Noldbach problem | PROGRAMMING | 1,000 | [
"brute force",
"math",
"number theory"
] | A. Noldbach problem | 2 | 64 | Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong. | The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000). | Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO. | [
"27 2\n",
"45 7\n"
] | [
"YES",
"NO"
] | In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | 0 | [
{
"input": "27 2",
"output": "YES"
},
{
"input": "45 7",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "15 1",
"output": "YES"
},
{
"input": "17 1",
"output": "YES"
},
{
"input": "34 5",
"output": "NO"
},
{
"input": "37 ... | 1,609,860,834 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 218 | 307,200 | import math
def is_prime(n):
if n == 2:
return True
if n < 2 or n % 2 == 0:
return False
limiter = math.isqrt(n)
for i in range(3, limiter + 1, 2):
if n % i == 0:
return False
return True
n, k = list(map(int, input().split(' ')))
primes, ans = [], 0
for i in range(n + 1):
if is_prime(i):
primes.append(i)
for i in range(len(primes) - 1):
temp = primes[i] + primes[i + 1] + 1
if is_prime(temp) and temp <= n:
ans += 1
print('YES' if ans >= k else 'NO')
'''
n = 27 k=2
primes={2, 3, 5, 7, 11, 13, 17, 19, 23}
2 + 3 + 1 = 6
3 + 5 + 1 = 9
5 + 7 + 1 = 13 V
7 + 11 + 1 = 19 V
11 + 13 + 1 = 25
13 + 17 + 1 = 31 X
'''
| Title: Noldbach problem
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
Input Specification:
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
Output Specification:
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
Demo Input:
['27 2\n', '45 7\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | ```python
import math
def is_prime(n):
if n == 2:
return True
if n < 2 or n % 2 == 0:
return False
limiter = math.isqrt(n)
for i in range(3, limiter + 1, 2):
if n % i == 0:
return False
return True
n, k = list(map(int, input().split(' ')))
primes, ans = [], 0
for i in range(n + 1):
if is_prime(i):
primes.append(i)
for i in range(len(primes) - 1):
temp = primes[i] + primes[i + 1] + 1
if is_prime(temp) and temp <= n:
ans += 1
print('YES' if ans >= k else 'NO')
'''
n = 27 k=2
primes={2, 3, 5, 7, 11, 13, 17, 19, 23}
2 + 3 + 1 = 6
3 + 5 + 1 = 9
5 + 7 + 1 = 13 V
7 + 11 + 1 = 19 V
11 + 13 + 1 = 25
13 + 17 + 1 = 31 X
'''
``` | 3.943211 |
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... | 1,587,727,036 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 218 | 0 | n, m = map(int, input().split())
pas = list(map(int, input().split()))
fail = list(map(int, input().split()))
k = max(pas)
t = min(pas)
k = max(k, 2*t)
p = min(fail)
print(k if p>k else -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())
pas = list(map(int, input().split()))
fail = list(map(int, input().split()))
k = max(pas)
t = min(pas)
k = max(k, 2*t)
p = min(fail)
print(k if p>k else -1)
``` | 3 | |
663 | A | Rebus | PROGRAMMING | 1,800 | [
"constructive algorithms",
"expression parsing",
"greedy",
"math"
] | null | null | You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer *n*. The goal is to replace each question mark with some positive integer from 1 to *n*, such that equality holds. | The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer *n* is positive and doesn't exceed 1<=000<=000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks. | The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to *n*. Follow the format given in the samples. | [
"? + ? - ? + ? + ? = 42\n",
"? - ? = 1\n",
"? = 1000000\n"
] | [
"Possible\n9 + 13 - 39 + 28 + 31 = 42\n",
"Impossible\n",
"Possible\n1000000 = 1000000\n"
] | none | 500 | [
{
"input": "? + ? - ? + ? + ? = 42",
"output": "Possible\n1 + 1 - 1 + 1 + 40 = 42"
},
{
"input": "? - ? = 1",
"output": "Impossible"
},
{
"input": "? = 1000000",
"output": "Possible\n1000000 = 1000000"
},
{
"input": "? + ? + ? + ? + ? + ? + ? + ? + ? + ? + ? + ? + ? + ? + ? +... | 1,461,009,804 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 61 | 4,608,000 | expr = input().split()
result_expr = None
if len(expr) == 3:
result_expr = "{0} = {0}".format(expr[-1])
else:
target_number = int(expr.pop())
expr.pop()
other_summands = ' '.join(expr[1:]).replace('?', '1')
first_summand = target_number - eval(other_summands)
if 0 < first_summand <= target_number:
result_expr = "{0} {1} = {2}".format(first_summand, other_summands, target_number)
if result_expr:
print("Possible")
print(result_expr)
else:
print("Impossible") | Title: Rebus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer *n*. The goal is to replace each question mark with some positive integer from 1 to *n*, such that equality holds.
Input Specification:
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer *n* is positive and doesn't exceed 1<=000<=000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output Specification:
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to *n*. Follow the format given in the samples.
Demo Input:
['? + ? - ? + ? + ? = 42\n', '? - ? = 1\n', '? = 1000000\n']
Demo Output:
['Possible\n9 + 13 - 39 + 28 + 31 = 42\n', 'Impossible\n', 'Possible\n1000000 = 1000000\n']
Note:
none | ```python
expr = input().split()
result_expr = None
if len(expr) == 3:
result_expr = "{0} = {0}".format(expr[-1])
else:
target_number = int(expr.pop())
expr.pop()
other_summands = ' '.join(expr[1:]).replace('?', '1')
first_summand = target_number - eval(other_summands)
if 0 < first_summand <= target_number:
result_expr = "{0} {1} = {2}".format(first_summand, other_summands, target_number)
if result_expr:
print("Possible")
print(result_expr)
else:
print("Impossible")
``` | 0 | |
151 | A | Soft Drinking | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make? | The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space. | Print a single integer — the number of toasts each friend can make. | [
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] | [
"2\n",
"3\n",
"0\n"
] | A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | 500 | [
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
... | 1,669,468,928 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 92 | 0 | inp = input()
arr = [int(item) for item in inp.split(" ")]
# print(arr)
n, k, l, c, d, p, nl, np = arr
ans = min(min(int((k*l)/nl), c*d), int(p/np))
print(int(ans/n))
| Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer — the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | ```python
inp = input()
arr = [int(item) for item in inp.split(" ")]
# print(arr)
n, k, l, c, d, p, nl, np = arr
ans = min(min(int((k*l)/nl), c*d), int(p/np))
print(int(ans/n))
``` | 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... | 1,538,431,888 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 280 | 0 | n = int(input())
sumx=0
sumy=0
sumz=0
for i in range(n):
coords = input().split()
sumx += int(coords[0])
sumy += int(coords[1])
sumz += int(coords[2])
if sumx == 0 and sumy == 0 and sumz == 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())
sumx=0
sumy=0
sumz=0
for i in range(n):
coords = input().split()
sumx += int(coords[0])
sumy += int(coords[1])
sumz += int(coords[2])
if sumx == 0 and sumy == 0 and sumz == 0:
print("YES")
else:
print("NO")
``` | 3.93 |
0 | none | none | none | 0 | [
"none"
] | null | null | Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not. | In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly. | For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower). | [
"6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n"
] | [
"Yes\nYes\nYes\nNo\nNo\nYes\n"
] | First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3. | 0 | [
{
"input": "6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000",
"output": "Yes\nYes\nYes\nNo\nNo\nYes"
},
{
"input": "3\n1 1\n8 27\n1000 1331",
"output": "Yes\nNo\nNo"
},
{
"input": "1\n12004 18012002",
"output": "Yes"
},
{
"input": "1\n3331 11095561",
"output": "Yes... | 1,501,429,913 | 4,613 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 7 | 1,000 | 4,608,000 | from math import pow
def gcd(x, y):
return x if y == 0 else gcd(y, x % y)
for _ in range(int(input())):
x, y = map(int, input().split())
t = x * y
k = gcd(x, y) ** 2
if round(pow(t, 1 / 3)) ** 3 == t and k % x == 0 and k % y == 0:
print("Yes")
else:
print("No")
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input Specification:
In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly.
