message stringlengths 2 59.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
t=int(input())
for i in range(t):
n=input()
print(len(n))
``` | instruction | 0 | 34,485 | 20 | 68,970 |
Yes | output | 1 | 34,485 | 20 | 68,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
import math
t = int(input())
for _ in range(t):
n = input()
n1 = len(n)
print(n1)
``` | instruction | 0 | 34,486 | 20 | 68,972 |
Yes | output | 1 | 34,486 | 20 | 68,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
#while n % 10 == 0:
# n //= 10
print(len(str(n)))
``` | instruction | 0 | 34,487 | 20 | 68,974 |
Yes | output | 1 | 34,487 | 20 | 68,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
testcases = int(input())
for t in range(testcases):
n = input()
for i in n[::-1]:
if i == '0':
n = n[:len(n)-1]
else:
break
print(len(n))
``` | instruction | 0 | 34,488 | 20 | 68,976 |
No | output | 1 | 34,488 | 20 | 68,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
import math
t=int(input())
while t>0 :
n=int(input())
check=True
while check :
if n%10==0 :
n=n/10
else :
check=False
count=int(math.log10(n))+1
print(count)
t-=1
``` | instruction | 0 | 34,489 | 20 | 68,978 |
No | output | 1 | 34,489 | 20 | 68,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
from math import log10
def answer(n):
return 1+int(log10(n))
for T in range(int(input())): print(answer(int(input())))
``` | instruction | 0 | 34,490 | 20 | 68,980 |
No | output | 1 | 34,490 | 20 | 68,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
num=int(input())
for _ in range(num):
string=str(input())
l=len(string)
c=0
for i in range(l-1,0,-1):
if string[i]=="0":
c+=1
else:
break
print(l-c)
``` | instruction | 0 | 34,491 | 20 | 68,982 |
No | output | 1 | 34,491 | 20 | 68,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,540 | 20 | 69,080 |
Tags: greedy, sortings
Correct Solution:
```
n=int(input())
s=input()
z=len(s)//2
s1=list(s[0:len(s)//2])
s2=list(s[len(s)//2:len(s)])
f=0
s1.sort(reverse=True)
s2.sort(reverse=True)
if all(s1[i]>s2[i] for i in range(z)):
f=1
s1.sort()
s2.sort()
if all(s1[i]<s2[i] for i in range(z)):
f=1
print('YES' if f else 'NO')
``` | output | 1 | 34,540 | 20 | 69,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,541 | 20 | 69,082 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
s = input()
l1, l2 = [int(i) for i in s[0:n]], [int(j) for j in s[n:]]
l1.sort()
l2.sort()
f = 0
if l1[0]>l2[0]:
for i in range(1,n):
if l1[i]<=l2[i]:
f=1
break
elif l1[0]<l2[0]:
for i in range(1,n):
if l1[i]>=l2[i]:
f=1
else:
f=1
if f==1:
print('NO')
else:
print('YES')
``` | output | 1 | 34,541 | 20 | 69,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,542 | 20 | 69,084 |
Tags: greedy, sortings
Correct Solution:
```
class CodeforcesTask160BSolution:
def __init__(self):
self.result = ''
self.n = 0
self.ticket = ''
def read_input(self):
self.n = int(input())
self.ticket = input()
def process_task(self):
p1 = [int(x) for x in self.ticket[self.n:]]
p2 = [int(x) for x in self.ticket[:self.n]]
p1.sort()
p2.sort()
bij = True
#print(p1, p2)
for x in range(self.n):
if p1[x] <= p2[x]:
bij = False
break
#print(bij)
if not bij:
bij = True
for x in range(self.n):
if p1[x] >= p2[x]:
bij = False
break
self.result = "YES" if bij else "NO"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask160BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 34,542 | 20 | 69,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,543 | 20 | 69,086 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
s = input()
n1 = [int(i) for i in s[:n]]
n2 = [int(i) for i in s[n:]]
n1.sort()
n2.sort()
for i in range(n):
if n1[i] == n2[i]:
print('NO')
break
if (n1[i] < n2[i]) ^ (n1[0] < n2[0]):
print('NO')
break
else:
print('YES')
``` | output | 1 | 34,543 | 20 | 69,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,544 | 20 | 69,088 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
digits = input()
# split digits into 2
first_half = digits[0:n]
second_half = digits[n:len(digits)]
# sort both halves
sorted_first_half = sorted(first_half)
sorted_second_half = sorted(second_half)
# cycle through both halves simultaneously to see if each digis is greater/less than corresponding
isGreater = True
isLess = True
for i in range(n):
if (sorted_first_half[i] <= sorted_second_half[i]):
isGreater = False
for i in range(n):
if (sorted_first_half[i] >= sorted_second_half[i]):
isLess = False
if(isGreater is True or isLess is True):
print("YES")
else:
print("NO")
``` | output | 1 | 34,544 | 20 | 69,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,545 | 20 | 69,090 |
Tags: greedy, sortings
Correct Solution:
```
n=int(input())
w=input()
z=zip(sorted(w[:n]),sorted(w[n:]))
print('YES' if abs(sum((x>y)-(x<y) for x,y in z))==n else 'NO')
``` | output | 1 | 34,545 | 20 | 69,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,546 | 20 | 69,092 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
num = list(map(int, list(input())))
def parse(num):
mid = len(num) // 2
beg = num[:mid]
end = num[mid:]
beg.sort()
end.sort()
if beg == end:
return "NO"
results = []
for i in range(len(beg)):
if beg[i] == end[i]:
return "NO"
elif beg[i] > end[i]:
results.append(True)
else:
results.append(False)
if len(list(set(results))) != 1:
return "NO"
if len(list(set(results))) == 1:
return "YES"
else:
return "NO"
print(parse(num))
``` | output | 1 | 34,546 | 20 | 69,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO | instruction | 0 | 34,547 | 20 | 69,094 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
tic = input()
a, b = list(tic[:n]), list(tic[n:])