Output Specification:
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n']
Demo Output:
['Yes\nYes\nYes\nNo\nNo\nYes\n']
Note:
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3. | ```python
from math import pow
def gcd(x, y):
return x if y == 0 else gcd(y, x % y)
for _ in range(int(input())):
x, y = map(int, input().split())
t = x * y
k = gcd(x, y) ** 2
if round(pow(t, 1 / 3)) ** 3 == t and k % x == 0 and k % y == 0:
print("Yes")
else:
print("No")
``` | 0 | |
703 | A | Mishka and Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her! | The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively. | If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line. | [
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] | [
"Mishka",
"Friendship is magic!^^",
"Chris"
] | In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris. | 500 | [
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 ... | 1,666,263,866 | 2,147,483,647 | Python 3 | OK | TESTS | 69 | 46 | 0 | a=int(input())
q=0
p=0
for i in range(a):
a,b=map(int,input().split())
q+=a>b
p+=b>a
if q==p:
print('Friendship is magic!^^')
else:
print('MCihsrhiksa'[q<p::2]) | Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris. | ```python
a=int(input())
q=0
p=0
for i in range(a):
a,b=map(int,input().split())
q+=a>b
p+=b>a
if q==p:
print('Friendship is magic!^^')
else:
print('MCihsrhiksa'[q<p::2])
``` | 3 | |
287 | A | IQ Test | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color. | Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black. | Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise. | [
"####\n.#..\n####\n....\n",
"####\n....\n####\n....\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | 500 | [
{
"input": "###.\n...#\n###.\n...#",
"output": "NO"
},
{
"input": ".##.\n#..#\n.##.\n#..#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.#.",
"output": "NO"
},
{
"input": "##..\n..##\n##..\n..##",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n.#.#\n.#.#",
"ou... | 1,699,897,170 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
bool all_eq(string s1,string s2)
{
for(int i=0;i<4;i++)
{
if(s1[i]==s2[i])
{
return true;
}
}
return false;
}
int main()
{
fast_io;
vector<string>a;
for(int i=0;i<4;i++)
{
string s;
cin>>s;
a.push_back(s);
}
cout << (all_eq(a[0], a[1]) || all_eq(a[2], a[3]) ? "YES\n" : "NO\n");
return 0;
}
| Title: IQ Test
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.
Input Specification:
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
Output Specification:
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
Demo Input:
['####\n.#..\n####\n....\n', '####\n....\n####\n....\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | ```python
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
bool all_eq(string s1,string s2)
{
for(int i=0;i<4;i++)
{
if(s1[i]==s2[i])
{
return true;
}
}
return false;
}
int main()
{
fast_io;
vector<string>a;
for(int i=0;i<4;i++)
{
string s;
cin>>s;
a.push_back(s);
}
cout << (all_eq(a[0], a[1]) || all_eq(a[2], a[3]) ? "YES\n" : "NO\n");
return 0;
}
``` | -1 | |
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... | 1,684,581,550 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 30 | 0 | def main():
size, a = map(int, input().split())
numbers = tuple(map(int, input().split()))
total = 0
a -= 1
for i in range(size):
x = abs(a - i)
t1 = -1
t2 = -1
if 0 <= a - x < size:
if numbers[a - x]:
t1 = 1
else:
t1 = 0
if 0 <= a + x < size:
if numbers[a + x]:
t1 = 1
else:
t1 = 0
if (t1 == -1 and t2 == 1) or (t1 == 1 and t2 == -1):
total += 1
elif t1 and t2:
total += 2
print(total)
if __name__ == '__main__':
main() | 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
def main():
size, a = map(int, input().split())
numbers = tuple(map(int, input().split()))
total = 0
a -= 1
for i in range(size):
x = abs(a - i)
t1 = -1
t2 = -1
if 0 <= a - x < size:
if numbers[a - x]:
t1 = 1
else:
t1 = 0
if 0 <= a + x < size:
if numbers[a + x]:
t1 = 1
else:
t1 = 0
if (t1 == -1 and t2 == 1) or (t1 == 1 and t2 == -1):
total += 1
elif t1 and t2:
total += 2
print(total)
if __name__ == '__main__':
main()
``` | 0 | |
305 | B | Continued Fractions | PROGRAMMING | 1,700 | [
"brute force",
"implementation",
"math"
] | null | null | A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal. | The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction.
The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction.
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. | Print "YES" if these fractions are equal and "NO" otherwise. | [
"9 4\n2\n2 4\n",
"9 4\n3\n2 3 1\n",
"9 4\n3\n1 2 4\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,000 | [
{
"input": "9 4\n2\n2 4",
"output": "YES"
},
{
"input": "9 4\n3\n2 3 1",
"output": "YES"
},
{
"input": "9 4\n3\n1 2 4",
"output": "NO"
},
{
"input": "39088169 24157817\n36\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2",
"output": "YES"
},
{
... | 1,370,119,584 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 187 | 5,222,400 | from fractions import gcd
p, q = [int(x) for x in input().split()]
n = int(input())
A = [int(x) for x in input().split()]
g = gcd(p, q)
p, q = p // g, q // g
poss = True
for a in A:
if q == 0 or p < a * q:
poss = False
break
p -= a * q
g = gcd(p, q)
p, q = q // g, p // g
print("YES" if poss and q == 0 else "NO")
| Title: Continued Fractions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal.
Input Specification:
The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction.
The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction.
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.
Output Specification:
Print "YES" if these fractions are equal and "NO" otherwise.
Demo Input:
['9 4\n2\n2 4\n', '9 4\n3\n2 3 1\n', '9 4\n3\n1 2 4\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
from fractions import gcd
p, q = [int(x) for x in input().split()]
n = int(input())
A = [int(x) for x in input().split()]
g = gcd(p, q)
p, q = p // g, q // g
poss = True
for a in A:
if q == 0 or p < a * q:
poss = False
break
p -= a * q
g = gcd(p, q)
p, q = q // g, p // g
print("YES" if poss and q == 0 else "NO")
``` | 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... | 1,669,530,665 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 81 | 124 | 0 | n = int(input())
list = []
for i in range (0,n):
ltemp = input().split()
list.append(ltemp)
x, y, z = 0, 0, 0
for i in list:
x += int(i[0])
y += int(i[1])
z += int(i[2])
if x == 0 and y == 0 and z == 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())
list = []
for i in range (0,n):
ltemp = input().split()
list.append(ltemp)
x, y, z = 0, 0, 0
for i in list:
x += int(i[0])
y += int(i[1])
z += int(i[2])
if x == 0 and y == 0 and z == 0:
print("YES")
else:
print("NO")
``` | 3.969 |
56 | B | Spoilt Permutation | PROGRAMMING | 1,300 | [
"implementation"
] | B. Spoilt Permutation | 2 | 256 | Vasya collects coins: he has exactly one coin for every year from 1 to *n*. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from *l* to *r* inclusively and put them in the reverse order. That is, he took a certain segment [*l*,<=*r*] and reversed it. At that the segment's endpoints did not coincide. For example, if *n*<==<=8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2,<=6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... *n* using exactly one segment reversal. If it is possible, find the segment itself. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000) which is the number of coins in Vasya's collection. The second line contains space-separated *n* integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to *n*, and every number is used exactly 1 time. | If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers *l* *r* (1<=≤<=*l*<=<<=*r*<=≤<=*n*) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... *n* the given one. | [
"8\n1 6 5 4 3 2 7 8\n",
"4\n2 3 4 1\n",
"4\n1 2 3 4\n"
] | [
"2 6\n",
"0 0\n",
"0 0\n"
] | none | 1,000 | [
{
"input": "8\n1 6 5 4 3 2 7 8",
"output": "2 6"
},
{
"input": "4\n2 3 4 1",
"output": "0 0"
},
{
"input": "4\n1 2 3 4",
"output": "0 0"
},
{
"input": "8\n1 3 2 4 6 5 7 8",
"output": "0 0"
},
{
"input": "8\n1 3 4 2 6 5 7 8",
"output": "0 0"
},
{
"input... | 1,591,806,956 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 7 | 654 | 10,342,400 | import os, sys
from io import IOBase, BytesIO
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log, pi, sin
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var, end="\n"): sys.stdout.write(str(var)+end)
def outarr(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
n = int(data())
arr = l()
left, right = 0, 0
cnt = 0
for i in range(1, n):
if arr[i]-arr[i-1] > 1:
if not left:
left = i + 1
elif not right:
right = i
else:
out("0 0")
exit()
outarr(str(left) + " " + str(right))
| Title: Spoilt Permutation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya collects coins: he has exactly one coin for every year from 1 to *n*. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from *l* to *r* inclusively and put them in the reverse order. That is, he took a certain segment [*l*,<=*r*] and reversed it. At that the segment's endpoints did not coincide. For example, if *n*<==<=8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2,<=6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... *n* using exactly one segment reversal. If it is possible, find the segment itself.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000) which is the number of coins in Vasya's collection. The second line contains space-separated *n* integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to *n*, and every number is used exactly 1 time.
Output Specification:
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers *l* *r* (1<=≤<=*l*<=<<=*r*<=≤<=*n*) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... *n* the given one.
Demo Input:
['8\n1 6 5 4 3 2 7 8\n', '4\n2 3 4 1\n', '4\n1 2 3 4\n']
Demo Output:
['2 6\n', '0 0\n', '0 0\n']
Note:
none | ```python
import os, sys
from io import IOBase, BytesIO
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log, pi, sin
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var, end="\n"): sys.stdout.write(str(var)+end)
def outarr(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
n = int(data())
arr = l()
left, right = 0, 0
cnt = 0
for i in range(1, n):
if arr[i]-arr[i-1] > 1:
if not left:
left = i + 1
elif not right:
right = i
else:
out("0 0")
exit()
outarr(str(left) + " " + str(right))
``` | 0 |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"outp... | 1,696,016,473 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 92 | 0 | def football(ch):
for i in range(len(ch)):
if ch[i:i+7] =='0000000' or ch[i:i+7]=='1111111':
return('YES')
return('NO')
ch=str(input())
print(football(ch)) | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
def football(ch):
for i in range(len(ch)):
if ch[i:i+7] =='0000000' or ch[i:i+7]=='1111111':
return('YES')
return('NO')
ch=str(input())
print(football(ch))
``` | 3.977 |
868 | C | Qualification Rounds | PROGRAMMING | 1,500 | [
"bitmasks",
"brute force",
"constructive algorithms",
"dp"
] | null | null | Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset! | The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise. | Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n",
"3 2\n1 0\n1 1\n0 1\n"
] | [
"NO\n",
"YES\n"
] | In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | 1,000 | [
{
"input": "5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0",
"output": "NO"
},
{
"input": "3 2\n1 0\n1 1\n0 1",
"output": "YES"
},
{
"input": "10 2\n1 0\n1 0\n0 0\n1 1\n0 0\n1 1\n0 0\n1 1\n0 1\n0 1",
"output": "YES"
},
{
"input": "10 3\n1 0 0\n0 1 1\n1 0 0\n0 1 0\n0 0 1\n1 0 1\n0 1 1... | 1,644,408,590 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 171 | 4,505,600 | from math import inf
from collections import *
import math, os, sys, heapq, bisect, random
from functools import lru_cache
from itertools import *
def inp(): return sys.stdin.readline().rstrip("\r\n")
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def inpu(): return int(inp())
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
M,M1=1000000007,998244353
def main():
how_much_noob_I_am = 1
#how_much_noob_I_am = inpu()
for __ in range(how_much_noob_I_am):
n,k = sep()
l=[lis() for i in range(n)]
cnt = [0]*k
for j in range(k):
for i in range(n):
if l[i][j]==1:
cnt[j]+=1
for i in range(k):
if cnt[i]>(n+1)//2:
print("NO")
break
else:
print("YES")
if __name__ == '__main__':
main()
| Title: Qualification Rounds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset!
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise.