a.sort()
b.sort()
if a[0] > b[0]:
a, b = b, a
for i in range(n):
if a[i] >= b[i]:
print("NO")
break
else:
print("YES")
``` | output | 1 | 34,547 | 20 | 69,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
n = int(input())
b = list(map(int, input()))
f, s = sorted(b[:n]), sorted(b[n:])
point = 0
t = f[0] < s[0]
while point < n and (f[point] < s[point]) == t and f[point] != s[point]:
point += 1
print('YES' if point == n else 'NO')
``` | instruction | 0 | 34,548 | 20 | 69,096 |
Yes | output | 1 | 34,548 | 20 | 69,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
n=int(input())
ticket=[int(x) for x in input()]
first=sorted(ticket[:n])
last=sorted(ticket[n:])
lucky=0
for i in range(n):
if first[i]>last[i]:
lucky+=1
elif first[i]<last[i]:
lucky-=1
if abs(lucky)==n:
print('YES')
else:
print('NO')
``` | instruction | 0 | 34,549 | 20 | 69,098 |
Yes | output | 1 | 34,549 | 20 | 69,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
n = int(input())
x = list(map(int, input()))
a, b = [], []
for i in range(n):
a.append(x[i])
b.append(x[n + i])
a.sort()
b.sort()
if a[0] == b[0]:
print('NO')
else:
if a[0] > b[0]:
for j in range(n):
if a[j] <= b[j]:
print('NO')
break
if j == n - 1:
print('YES')
else:
for j in range(n):
if a[j] >= b[j]:
print('NO')
break
if j == n - 1:
print('YES')
``` | instruction | 0 | 34,550 | 20 | 69,100 |
Yes | output | 1 | 34,550 | 20 | 69,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
n=int(input())
a=input()
b=[]
c=[]
for i in range(2*n):
b.append(int(a[i]))
c.append(int(a[i]))
b_1=b[0:n]
b_2=b[n:]
c_1=c[0:n]
c_2=c[n:]
great_1=0
less_1=0
for i in range(n):
m_1=min(b_1)
m_2=min(b_2)
if m_1 < m_2:
less_1+=1
del b_1[b_1.index(m_1)]
del b_2[b_2.index(m_2)]
else:
break
if less_1==n:
print("YES")
exit()
for i in range(n):
m_1=max(c_1)
m_2=max(c_2)
if m_1 > m_2:
great_1+=1
del c_1[c_1.index(m_1)]
del c_2[c_2.index(m_2)]
else:
break
if great_1 == n:
print("YES")
exit()
print("NO")
``` | instruction | 0 | 34,551 | 20 | 69,102 |
Yes | output | 1 | 34,551 | 20 | 69,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
import math
st=''
def func(n,s):
s1,s2=sorted(s[:n]),sorted(s[n:])
if s1[0]<s2[0]:
for i in range(n):
if s1[i]>s2[i]:
return 'NO'
return 'YES'
elif s1[0]>s2[0]:
for i in range(n):
if s1[i]<s2[i]:
return 'NO'
return 'YES'
else:
return 'NO'
for _ in range(1):#int(input())):
#n,k=map(int,input().split())
n = int(input())
#l1=[]
#inp=input().split()
s=input()
#l1=list(map(int,input().split()))
#l2 = list(map(int, input().split()))
#l1=input().split()
#l2=input().split()
st+=str(func(n,s))+'\n'
print(st)
``` | instruction | 0 | 34,552 | 20 | 69,104 |
No | output | 1 | 34,552 | 20 | 69,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
n = int(input())
num = list(map(int, list(input())))
def parse(num):
mid = len(num) // 2
beg = num[:mid]
end = num[mid:]
beg.sort()
end.sort()
results = []
for i in range(len(beg)):
if len(list(set(results))) != 1:
return "NO"
if beg[i] > end[i]:
results.append(True)
else:
results.append(False)
if len(list(set(results))) == 1:
return "Yes"
else:
return "NO"
print(parse(num))
``` | instruction | 0 | 34,553 | 20 | 69,106 |
No | output | 1 | 34,553 | 20 | 69,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
n=int(input())
number=int(input())
def num(number):
nums=[]
count=0
while number:
val=number%10
nums.append(val)
number//=10
count += 1
if count<n*2:
x=(n*2)-count
while x!=0:
nums.append(0)
x-=1
return nums
nums=num(number)
firsthalf=nums[:n]
sechalf=nums[n:]
firsthalf.sort()
sechalf.sort()
count=0
for i in range(n):
if int(firsthalf[i])>int(sechalf[i]):
count+=1
else:
count-=1
if count==0:
print('NO')
else:
print('YES')
``` | instruction | 0 | 34,554 | 20 | 69,108 |
No | output | 1 | 34,554 | 20 | 69,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Each of you probably has your personal experience of riding public transportation and buying tickets. After a person buys a ticket (which traditionally has an even number of digits), he usually checks whether the ticket is lucky. Let us remind you that a ticket is lucky if the sum of digits in its first half matches the sum of digits in its second half.
But of course, not every ticket can be lucky. Far from it! Moreover, sometimes one look at a ticket can be enough to say right away that the ticket is not lucky. So, let's consider the following unluckiness criterion that can definitely determine an unlucky ticket. We'll say that a ticket is definitely unlucky if each digit from the first half corresponds to some digit from the second half so that each digit from the first half is strictly less than the corresponding digit from the second one or each digit from the first half is strictly more than the corresponding digit from the second one. Each digit should be used exactly once in the comparisons. In other words, there is such bijective correspondence between the digits of the first and the second half of the ticket, that either each digit of the first half turns out strictly less than the corresponding digit of the second half or each digit of the first half turns out strictly more than the corresponding digit from the second half.
For example, ticket 2421 meets the following unluckiness criterion and will not be considered lucky (the sought correspondence is 2 > 1 and 4 > 2), ticket 0135 also meets the criterion (the sought correspondence is 0 < 3 and 1 < 5), and ticket 3754 does not meet the criterion.
You have a ticket in your hands, it contains 2n digits. Your task is to check whether it meets the unluckiness criterion.