Output Specification:
Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n', '3 2\n1 0\n1 1\n0 1\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | ```python
from math import inf
from collections import *
import math, os, sys, heapq, bisect, random
from functools import lru_cache
from itertools import *
def inp(): return sys.stdin.readline().rstrip("\r\n")
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def inpu(): return int(inp())
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
M,M1=1000000007,998244353
def main():
how_much_noob_I_am = 1
#how_much_noob_I_am = inpu()
for __ in range(how_much_noob_I_am):
n,k = sep()
l=[lis() for i in range(n)]
cnt = [0]*k
for j in range(k):
for i in range(n):
if l[i][j]==1:
cnt[j]+=1
for i in range(k):
if cnt[i]>(n+1)//2:
print("NO")
break
else:
print("YES")
if __name__ == '__main__':
main()
``` | 0 | |
155 | A | I_love_\%username\% | PROGRAMMING | 800 | [
"brute force"
] | null | null | Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. | Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. | [
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] | [
"2\n",
"4\n"
] | In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | 500 | [
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
... | 1,675,172,939 | 2,147,483,647 | PyPy 3 | OK | TESTS | 37 | 186 | 0 | n=int(input())
nums=list(map(int,input().split()))
worst,best,ans=0,0,0
worst,best=nums[0],nums[0]
for item in nums[1:n]:
if item >best:
best=item
ans+=1
elif item<worst:
worst=item
ans+=1
print(ans)
| Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | ```python
n=int(input())
nums=list(map(int,input().split()))
worst,best,ans=0,0,0
worst,best=nums[0],nums[0]
for item in nums[1:n]:
if item >best:
best=item
ans+=1
elif item<worst:
worst=item
ans+=1
print(ans)
``` | 3 | |
515 | C | Drazil and Factorial | PROGRAMMING | 1,400 | [
"greedy",
"math",
"sortings"
] | null | null | Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes. | Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. | [
"4\n1234\n",
"3\n555\n"
] | [
"33222\n",
"555\n"
] | In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "4\n1234",
"output": "33222"
},
{
"input": "3\n555",
"output": "555"
},
{
"input": "15\n012345781234578",
"output": "7777553333222222222222"
},
{
"input": "1\n8",
"output": "7222"
},
{
"input": "10\n1413472614",
"output": "75333332222222"
},
{
... | 1,636,992,233 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | n = int(input())
A = input()
A= list(A)
A.sort()
a4 = A.count('4')
a8 = A.count('8')
ss = ''
for i in range(n-1,-1,-1):
if A[i] not in ['1','4','8']:
ss += A[i]
ss += '223'*a4 + '2227'*a8
ss = list(ss)
ss.sort()
s = ''.join(ss)
print(s[::-1])
| Title: Drazil and Factorial
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
Output Specification:
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Demo Input:
['4\n1234\n', '3\n555\n']
Demo Output:
['33222\n', '555\n']
Note:
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
n = int(input())
A = input()
A= list(A)
A.sort()
a4 = A.count('4')
a8 = A.count('8')
ss = ''
for i in range(n-1,-1,-1):
if A[i] not in ['1','4','8']:
ss += A[i]
ss += '223'*a4 + '2227'*a8
ss = list(ss)
ss.sort()
s = ''.join(ss)
print(s[::-1])
``` | 0 | |
515 | C | Drazil and Factorial | PROGRAMMING | 1,400 | [
"greedy",
"math",
"sortings"
] | null | null | Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes. | Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. | [
"4\n1234\n",
"3\n555\n"
] | [
"33222\n",
"555\n"
] | In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "4\n1234",
"output": "33222"
},
{
"input": "3\n555",
"output": "555"
},
{
"input": "15\n012345781234578",
"output": "7777553333222222222222"
},
{
"input": "1\n8",
"output": "7222"
},
{
"input": "10\n1413472614",
"output": "75333332222222"
},
{
... | 1,616,581,163 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 0 | def fact(n):
prod1=1
for i in range(2,n+1):
prod1*=i;
return prod1
t=int(input())
prod=1
while t>0:
rem=t%10
prod*=fact(rem)
t=t/10
print(prod) | Title: Drazil and Factorial
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
Output Specification:
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Demo Input:
['4\n1234\n', '3\n555\n']
Demo Output:
['33222\n', '555\n']
Note:
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
def fact(n):
prod1=1
for i in range(2,n+1):
prod1*=i;
return prod1
t=int(input())
prod=1
while t>0:
rem=t%10
prod*=fact(rem)
t=t/10
print(prod)
``` | -1 | |
727 | C | Guess the Array | PROGRAMMING | 1,400 | [
"constructive algorithms",
"interactive",
"math"
] | null | null | This is an interactive problem. You should use flush operation after each printed line. For example, in C++ you should use fflush(stdout), in Java you should use System.out.flush(), and in Pascal — flush(output).
In this problem you should guess an array *a* which is unknown for you. The only information you have initially is the length *n* of the array *a*.
The only allowed action is to ask the sum of two elements by their indices. Formally, you can print two indices *i* and *j* (the indices should be distinct). Then your program should read the response: the single integer equals to *a**i*<=+<=*a**j*.
It is easy to prove that it is always possible to guess the array using at most *n* requests.
Write a program that will guess the array *a* by making at most *n* requests. | none | none | [
"5\n \n9\n \n7\n \n9\n \n11\n \n6\n "
] | [
"? 1 5\n \n? 2 3\n \n? 4 1\n \n? 5 2\n \n? 3 4\n \n! 4 6 1 5 5"
] | The format of a test to make a hack is:
- The first line contains an integer number *n* (3 ≤ *n* ≤ 5000) — the length of the array.- The second line contains *n* numbers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>) — the elements of the array to guess. | 1,500 | [
{
"input": "5\n4 6 1 5 5",
"output": "5 out of 5"
},
{
"input": "3\n1 1 1",
"output": "3 out of 3"
},
{
"input": "4\n100 1 100 1",
"output": "4 out of 4"
},
{
"input": "10\n9 5 10 7 4 4 8 5 10 5",
"output": "10 out of 10"
},
{
"input": "3\n1 1 1",
"output": "3... | 1,612,812,781 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 307,200 | import sys
n = int(input())
print("?" , 1 , n)
sys.stdout.flush()
x = int(input())
print("?" , 1 , n - 1)
sys.stdout.flush()
y = int(input())
print("?" ,n -1, n)
sys.stdout.flush()
z = int(input())
A = [0] * n
A[n-1] = (x- y + z)//2
A[0] = x - A[n - 1]
A[n - 2] = y - A[0]
for i in range(2 , n -1):
print("?" , 1 , i)
sys.stdout.flush()
x = int(input())
A[i- 1] = x - A[0]
print(*A)
| Title: Guess the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This is an interactive problem. You should use flush operation after each printed line. For example, in C++ you should use fflush(stdout), in Java you should use System.out.flush(), and in Pascal — flush(output).
In this problem you should guess an array *a* which is unknown for you. The only information you have initially is the length *n* of the array *a*.