Input
The first line contains an integer n (1 ≤ n ≤ 100). The second line contains a string that consists of 2n digits and defines your ticket.
Output
In the first line print "YES" if the ticket meets the unluckiness criterion. Otherwise, print "NO" (without the quotes).
Examples
Input
2
2421
Output
YES
Input
2
0135
Output
YES
Input
2
3754
Output
NO
Submitted Solution:
```
n=int(input())
nums=list(str(input()))
"""def num(number):
nums=[]
while number:
val=number%10
nums.append(val)
number//=10
return nums
nums=num(number)"""
firsthalf=nums[:n]
sechalf=nums[n:]
firsthalf.sort()
sechalf.sort()
count=0
for i in range(n):
if int(firsthalf[i])>int(sechalf[i]):
count+=1
elif int(firsthalf[i])<int(sechalf[i]):
count-=1
if count==0:
print('NO')
else:
print('YES')
``` | instruction | 0 | 34,555 | 20 | 69,110 |
No | output | 1 | 34,555 | 20 | 69,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base b caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length n without leading zeros in this number system. Each page in Nick's notepad has enough space for c numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page.
Input
The only input line contains three space-separated integers b, n and c (2 ≤ b < 10106, 1 ≤ n < 10106, 1 ≤ c ≤ 109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros.
Output
In the only line output the amount of numbers written on the same page as the last number.
Examples
Input
2 3 3
Output
1
Input
2 3 4
Output
4
Note
In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page.
Submitted Solution:
```
b,n,c=map(int,input().split())
x=((b-1)*pow(b,n-1,c))
if x==0:
print(c)
else:
print(x)
``` | instruction | 0 | 34,556 | 20 | 69,112 |
No | output | 1 | 34,556 | 20 | 69,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base b caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length n without leading zeros in this number system. Each page in Nick's notepad has enough space for c numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page.
Input
The only input line contains three space-separated integers b, n and c (2 ≤ b < 10106, 1 ≤ n < 10106, 1 ≤ c ≤ 109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros.
Output
In the only line output the amount of numbers written on the same page as the last number.
Examples
Input
2 3 3
Output
1
Input
2 3 4
Output
4
Note
In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page.
Submitted Solution:
```
def expo(b,c,h):
if(c==1 or c==0):
return pow(b,c)%h
if(c%2==0):
d=expo(b,c/2,h)
return (d*d)%h
else:
return ((b%h)*expo(b,c-1,h))%h
a=list(map(int,input().split()))
k=(((a[0]-1)%a[2])*expo(a[0],a[1]-1,a[2]))%a[2]
if(k!=0):
print(k)
else:
print(a[2])
``` | instruction | 0 | 34,557 | 20 | 69,114 |
No | output | 1 | 34,557 | 20 | 69,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base b caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length n without leading zeros in this number system. Each page in Nick's notepad has enough space for c numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page.
Input
The only input line contains three space-separated integers b, n and c (2 ≤ b < 10106, 1 ≤ n < 10106, 1 ≤ c ≤ 109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros.
Output
In the only line output the amount of numbers written on the same page as the last number.
Examples
Input
2 3 3
Output
1
Input
2 3 4
Output
4
Note
In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def getInt(): return int(input())
def getVars(): return map(int, input().split())
def getList(): return list(map(int, input().split()))
def getStr(): return input().strip()
## -------------------------------
def addDictList(d, key, val):
if key not in d: d[key] = []
d[key].append(val)
def addDictInt(d, key, val):
if key not in d: d[key] = 0
d[key] = val
def addDictCount(d, key):
if key not in d: d[key] = 0
d[key] += 1
def addDictSum(d, key, val):
if key not in d: d[key] = 0
d[key] += val
## -------------------------------
b, n, c = getVars()
ost = (b**n)%c
if c == 0:
ost = c
print(ost)
``` | instruction | 0 | 34,558 | 20 | 69,116 |
No | output | 1 | 34,558 | 20 | 69,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base b caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length n without leading zeros in this number system. Each page in Nick's notepad has enough space for c numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page.
Input
The only input line contains three space-separated integers b, n and c (2 ≤ b < 10106, 1 ≤ n < 10106, 1 ≤ c ≤ 109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros.
Output
In the only line output the amount of numbers written on the same page as the last number.
Examples
Input
2 3 3
Output
1
Input
2 3 4
Output
4
Note
In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page.
Submitted Solution:
```
__author__ = 'Darren'
# The answer is b^(n-1) * (b-1) mod c
# TLE (Python)
def solve():
b, n, c = map(int, input().split())
base, exp = b % c, n - 1
ans, u = 1, base
while exp:
ans = (ans * u ** (exp % 10)) % c
u = u ** 10 % c
exp //= 10
ans = ans * (b - 1) % c
if ans:
print(ans)
else:
print(c)
def solve1():
b, n, c = map(int, input().split())
p = phi(c)
ans = pow_mod(b % c, (n - 1) % p + p, c) * ((b - 1) % c) % c
if ans:
print(ans)
else:
print(c)
def phi(n) -> int:
result = n
x = 2
while x * x <= n:
if n % x == 0:
while n % x == 0:
n //= x
result -= result // x
x += 1
if n > 1:
result -= result // n
return result
def pow_mod(a, n, m):
result = 1
a %= m
while n:
if n & 1:
result = (result * a) % m
a = (a * a) % m
n >>= 1
return result % m
if __name__ == '__main__':
# solve()
solve1()
``` | instruction | 0 | 34,559 | 20 | 69,118 |
No | output | 1 | 34,559 | 20 | 69,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.
Input
The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers li and ri (1 ≤ li ≤ ri ≤ 9 ·1018).
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Output
Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from li to ri, inclusively).
Examples
Input
1
1 9
Output
9
Input
1
12 15
Output
2
Submitted Solution:
```
def digits(num):
a = []
while num != 0:
a.append(num%10)
num /= 10
return a
t = int(input())
for i in range(1,t+1):
b = []
count = 0
m,n = input().split()
m = int(m)
n = int(n)
z = 0
for i in range(m,n+1):
b = digits(i)
for j in b:
if i%j == 0:
z = 1
else:
z = 0
break
if z == 1:
count += 1
print(count)
``` | instruction | 0 | 34,713 | 20 | 69,426 |
No | output | 1 | 34,713 | 20 | 69,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.
Input
The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers li and ri (1 ≤ li ≤ ri ≤ 9 ·1018).
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Output
Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from li to ri, inclusively).
Examples
Input
1
1 9
Output
9
Input
1
12 15
Output
2
Submitted Solution:
```
def BF(x):
temp=x
count=0
while(x>0):
r=x%10
if(r!=0 and (temp%r)==0):
count+=1
x=x//10
if(count==len(str(temp))):
return True
n=int(input())
for i in range(0,n):
l,r=map(int,input().split())
count=0
for i in range(l,r+1):
if(BF(i)):
count+=1
print(count)
``` | instruction | 0 | 34,714 | 20 | 69,428 |
No | output | 1 | 34,714 | 20 | 69,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.
Input
The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers li and ri (1 ≤ li ≤ ri ≤ 9 ·1018).
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Output
Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from li to ri, inclusively).
Examples
Input
1
1 9
Output
9
Input
1
12 15
Output
2
Submitted Solution:
```
t = int(input())
l,r = input().split()
l = int(l)
r = int(r)
count = 0
for i in range(l,r+1):
j = str(i)
j = list(j)
j = [int(x) for x in j]
for k in j:
if k != 0:
if i%k != 0:
break
else:
count += 1
print(count)
``` | instruction | 0 | 34,715 | 20 | 69,430 |
No | output | 1 | 34,715 | 20 | 69,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.
Given a number X, compute <image> modulo 109 + 7.
Input
The first line of input will contain the integer X (1 ≤ X ≤ 10700).
Output
Print a single integer, the answer to the question.
Examples
Input
21
Output
195
Input
345342
Output
390548434
Note
The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195. | instruction | 0 | 34,860 | 20 | 69,720 |
Tags: dp, math
Correct Solution:
```
# Problem G
num = input()
num_list = []
for i in range(len(num)):
num_list.append(int(num[i]))
myMod = (10 ** 9) + 7
length = len(num_list)
f = [0] * (length + 1)
t = [1] * (length + 1)
for i in range(length):
f[i+1] = (f[i] * 10 + 1) % myMod
t[i+1] = (t[i] * 10) % myMod
ans = 0
for i in range(1, 10):
dp = [0] * (length + 1)
for j in range(length):
dp[j+1] = (dp[j] * i + (10 - i) * (dp[j] * 10 + t[j])) % myMod
c = 0
ctr = 0
for k in num_list:
z = min(i, k)
o = k - z
ans += o * (dp[length-1-ctr] * t[c+1] + f[c+1] * t[length-1-ctr]) % myMod
ans += z * (dp[length-1-ctr] * t[c] + f[c] * t[length-1-ctr]) % myMod
ans %= myMod
c += k >= i
ctr += 1
ans += f[c]
if ans >= myMod:
ans -= myMod
print(ans)
``` | output | 1 | 34,860 | 20 | 69,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.
Given a number X, compute <image> modulo 109 + 7.
Input
The first line of input will contain the integer X (1 ≤ X ≤ 10700).
Output
Print a single integer, the answer to the question.
Examples
Input
21
Output
195
Input
345342
Output
390548434
Note
The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195.
Submitted Solution:
```
n=int(input())
sum=0
for i in range(1,n+1):
t=str(i)
m=sorted(t)
g=[str(j) for j in m]
j=int("".join(g))
sum=sum+j
print(sum)
``` | instruction | 0 | 34,861 | 20 | 69,722 |
No | output | 1 | 34,861 | 20 | 69,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.
Given a number X, compute <image> modulo 109 + 7.
Input
The first line of input will contain the integer X (1 ≤ X ≤ 10700).
Output
Print a single integer, the answer to the question.
Examples
Input
21
Output
195
Input
345342
Output
390548434
Note
The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195.
Submitted Solution:
```
x = input()
summ = 0
le = len(str(x))
for i in range(1,int(x)+1):
li = list(str(i))
li.sort()
s = map(str, li)
s = ''.join(s)
s = int(s)
summ+= s
print(summ%(10e9+7))
``` | instruction | 0 | 34,862 | 20 | 69,724 |
No | output | 1 | 34,862 | 20 | 69,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.
Given a number X, compute <image> modulo 109 + 7.
Input
The first line of input will contain the integer X (1 ≤ X ≤ 10700).
Output
Print a single integer, the answer to the question.
Examples
Input
21
Output
195
Input
345342
Output
390548434
Note
The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195.
Submitted Solution:
```
n = int(input())
s = 0
for i in range(1,n+1):
x = list(str(i))
# print(x)
l = sorted(x)
n = int(''.join(map(str,l)))
s += n
print(s)
``` | instruction | 0 | 34,863 | 20 | 69,726 |
No | output | 1 | 34,863 | 20 | 69,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.
Given a number X, compute <image> modulo 109 + 7.
Input
The first line of input will contain the integer X (1 ≤ X ≤ 10700).
Output
Print a single integer, the answer to the question.
Examples
Input
21
Output
195
Input
345342
Output
390548434
Note
The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195.
Submitted Solution:
```
m=int(input())
sum=0
for j in range(m+1):
i=str(j)
r=[]
for p in range(len(i)):
r.append(i[p])
r.sort()
m=''
for p in range(len(r)):
m=m+(r[p])
i=int(m)
sum=sum+i
print(sum)
``` | instruction | 0 | 34,864 | 20 | 69,728 |
No | output | 1 | 34,864 | 20 | 69,729 |
Provide a correct Python 3 solution for this coding contest problem.
Professor Abacus has just built a new computing engine for making numerical tables. It was designed to calculate the values of a polynomial function in one variable at several points at a time. With the polynomial function f(x) = x2 + 2x + 1, for instance, a possible expected calculation result is 1 (= f(0)), 4 (= f(1)), 9 (= f(2)), 16 (= f(3)), and 25 (= f(4)).