The only allowed action is to ask the sum of two elements by their indices. Formally, you can print two indices *i* and *j* (the indices should be distinct). Then your program should read the response: the single integer equals to *a**i*<=+<=*a**j*.
It is easy to prove that it is always possible to guess the array using at most *n* requests.
Write a program that will guess the array *a* by making at most *n* requests.
Input Specification:
none
Output Specification:
none
Demo Input:
['5\n\xa0\n9\n\xa0\n7\n\xa0\n9\n\xa0\n11\n\xa0\n6\n\xa0']
Demo Output:
['? 1 5\n\xa0\n? 2 3\n\xa0\n? 4 1\n\xa0\n? 5 2\n\xa0\n? 3 4\n\xa0\n! 4 6 1 5 5']
Note:
The format of a test to make a hack is:
- The first line contains an integer number *n* (3 ≤ *n* ≤ 5000) — the length of the array.- The second line contains *n* numbers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>) — the elements of the array to guess. | ```python
import sys
n = int(input())
print("?" , 1 , n)
sys.stdout.flush()
x = int(input())
print("?" , 1 , n - 1)
sys.stdout.flush()
y = int(input())
print("?" ,n -1, n)
sys.stdout.flush()
z = int(input())
A = [0] * n
A[n-1] = (x- y + z)//2
A[0] = x - A[n - 1]
A[n - 2] = y - A[0]
for i in range(2 , n -1):
print("?" , 1 , i)
sys.stdout.flush()
x = int(input())
A[i- 1] = x - A[0]
print(*A)
``` | 0 | |
401 | C | Team | PROGRAMMING | 1,400 | [
"constructive algorithms",
"greedy",
"implementation"
] | null | null | Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork.
For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that:
- there wouldn't be a pair of any side-adjacent cards with zeroes in a row; - there wouldn't be a group of three consecutive cards containing numbers one.
Today Vanya brought *n* cards with zeroes and *m* cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way. | The first line contains two integers: *n* (1<=≤<=*n*<=≤<=106) — the number of cards containing number 0; *m* (1<=≤<=*m*<=≤<=106) — the number of cards containing number 1. | In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1. | [
"1 2\n",
"4 8\n",
"4 10\n",
"1 5\n"
] | [
"101\n",
"110110110101\n",
"11011011011011\n",
"-1\n"
] | none | 1,500 | [
{
"input": "1 2",
"output": "101"
},
{
"input": "4 8",
"output": "110110110101"
},
{
"input": "4 10",
"output": "11011011011011"
},
{
"input": "1 5",
"output": "-1"
},
{
"input": "3 4",
"output": "1010101"
},
{
"input": "3 10",
"output": "-1"
},
... | 1,622,708,970 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 109 | 0 | n,m=map(int,input().split())
if n>=m:
if n-m<=1:
k=0
for i in range(n+m):
if k==0:
print("0",end="")
else:
print("1",end="")
k=k^1
else:
print(-1)
else:
x=m//2
if m%2!=0:
print(-1)
else:
if abs(x-n)>1:
print(-1)
else:
if n>x:
k=0
for i in range(n+x):
if k==0:
print("0",end="")
else:
print("11",end="")
k=k^1
else:
k=1
for i in range(n+x):
if k==1:
print("11",end="")
else:
print("0",end="")
k=k^1
| Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork.
For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that:
- there wouldn't be a pair of any side-adjacent cards with zeroes in a row; - there wouldn't be a group of three consecutive cards containing numbers one.
Today Vanya brought *n* cards with zeroes and *m* cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way.
Input Specification:
The first line contains two integers: *n* (1<=≤<=*n*<=≤<=106) — the number of cards containing number 0; *m* (1<=≤<=*m*<=≤<=106) — the number of cards containing number 1.
Output Specification:
In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1.
Demo Input:
['1 2\n', '4 8\n', '4 10\n', '1 5\n']
Demo Output:
['101\n', '110110110101\n', '11011011011011\n', '-1\n']
Note:
none | ```python
n,m=map(int,input().split())
if n>=m:
if n-m<=1:
k=0
for i in range(n+m):
if k==0:
print("0",end="")
else:
print("1",end="")
k=k^1
else:
print(-1)
else:
x=m//2
if m%2!=0:
print(-1)
else:
if abs(x-n)>1:
print(-1)
else:
if n>x:
k=0
for i in range(n+x):
if k==0:
print("0",end="")
else:
print("11",end="")
k=k^1
else:
k=1
for i in range(n+x):
if k==1:
print("11",end="")
else:
print("0",end="")
k=k^1
``` | 0 | |
349 | A | Cinema Line | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line? | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line. | Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO". | [
"4\n25 25 50 50\n",
"2\n25 100\n",
"4\n50 50 25 25\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "4\n25 25 50 50",
"output": "YES"
},
{
"input": "2\n25 100",
"output": "NO"
},
{
"input": "4\n50 50 25 25",
"output": "NO"
},
{
"input": "3\n25 50 100",
"output": "NO"
},
{
"input": "10\n25 25 25 25 25 25 25 25 25 25",
"output": "YES"
},
{
"... | 1,626,452,418 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 218 | 12,288,000 | n = int(input())
arr = list(map(int, input().split()))
change = 0
cond = True
for el in arr:
dif = el - 25
change -= dif
if change < 0:
cond = False
break
change += 25
if cond:
print('YES')
else:
print('NO') | Title: Cinema Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
Output Specification:
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
Demo Input:
['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
n = int(input())
arr = list(map(int, input().split()))
change = 0
cond = True
for el in arr:
dif = el - 25
change -= dif
if change < 0:
cond = False
break
change += 25
if cond:
print('YES')
else:
print('NO')
``` | 0 | |
1,006 | C | Three Parts of the Array | PROGRAMMING | 1,200 | [
"binary search",
"data structures",
"two pointers"
] | null | null | You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible. | The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$. | Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$). | [
"5\n1 3 1 1 4\n",
"5\n1 3 2 1 4\n",
"3\n4 1 2\n"
] | [
"5\n",
"4\n",
"0\n"
] | In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$. | 0 | [
{
"input": "5\n1 3 1 1 4",
"output": "5"
},
{
"input": "5\n1 3 2 1 4",
"output": "4"
},
{
"input": "3\n4 1 2",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "5\n1 3 5 4 5",
"output": ... | 1,617,432,648 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 11 | 296 | 25,395,200 | n = int(input())
arr = list(map(int, input().split()))
i = 0
j = n - 1
b = 0
while i < j - 1:
if arr[i] < arr[j]:
i += 1
arr[i] += arr[i - 1]
elif arr[j] < arr[i]:
j -= 1
arr[j] += arr[j + 1]
else:
b = arr[i]
i += 1
arr[i] += arr[i - 1]
j -= 1
arr[j] += arr[j + 1]
if arr[i] == arr[j] and i != j:
b = arr[i]
print(b) | Title: Three Parts of the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$.