It is a pity, however, the engine seemingly has faulty components and exactly one value among those calculated simultaneously is always wrong. With the same polynomial function as above, it can, for instance, output 1, 4, 12, 16, and 25 instead of 1, 4, 9, 16, and 25.
You are requested to help the professor identify the faulty components. As the first step, you should write a program that scans calculation results of the engine and finds the wrong values.
Input
The input is a sequence of datasets, each representing a calculation result in the following format.
d
v0
v1
...
vd+2
Here, d in the first line is a positive integer that represents the degree of the polynomial, namely, the highest exponent of the variable. For instance, the degree of 4x5 + 3x + 0.5 is five and that of 2.4x + 3.8 is one. d is at most five.
The following d + 3 lines contain the calculation result of f(0), f(1), ... , and f(d + 2) in this order, where f is the polynomial function. Each of the lines contains a decimal fraction between -100.0 and 100.0, exclusive.
You can assume that the wrong value, which is exactly one of f(0), f(1), ... , and f(d+2), has an error greater than 1.0. Since rounding errors are inevitable, the other values may also have errors but they are small and never exceed 10-6.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output i in a line when vi is wrong.
Example
Input
2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0
Output
2
1
1
6 | instruction | 0 | 35,100 | 20 | 70,200 |
"Correct Solution:
```
from functools import reduce
EPSILON = 1e-5
d = 0
Vs = []
times = lambda x,y: x*y
def interpolate(x, e):
return sum(
reduce(
times, [(x-j)/(i-j) for j in range(len(Vs)) if j not in [i,x,e]]
)*v for i,v in enumerate(Vs) if i not in [x,e])
def outlier(e):
for i,v in enumerate(Vs):
if i==e: continue
if abs(interpolate(i,e) - v) < EPSILON: return False
return True
def solve():
for i in range(d+3):
if not outlier(i): return i
return -1
while True:
d = int(input())
if d==0: break
Vs = [float(input()) for i in range(d+3)]
print(solve())
``` | output | 1 | 35,100 | 20 | 70,201 |
Provide a correct Python 3 solution for this coding contest problem.
Professor Abacus has just built a new computing engine for making numerical tables. It was designed to calculate the values of a polynomial function in one variable at several points at a time. With the polynomial function f(x) = x2 + 2x + 1, for instance, a possible expected calculation result is 1 (= f(0)), 4 (= f(1)), 9 (= f(2)), 16 (= f(3)), and 25 (= f(4)).
It is a pity, however, the engine seemingly has faulty components and exactly one value among those calculated simultaneously is always wrong. With the same polynomial function as above, it can, for instance, output 1, 4, 12, 16, and 25 instead of 1, 4, 9, 16, and 25.
You are requested to help the professor identify the faulty components. As the first step, you should write a program that scans calculation results of the engine and finds the wrong values.
Input
The input is a sequence of datasets, each representing a calculation result in the following format.
d
v0
v1
...
vd+2
Here, d in the first line is a positive integer that represents the degree of the polynomial, namely, the highest exponent of the variable. For instance, the degree of 4x5 + 3x + 0.5 is five and that of 2.4x + 3.8 is one. d is at most five.
The following d + 3 lines contain the calculation result of f(0), f(1), ... , and f(d + 2) in this order, where f is the polynomial function. Each of the lines contains a decimal fraction between -100.0 and 100.0, exclusive.
You can assume that the wrong value, which is exactly one of f(0), f(1), ... , and f(d+2), has an error greater than 1.0. Since rounding errors are inevitable, the other values may also have errors but they are small and never exceed 10-6.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output i in a line when vi is wrong.
Example
Input
2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0
Output
2
1
1
6 | instruction | 0 | 35,101 | 20 | 70,202 |
"Correct Solution:
```
from itertools import combinations
def build(X, Y):
A = []
for x in X:
res = 1
for xi in X:
if x == xi:
continue
res *= x - xi
A.append(Y[x] / res)
return A
def calc(X, A, x):
base = 1
for xi in X:
base *= i - xi
return sum(base / (i - x) * a for x, a in zip(X, A))
while 1:
d = int(input())
if d == 0:
break
N = d+3
Y = [float(input()) for i in range(N)]
cnts = [0]*N
for X in combinations(range(N), d+1):
U = [0]*N
for x in X:
U[x] = 1
A = build(X, Y)
for i in range(N):
if U[i]:
continue
res = calc(X, A, i)
if abs(Y[i] - res) > 0.5:
cnts[i] += 1
print(cnts.index(max(cnts)))
``` | output | 1 | 35,101 | 20 | 70,203 |
Provide a correct Python 3 solution for this coding contest problem.
Professor Abacus has just built a new computing engine for making numerical tables. It was designed to calculate the values of a polynomial function in one variable at several points at a time. With the polynomial function f(x) = x2 + 2x + 1, for instance, a possible expected calculation result is 1 (= f(0)), 4 (= f(1)), 9 (= f(2)), 16 (= f(3)), and 25 (= f(4)).
It is a pity, however, the engine seemingly has faulty components and exactly one value among those calculated simultaneously is always wrong. With the same polynomial function as above, it can, for instance, output 1, 4, 12, 16, and 25 instead of 1, 4, 9, 16, and 25.
You are requested to help the professor identify the faulty components. As the first step, you should write a program that scans calculation results of the engine and finds the wrong values.
Input
The input is a sequence of datasets, each representing a calculation result in the following format.
d
v0
v1
...
vd+2
Here, d in the first line is a positive integer that represents the degree of the polynomial, namely, the highest exponent of the variable. For instance, the degree of 4x5 + 3x + 0.5 is five and that of 2.4x + 3.8 is one. d is at most five.
The following d + 3 lines contain the calculation result of f(0), f(1), ... , and f(d + 2) in this order, where f is the polynomial function. Each of the lines contains a decimal fraction between -100.0 and 100.0, exclusive.
You can assume that the wrong value, which is exactly one of f(0), f(1), ... , and f(d+2), has an error greater than 1.0. Since rounding errors are inevitable, the other values may also have errors but they are small and never exceed 10-6.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output i in a line when vi is wrong.
Example
Input
2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0
Output
2
1
1
6 | instruction | 0 | 35,102 | 20 | 70,204 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f(n):
a = [F() for _ in range(n+3)]
kd = {}
for i in range(n+3):
k = [i**c for c in range(n,-1,-1)]
kd[(i,)] = [a[i], k]
for _ in range(n):
nd = {}
kl = sorted(list(kd.keys()))
for i in range(len(kl)):
k1 = kl[i]
t1 = kd[k1]
if t1[-1][-1] == 0:
nd[k1] = [t1[0], t1[1][:-1]]
continue
for j in range(i+1,len(kl)):
k2 = kl[j]
sf = 0
for c in k1:
if c not in k2:
sf += 1
if sf != 1:
continue
t2 = kd[k2]
if t2[-1][-1] == 0:
continue
w = t2[-1][-1] / t1[-1][-1]
mt = t2[0] - t1[0] * w
ma = [c2 - c1*w for c1,c2 in zip(t1[-1][:-1],t2[-1][:-1])]
mk = tuple(sorted(set(k1) | set(k2)))
nd[mk] = [mt, ma]
kd = nd
msa = inf
msi = -1
kl = sorted(list(kd.keys()))
# for k in kl:
# print(k, kd[k], kd[k][0] / kd[k][1][0] if abs(kd[k][1][0]) > eps else 'none')
for i in range(n+3):
mi = inf
ma = -inf
for k in kl:
if i in k or abs(kd[k][1][0]) < eps:
# print('not int', i, k)
continue
t = kd[k][0] / kd[k][1][0]
# print('in', i,t,k)
if mi > t:
mi = t
if ma < t:
ma = t
sa = ma - mi
# print(i,ma,mi,sa)
if msa > sa:
msa = sa
msi = i
return msi
while 1:
n = I()
if n == 0:
break
rr.append(f(n))
# print('rr', rr[-1])
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 35,102 | 20 | 70,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Abacus has just built a new computing engine for making numerical tables. It was designed to calculate the values of a polynomial function in one variable at several points at a time. With the polynomial function f(x) = x2 + 2x + 1, for instance, a possible expected calculation result is 1 (= f(0)), 4 (= f(1)), 9 (= f(2)), 16 (= f(3)), and 25 (= f(4)).
It is a pity, however, the engine seemingly has faulty components and exactly one value among those calculated simultaneously is always wrong. With the same polynomial function as above, it can, for instance, output 1, 4, 12, 16, and 25 instead of 1, 4, 9, 16, and 25.
You are requested to help the professor identify the faulty components. As the first step, you should write a program that scans calculation results of the engine and finds the wrong values.
Input
The input is a sequence of datasets, each representing a calculation result in the following format.
d
v0
v1
...
vd+2
Here, d in the first line is a positive integer that represents the degree of the polynomial, namely, the highest exponent of the variable. For instance, the degree of 4x5 + 3x + 0.5 is five and that of 2.4x + 3.8 is one. d is at most five.
The following d + 3 lines contain the calculation result of f(0), f(1), ... , and f(d + 2) in this order, where f is the polynomial function. Each of the lines contains a decimal fraction between -100.0 and 100.0, exclusive.
You can assume that the wrong value, which is exactly one of f(0), f(1), ... , and f(d+2), has an error greater than 1.0. Since rounding errors are inevitable, the other values may also have errors but they are small and never exceed 10-6.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output i in a line when vi is wrong.
Example
Input
2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0
Output
2
1
1
6
Submitted Solution:
```
import numpy as np
``` | instruction | 0 | 35,103 | 20 | 70,206 |
No | output | 1 | 35,103 | 20 | 70,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Abacus has just built a new computing engine for making numerical tables. It was designed to calculate the values of a polynomial function in one variable at several points at a time. With the polynomial function f(x) = x2 + 2x + 1, for instance, a possible expected calculation result is 1 (= f(0)), 4 (= f(1)), 9 (= f(2)), 16 (= f(3)), and 25 (= f(4)).
It is a pity, however, the engine seemingly has faulty components and exactly one value among those calculated simultaneously is always wrong. With the same polynomial function as above, it can, for instance, output 1, 4, 12, 16, and 25 instead of 1, 4, 9, 16, and 25.
You are requested to help the professor identify the faulty components. As the first step, you should write a program that scans calculation results of the engine and finds the wrong values.
Input
The input is a sequence of datasets, each representing a calculation result in the following format.
d
v0
v1
...
vd+2
Here, d in the first line is a positive integer that represents the degree of the polynomial, namely, the highest exponent of the variable. For instance, the degree of 4x5 + 3x + 0.5 is five and that of 2.4x + 3.8 is one. d is at most five.
The following d + 3 lines contain the calculation result of f(0), f(1), ... , and f(d + 2) in this order, where f is the polynomial function. Each of the lines contains a decimal fraction between -100.0 and 100.0, exclusive.
You can assume that the wrong value, which is exactly one of f(0), f(1), ... , and f(d+2), has an error greater than 1.0. Since rounding errors are inevitable, the other values may also have errors but they are small and never exceed 10-6.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output i in a line when vi is wrong.
Example
Input
2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0
Output
2
1
1
6
Submitted Solution:
```
from functools import reduce
EPSILON = 2e-6
d = 0
Vs = []
times = lambda x,y: x*y
def interpolate(x, e):
return sum(
reduce(
times, [(x-j)/(i-j) for j in range(len(Vs)) if j not in [i,x,e]]
)*v for i,v in enumerate(Vs) if i not in [x,e])
def outlier(e):
for i,v in enumerate(Vs):
if i==e: continue
if abs(interpolate(i,e) - v) < EPSILON: return False
return True
def solve():
for i in range(d+3):
if not outlier(i): return i
return -1
while True:
d = int(input())
if d==0: break
Vs = [float(input()) for i in range(d+3)]
print(solve())
``` | instruction | 0 | 35,104 | 20 | 70,208 |
No | output | 1 | 35,104 | 20 | 70,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC's math teacher always gives the class some interesting math problems so that they don't get bored. Today the problem is as follows. Given an integer n, you can perform the following operations zero or more times:
* mul x: multiplies n by x (where x is an arbitrary positive integer).