Output Specification:
Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
Demo Input:
['5\n1 3 1 1 4\n', '5\n1 3 2 1 4\n', '3\n4 1 2\n']
Demo Output:
['5\n', '4\n', '0\n']
Note:
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$. | ```python
n = int(input())
arr = list(map(int, input().split()))
i = 0
j = n - 1
b = 0
while i < j - 1:
if arr[i] < arr[j]:
i += 1
arr[i] += arr[i - 1]
elif arr[j] < arr[i]:
j -= 1
arr[j] += arr[j + 1]
else:
b = arr[i]
i += 1
arr[i] += arr[i - 1]
j -= 1
arr[j] += arr[j + 1]
if arr[i] == arr[j] and i != j:
b = arr[i]
print(b)
``` | 0 | |
322 | A | Ciel and Dancing | PROGRAMMING | 1,000 | [
"greedy"
] | null | null | Fox Ciel and her friends are in a dancing room. There are *n* boys and *m* girls here, and they never danced before. There will be some songs, during each song, there must be exactly one boy and one girl are dancing. Besides, there is a special rule:
- either the boy in the dancing pair must dance for the first time (so, he didn't dance with anyone before); - or the girl in the dancing pair must dance for the first time.
Help Fox Ciel to make a schedule that they can dance as many songs as possible. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of boys and girls in the dancing room. | In the first line print *k* — the number of songs during which they can dance. Then in the following *k* lines, print the indexes of boys and girls dancing during songs chronologically. You can assume that the boys are indexed from 1 to *n*, and the girls are indexed from 1 to *m*. | [
"2 1\n",
"2 2\n"
] | [
"2\n1 1\n2 1\n",
"3\n1 1\n1 2\n2 2\n"
] | In test case 1, there are 2 boys and 1 girl. We can have 2 dances: the 1st boy and 1st girl (during the first song), the 2nd boy and 1st girl (during the second song).
And in test case 2, we have 2 boys with 2 girls, the answer is 3. | 500 | [
{
"input": "2 1",
"output": "2\n1 1\n2 1"
},
{
"input": "2 2",
"output": "3\n1 1\n1 2\n2 2"
},
{
"input": "1 1",
"output": "1\n1 1"
},
{
"input": "2 3",
"output": "4\n1 1\n1 2\n1 3\n2 3"
},
{
"input": "4 4",
"output": "7\n1 1\n1 2\n1 3\n1 4\n4 4\n3 4\n2 4"
}... | 1,632,322,604 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 124 | 6,963,200 | n, m = list(map(int, input().split()))
combinations = set()
boysDanced = set()
girlsDanced = set()
cnt = 0
arr = []
for i in range(1, n+1):
for j in range(1, m+1):
if (i, j) not in combinations:
if i not in boysDanced or j in girlsDanced:
arr.append((i, j))
cnt +=1
combinations.add((i, j))
boysDanced.add(i)
girlsDanced.add(j)
print(cnt)
for b, g in arr:
print(b, g) | Title: Ciel and Dancing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel and her friends are in a dancing room. There are *n* boys and *m* girls here, and they never danced before. There will be some songs, during each song, there must be exactly one boy and one girl are dancing. Besides, there is a special rule:
- either the boy in the dancing pair must dance for the first time (so, he didn't dance with anyone before); - or the girl in the dancing pair must dance for the first time.
Help Fox Ciel to make a schedule that they can dance as many songs as possible.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of boys and girls in the dancing room.
Output Specification:
In the first line print *k* — the number of songs during which they can dance. Then in the following *k* lines, print the indexes of boys and girls dancing during songs chronologically. You can assume that the boys are indexed from 1 to *n*, and the girls are indexed from 1 to *m*.
Demo Input:
['2 1\n', '2 2\n']
Demo Output:
['2\n1 1\n2 1\n', '3\n1 1\n1 2\n2 2\n']
Note:
In test case 1, there are 2 boys and 1 girl. We can have 2 dances: the 1st boy and 1st girl (during the first song), the 2nd boy and 1st girl (during the second song).
And in test case 2, we have 2 boys with 2 girls, the answer is 3. | ```python
n, m = list(map(int, input().split()))
combinations = set()
boysDanced = set()
girlsDanced = set()
cnt = 0
arr = []
for i in range(1, n+1):
for j in range(1, m+1):
if (i, j) not in combinations:
if i not in boysDanced or j in girlsDanced:
arr.append((i, j))
cnt +=1
combinations.add((i, j))
boysDanced.add(i)
girlsDanced.add(j)
print(cnt)
for b, g in arr:
print(b, g)
``` | 0 | |
141 | A | Amusing Joke | PROGRAMMING | 800 | [
"implementation",
"sortings",
"strings"
] | null | null | So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning. | The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100. | Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes. | [
"SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n",
"PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n",
"BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L". | 500 | [
{
"input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS",
"output": "YES"
},
{
"input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI",
"output": "NO"
},
{
"input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER",
"output": "NO"
},
{
"input": "B\nA\nAB",
"output": ... | 1,696,243,171 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | a = list(input())
b = list(input())
c = list(input())
d = a + b
d.sort()
c.sort()
if c == d:
print('yes')
else:
print('no') | Title: Amusing Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
Input Specification:
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Output Specification:
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
Demo Input:
['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L". | ```python
a = list(input())
b = list(input())
c = list(input())
d = a + b
d.sort()
c.sort()
if c == d:
print('yes')
else:
print('no')
``` | 0 | |
721 | B | Passwords | PROGRAMMING | 1,100 | [
"implementation",
"math",
"sortings",
"strings"
] | null | null | Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords. | Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. | [
"5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n",
"4 100\n11\n22\n1\n2\n22\n"
] | [
"1 15\n",
"3 4\n"
] | Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. | 1,000 | [
{
"input": "5 2\ncba\nabc\nbb1\nabC\nABC\nabc",
"output": "1 15"
},
{
"input": "4 100\n11\n22\n1\n2\n22",
"output": "3 4"
},
{
"input": "1 1\na1\na1",
"output": "1 1"
},
{
"input": "1 100\na1\na1",
"output": "1 1"
},
{
"input": "2 1\nabc\nAbc\nAbc",
"output": ... | 1,622,097,211 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 77 | 0 | n, k = map(int,input().split())
l = []
for _ in range(n+1):
l.append(input())
pwd = l[-1]
l.pop()
sl = sorted(l)
tot = 0
smaller_than_pwd = []
for e in sl:
if len(e) < len(pwd):
smaller_than_pwd.append(e)
tot += 1
if len(e) == len(pwd):
tot += 1
worst = ((n-1)//k) * 5 + tot
best = len(smaller_than_pwd) + (len(smaller_than_pwd)//k) * 5 + 1
print(best, end = ' ')
print(worst)
| Title: Passwords
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
Output Specification:
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Demo Input:
['5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n', '4 100\n11\n22\n1\n2\n22\n']
Demo Output:
['1 15\n', '3 4\n']
Note:
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. | ```python
n, k = map(int,input().split())
l = []
for _ in range(n+1):
l.append(input())
pwd = l[-1]
l.pop()
sl = sorted(l)
tot = 0
smaller_than_pwd = []
for e in sl:
if len(e) < len(pwd):
smaller_than_pwd.append(e)
tot += 1
if len(e) == len(pwd):
tot += 1
worst = ((n-1)//k) * 5 + tot
best = len(smaller_than_pwd) + (len(smaller_than_pwd)//k) * 5 + 1
print(best, end = ' ')
print(worst)
``` | 0 | |
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "... | 1,662,904,131 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 248 | 1,945,600 | import math as mt
from collections import defaultdict,deque
from re import I
import sys
from bisect import bisect_right as b_r
from bisect import bisect_left as b_l
from os import path
from heapq import *
input=sys.stdin.readline
mod=1000000007
INT_MAX = sys.maxsize-1
INT_MIN = -sys.maxsize
# if(path.exists('input.txt')):
# sys.stdin = open('input.txt','r')
# sys.stdout = open('output.txt','w')
def myyy__answer():
n,m=map(int,input().split())
i=1
while(m>=i):
m-=i
i=(i%n)+1
# print(i,m)
return m
if __name__ == "__main__":
# for _ in range(int(input())):
print(myyy__answer()) | Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
import math as mt
from collections import defaultdict,deque
from re import I
import sys
from bisect import bisect_right as b_r
from bisect import bisect_left as b_l
from os import path
from heapq import *
input=sys.stdin.readline
mod=1000000007
INT_MAX = sys.maxsize-1
INT_MIN = -sys.maxsize
# if(path.exists('input.txt')):
# sys.stdin = open('input.txt','r')
# sys.stdout = open('output.txt','w')
def myyy__answer():
n,m=map(int,input().split())
i=1
while(m>=i):
m-=i
i=(i%n)+1
# print(i,m)
return m
if __name__ == "__main__":
# for _ in range(int(input())):
print(myyy__answer())
``` | 3.934376 |
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... | 1,525,210,984 | 2,147,483,647 | Python 3 | OK | TESTS | 14 | 77 | 7,065,600 | input()
s = input()
n = 0
if s=="0":
print(s)
else:
for i in s:
if i=="0":
n = n+1
print ("1"+"0"*n) | 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()
s = input()
n = 0
if s=="0":
print(s)
else:
for i in s:
if i=="0":
n = n+1
print ("1"+"0"*n)
``` | 3 | |
581 | A | Vasya the Hipster | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him? | The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got. | Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day. | [
"3 1\n",
"2 3\n",
"7 3\n"
] | [
"1 1\n",
"2 0\n",
"3 2\n"
] | In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | 500 | [
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input":... | 1,679,595,118 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 46 | 0 | a,b=map(int,input().split())
cb=0
cr=0
if a>b:
cb+=b
cr=(a-b)//2
print(cb,cr)
else:
cr+=a
cb=(b-a)//2
print(cr,cb)
| Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | ```python
a,b=map(int,input().split())
cb=0
cr=0
if a>b:
cb+=b
cr=(a-b)//2
print(cb,cr)
else:
cr+=a
cb=(b-a)//2
print(cr,cb)
``` | 3 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,617,290,299 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 122 | 0 | size= input('')
size= size.split(' ')
area= int(size[0])*int(size[1])
print(area/2)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
size= input('')
size= size.split(' ')
area= int(size[0])*int(size[1])
print(area/2)
``` | 0 |
4 | A | Watermelon | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Watermelon | 1 | 64 | One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. | The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. | Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. | [
"8\n"
] | [
"YES\n"
] | For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | 0 | [
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"... | 1,696,526,464 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | def sol(value):
value = int(value)
if !(value % 2):
return "YES"
else:
return "NO" | Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | ```python
def sol(value):
value = int(value)
if !(value % 2):
return "YES"
else:
return "NO"
``` | -1 |
304 | A | Pythagorean Theorem II | PROGRAMMING | 1,200 | [
"brute force",
"math"
] | null | null | In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:
In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
The theorem can be written as an equation relating the lengths of the sides *a*, *b* and *c*, often called the Pythagorean equation:
where *c* represents the length of the hypotenuse, and *a* and *b* represent the lengths of the other two sides.
Given *n*, your task is to count how many right-angled triangles with side-lengths *a*, *b* and *c* that satisfied an inequality 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*. | The only line contains one integer *n* (1<=≤<=*n*<=≤<=104) as we mentioned above. | Print a single integer — the answer to the problem. | [
"5\n",
"74\n"
] | [
"1\n",
"35\n"
] | none | 500 | [
{
"input": "5",
"output": "1"
},
{
"input": "74",
"output": "35"
},
{
"input": "1000",
"output": "881"
},
{
"input": "586",
"output": "472"
},
{
"input": "2",
"output": "0"
},
{
"input": "362",
"output": "258"
},
{
"input": "778",
"outp... | 1,623,990,617 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int count = 0;
for (int i=1; i<n-1; i++){
for (int j=i+1; j<n; j++){
for (int k=j+1; k<n+1; k++){
if(i*i+j*j == k*k){
count++;
}
}
}
}
cout<<count;
}
| Title: Pythagorean Theorem II
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:
In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
The theorem can be written as an equation relating the lengths of the sides *a*, *b* and *c*, often called the Pythagorean equation:
where *c* represents the length of the hypotenuse, and *a* and *b* represent the lengths of the other two sides.
Given *n*, your task is to count how many right-angled triangles with side-lengths *a*, *b* and *c* that satisfied an inequality 1<=≤<=*a*<=≤<=*b*<=≤<=*c*<=≤<=*n*.