* sqrt: replaces n with √{n} (to apply this operation, √{n} must be an integer).
You can perform these operations as many times as you like. What is the minimum value of n, that can be achieved and what is the minimum number of operations, to achieve that minimum value?
Apparently, no one in the class knows the answer to this problem, maybe you can help them?
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^6) — the initial number.
Output
Print two integers: the minimum integer n that can be achieved using the described operations and the minimum number of operations required.
Examples
Input
20
Output
10 2
Input
5184
Output
6 4
Note
In the first example, you can apply the operation mul 5 to get 100 and then sqrt to get 10.
In the second example, you can first apply sqrt to get 72, then mul 18 to get 1296 and finally two more sqrt and you get 6.
Note, that even if the initial value of n is less or equal 10^6, it can still become greater than 10^6 after applying one or more operations. | instruction | 0 | 35,155 | 20 | 70,310 |
Tags: greedy, math, number theory
Correct Solution:
```
from math import ceil, log
d=dict()
def pf(n):
i=2
while(i*i<= n):
while(n%i==0):
d[i]=d.get(i,0)+1
n=n//i
i+=1
if(n!=1):
d[n]=d.get(n,0)+1
n=int(input())
if(n==1):
print(1,0)
else:
pf(n)
m= max(d.values())
steps=0
flag=False
for i in d.values():
if( i!=m):
steps+=1
flag=True
break
x= log(m,2)
if(flag or x.is_integer()):
steps+=ceil(x)
else:
steps+=ceil(x)+1
ans=1
for i in d.keys():
ans = ans * i
print(ans,steps)
``` | output | 1 | 35,155 | 20 | 70,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC's math teacher always gives the class some interesting math problems so that they don't get bored. Today the problem is as follows. Given an integer n, you can perform the following operations zero or more times:
* mul x: multiplies n by x (where x is an arbitrary positive integer).
* sqrt: replaces n with √{n} (to apply this operation, √{n} must be an integer).
You can perform these operations as many times as you like. What is the minimum value of n, that can be achieved and what is the minimum number of operations, to achieve that minimum value?
Apparently, no one in the class knows the answer to this problem, maybe you can help them?
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^6) — the initial number.
Output
Print two integers: the minimum integer n that can be achieved using the described operations and the minimum number of operations required.
Examples
Input
20
Output
10 2
Input
5184
Output
6 4
Note
In the first example, you can apply the operation mul 5 to get 100 and then sqrt to get 10.
In the second example, you can first apply sqrt to get 72, then mul 18 to get 1296 and finally two more sqrt and you get 6.
Note, that even if the initial value of n is less or equal 10^6, it can still become greater than 10^6 after applying one or more operations. | instruction | 0 | 35,157 | 20 | 70,314 |
Tags: greedy, math, number theory
Correct Solution:
```
from math import log
from math import ceil
A=int(input())
a=A
m=[]
i=2
def Z(x):
return round(x)==x
b=round(a**0.5 + 1)
while i<=b:
if a%i==0:
m+=[[i,0],]
while a%i==0:
m[-1][1]+=1
a//=i
i+=1
if a!=1:
m+=[[a,1],]
f=True
i=0
while i<len(m):
if m[i][1]!=1:
f=False
i+=1
if f:
print(A,0)
else:
t=True
p=1
i=0
while i<len(m):
p*=m[i][0]
i+=1
print(p,end=' ')
f=Z(log(m[0][1],2))
if f:
i = 1
while i < len(m):
if m[i][1]!=m[0][1]:
f=False
i+=1
if f:
print(round(log(m[0][1],2)))
t = False
if t:
maxm1=max(m,key=lambda x:x[1])[1]
print(1+ceil(log(maxm1,2)))
``` | output | 1 | 35,157 | 20 | 70,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
JATC's math teacher always gives the class some interesting math problems so that they don't get bored. Today the problem is as follows. Given an integer n, you can perform the following operations zero or more times:
* mul x: multiplies n by x (where x is an arbitrary positive integer).
* sqrt: replaces n with √{n} (to apply this operation, √{n} must be an integer).
You can perform these operations as many times as you like. What is the minimum value of n, that can be achieved and what is the minimum number of operations, to achieve that minimum value?
Apparently, no one in the class knows the answer to this problem, maybe you can help them?
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^6) — the initial number.
Output
Print two integers: the minimum integer n that can be achieved using the described operations and the minimum number of operations required.
Examples
Input
20
Output
10 2
Input
5184
Output
6 4
Note
In the first example, you can apply the operation mul 5 to get 100 and then sqrt to get 10.
In the second example, you can first apply sqrt to get 72, then mul 18 to get 1296 and finally two more sqrt and you get 6.
Note, that even if the initial value of n is less or equal 10^6, it can still become greater than 10^6 after applying one or more operations. | instruction | 0 | 35,160 | 20 | 70,320 |
Tags: greedy, math, number theory
Correct Solution:
```
import sys
from math import sqrt, log, ceil
input_file = sys.stdin
n = int(input_file.readline())
def factor(n):
lst = []
prod = 1
for i in range(2, n+1):
if n % i == 0:
prod *= i
lst.append(0)
while n % i == 0:
lst[-1] += 1
n /= i
if n < i:
break
return lst, prod
if n == 1:
print(1, 0)
else:
lst, ans = factor(n)
maxi, mini = max(lst), min(lst)
if maxi == mini and log(maxi, 2) == int(log(maxi, 2)):
print(ans, int(log(maxi, 2)))
else:
print(ans, int(ceil(log(maxi, 2)) + 1))
``` | output | 1 | 35,160 | 20 | 70,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading zeroes. Let's call it d. The numeration starts with 1, starting from the most significant digit. Petya wants to perform the following operation k times: find the minimum x (1 ≤ x < n) such that dx = 4 and dx + 1 = 7, if x is odd, then to assign dx = dx + 1 = 4, otherwise to assign dx = dx + 1 = 7. Note that if no x was found, then the operation counts as completed and the array doesn't change at all.