Input Specification:
The only line contains one integer *n* (1<=≤<=*n*<=≤<=104) as we mentioned above.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['5\n', '74\n']
Demo Output:
['1\n', '35\n']
Note:
none | ```python
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int count = 0;
for (int i=1; i<n-1; i++){
for (int j=i+1; j<n; j++){
for (int k=j+1; k<n+1; k++){
if(i*i+j*j == k*k){
count++;
}
}
}
}
cout<<count;
}
``` | -1 | |
616 | A | Comparing Two Long Integers | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input(). | The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits. | Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=". | [
"9\n10\n",
"11\n10\n",
"00012345\n12345\n",
"0123\n9\n",
"0123\n111\n"
] | [
"<\n",
">\n",
"=\n",
">\n",
">\n"
] | none | 0 | [
{
"input": "9\n10",
"output": "<"
},
{
"input": "11\n10",
"output": ">"
},
{
"input": "00012345\n12345",
"output": "="
},
{
"input": "0123\n9",
"output": ">"
},
{
"input": "0123\n111",
"output": ">"
},
{
"input": "9\n9",
"output": "="
},
{
... | 1,662,789,931 | 2,147,483,647 | PyPy 3 | OK | TESTS | 120 | 343 | 8,908,800 | import sys
x= sys.stdin.readline().strip()
y= sys.stdin.readline().strip()
ind = 0
while ind <len(x):
if x[ind]!="0":
break
ind+=1
x = x[ind:]
ind = 0
while ind <len(y):
if y[ind]!="0":
break
ind+=1
y = y[ind:]
if len(x)>len(y):
sys.stdout.write(">\n")
elif len(y)>len(x):
sys.stdout.write("<\n")
else:
f = 0
for i in range(len(x)):
if int(x[i]) > int(y[i]):
sys.stdout.write(">\n")
f = 2
break
elif int(y[i]) > int(x[i]):
sys.stdout.write("<\n")
f = 2
break
if f == 0:
sys.stdout.write("=\n") | Title: Comparing Two Long Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input().
Input Specification:
The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits.
Output Specification:
Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=".
Demo Input:
['9\n10\n', '11\n10\n', '00012345\n12345\n', '0123\n9\n', '0123\n111\n']
Demo Output:
['<\n', '>\n', '=\n', '>\n', '>\n']
Note:
none | ```python
import sys
x= sys.stdin.readline().strip()
y= sys.stdin.readline().strip()
ind = 0
while ind <len(x):
if x[ind]!="0":
break
ind+=1
x = x[ind:]
ind = 0
while ind <len(y):
if y[ind]!="0":
break
ind+=1
y = y[ind:]
if len(x)>len(y):
sys.stdout.write(">\n")
elif len(y)>len(x):
sys.stdout.write("<\n")
else:
f = 0
for i in range(len(x)):
if int(x[i]) > int(y[i]):
sys.stdout.write(">\n")
f = 2
break
elif int(y[i]) > int(x[i]):
sys.stdout.write("<\n")
f = 2
break
if f == 0:
sys.stdout.write("=\n")
``` | 3 | |
612 | A | The Text Splitting | PROGRAMMING | 1,300 | [
"brute force",
"implementation",
"strings"
] | null | null | You are given the string *s* of length *n* and the numbers *p*,<=*q*. Split the string *s* to pieces of length *p* and *q*.
For example, the string "Hello" for *p*<==<=2, *q*<==<=3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".
Note it is allowed to split the string *s* to the strings only of length *p* or to the strings only of length *q* (see the second sample test). | The first line contains three positive integers *n*,<=*p*,<=*q* (1<=≤<=*p*,<=*q*<=≤<=*n*<=≤<=100).
The second line contains the string *s* consists of lowercase and uppercase latin letters and digits. | If it's impossible to split the string *s* to the strings of length *p* and *q* print the only number "-1".
Otherwise in the first line print integer *k* — the number of strings in partition of *s*.
Each of the next *k* lines should contain the strings in partition. Each string should be of the length *p* or *q*. The string should be in order of their appearing in string *s* — from left to right.
If there are several solutions print any of them. | [
"5 2 3\nHello\n",
"10 9 5\nCodeforces\n",
"6 4 5\nPrivet\n",
"8 1 1\nabacabac\n"
] | [
"2\nHe\nllo\n",
"2\nCodef\norces\n",
"-1\n",
"8\na\nb\na\nc\na\nb\na\nc\n"
] | none | 0 | [
{
"input": "5 2 3\nHello",
"output": "2\nHe\nllo"
},
{
"input": "10 9 5\nCodeforces",
"output": "2\nCodef\norces"
},
{
"input": "6 4 5\nPrivet",
"output": "-1"
},
{
"input": "8 1 1\nabacabac",
"output": "8\na\nb\na\nc\na\nb\na\nc"
},
{
"input": "1 1 1\n1",
"ou... | 1,451,061,621 | 6,021 | Python 3 | WRONG_ANSWER | TESTS | 5 | 62 | 0 | def main():
n, p, q = map(int, input().split())
s = str(input())
l = []
for i in range(n):
for j in range(i, n):
sl = s[i:j + 1]
if len(sl) == p or len(sl) == q:
l.append(sl)
for i in range(len(l)):
for j in range(i, len(l)):
if l[i] + l[j] == s:
print(2, l[i], l[j], sep="\n")
return
if p == q == 1:
print(n)
for elem in s:
print(elem)
return
print(-1)
if __name__ == "__main__":
main() | Title: The Text Splitting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given the string *s* of length *n* and the numbers *p*,<=*q*. Split the string *s* to pieces of length *p* and *q*.
For example, the string "Hello" for *p*<==<=2, *q*<==<=3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".
Note it is allowed to split the string *s* to the strings only of length *p* or to the strings only of length *q* (see the second sample test).
Input Specification:
The first line contains three positive integers *n*,<=*p*,<=*q* (1<=≤<=*p*,<=*q*<=≤<=*n*<=≤<=100).
The second line contains the string *s* consists of lowercase and uppercase latin letters and digits.
Output Specification:
If it's impossible to split the string *s* to the strings of length *p* and *q* print the only number "-1".
Otherwise in the first line print integer *k* — the number of strings in partition of *s*.
Each of the next *k* lines should contain the strings in partition. Each string should be of the length *p* or *q*. The string should be in order of their appearing in string *s* — from left to right.
If there are several solutions print any of them.
Demo Input:
['5 2 3\nHello\n', '10 9 5\nCodeforces\n', '6 4 5\nPrivet\n', '8 1 1\nabacabac\n']
Demo Output:
['2\nHe\nllo\n', '2\nCodef\norces\n', '-1\n', '8\na\nb\na\nc\na\nb\na\nc\n']
Note:
none | ```python
def main():
n, p, q = map(int, input().split())
s = str(input())
l = []
for i in range(n):
for j in range(i, n):
sl = s[i:j + 1]
if len(sl) == p or len(sl) == q:
l.append(sl)
for i in range(len(l)):
for j in range(i, len(l)):
if l[i] + l[j] == s:
print(2, l[i], l[j], sep="\n")
return
if p == q == 1:
print(n)
for elem in s:
print(elem)
return
print(-1)
if __name__ == "__main__":
main()
``` | 0 | |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
... | 1,675,175,111 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 46 | 0 | k, r = map(int, input().split())
i = 1
while (k * i) % 10 != 0 and (k * i) % 10 != r:
i += 1
print(i) | Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
k, r = map(int, input().split())
i = 1
while (k * i) % 10 != 0 and (k * i) % 10 != r:
i += 1
print(i)
``` | 3 |
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.