You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of digits d, starting with d1. It is guaranteed that the first digit of the number does not equal zero.
Output
In the single line print the result without spaces — the number after the k operations are fulfilled.
Examples
Input
7 4
4727447
Output
4427477
Input
4 2
4478
Output
4478
Note
In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.
In the second sample: 4478 → 4778 → 4478. | instruction | 0 | 35,252 | 20 | 70,504 |
Tags: strings
Correct Solution:
```
"""
Author - Satwik Tiwari .
2nd NOV , 2020 - Monday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
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 graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def solve(case):
n,m = sep()
a = list(inp())
for i in range(n):
a[i] = int(a[i])
# ind = -1
# for i in range(n-2):
# if(a[i] == 4 and a[i+1] == 4 and a[i+2] == 7):
# ind = i
# break
for i in range(n-1):
if(m==0):
break
if(a[i] == 4 and a[i+1] == 7):
if(i>0 and a[i-1] == 4 and i%2==1):
if(m%2==1):
a[i] = 7
break
if(i%2==0):
a[i+1] = 4
m-=1
else:
a[i] = 7
m-=1
print(''.join(str(a[i]) for i in range(n)))
testcase(1)
# testcase(int(inp()))
``` | output | 1 | 35,252 | 20 | 70,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading zeroes. Let's call it d. The numeration starts with 1, starting from the most significant digit. Petya wants to perform the following operation k times: find the minimum x (1 ≤ x < n) such that dx = 4 and dx + 1 = 7, if x is odd, then to assign dx = dx + 1 = 4, otherwise to assign dx = dx + 1 = 7. Note that if no x was found, then the operation counts as completed and the array doesn't change at all.
You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of digits d, starting with d1. It is guaranteed that the first digit of the number does not equal zero.
Output
In the single line print the result without spaces — the number after the k operations are fulfilled.
Examples
Input
7 4
4727447
Output
4427477
Input
4 2
4478
Output
4478
Note
In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.
In the second sample: 4478 → 4778 → 4478. | instruction | 0 | 35,253 | 20 | 70,506 |
Tags: strings
Correct Solution:
```
# 447 or 477 start with odd
n, k = map(int, input().split())
s = input()
s = list(s)
for i in range(n - 1):
if not k:
break
if s[i] != '4':
continue
tt = ''.join(s[i:i + 3])
if tt in ('447', '477') and i % 2 == 0:
if k % 2 == 1 and tt == '447':
s[i + 1] = '7'
if k % 2 == 1 and tt == '477':
s[i + 1] = '4'
break
if s[i] == '4' and s[i + 1] == '7':
if i % 2 == 0:
s[i + 1] = '4'
else:
s[i] = '7'
k -= 1
print(''.join(s))
``` | output | 1 | 35,253 | 20 | 70,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading zeroes. Let's call it d. The numeration starts with 1, starting from the most significant digit. Petya wants to perform the following operation k times: find the minimum x (1 ≤ x < n) such that dx = 4 and dx + 1 = 7, if x is odd, then to assign dx = dx + 1 = 4, otherwise to assign dx = dx + 1 = 7. Note that if no x was found, then the operation counts as completed and the array doesn't change at all.
You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of digits d, starting with d1. It is guaranteed that the first digit of the number does not equal zero.
Output
In the single line print the result without spaces — the number after the k operations are fulfilled.
Examples
Input
7 4
4727447
Output
4427477
Input
4 2
4478
Output
4478
Note
In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.
In the second sample: 4478 → 4778 → 4478. | instruction | 0 | 35,254 | 20 | 70,508 |
Tags: strings
Correct Solution:
```
n, k = map(int, input().split())
s = list(input())
cnt = 0
for i in range(n - 1):
if cnt == k:
break
if s[i] == '4' and s[i + 1] == '7':
if i & 1:
if s[i - 1] == '4':
if (cnt - k) & 1:
s[i] = '7'
print(''.join(s))
exit()
s[i] = '7'
else:
if i != n - 2:
if s[i + 2] == '7':
if (cnt - k) & 1:
s[i + 1] = '4'
print(''.join(s))
exit()
s[i + 1] = '4'
cnt += 1
print(''.join(s))
``` | output | 1 | 35,254 | 20 | 70,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading zeroes. Let's call it d. The numeration starts with 1, starting from the most significant digit. Petya wants to perform the following operation k times: find the minimum x (1 ≤ x < n) such that dx = 4 and dx + 1 = 7, if x is odd, then to assign dx = dx + 1 = 4, otherwise to assign dx = dx + 1 = 7. Note that if no x was found, then the operation counts as completed and the array doesn't change at all.
You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of digits d, starting with d1. It is guaranteed that the first digit of the number does not equal zero.
Output
In the single line print the result without spaces — the number after the k operations are fulfilled.
Examples
Input
7 4
4727447
Output
4427477
Input
4 2
4478
Output
4478
Note
In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.
In the second sample: 4478 → 4778 → 4478. | instruction | 0 | 35,255 | 20 | 70,510 |
Tags: strings
Correct Solution:
```
n, m = map(int, input().split())
luckstring = list(input())
start = 0
idx = 0
while m:
idx = start
while idx < len(luckstring) - 1:
if luckstring[idx] == '4' and luckstring[idx + 1] == '7':
break
idx += 1
if idx == len(luckstring) - 1:
break
if (idx & 1) == 0:
if(idx + 2 < len(luckstring) and luckstring[idx+2] == '7'):
if (m & 1) == 1:
luckstring[idx+1] = '4'
m = 0
break
luckstring[idx + 1]='4'
start = idx + 1
else:
if(idx - 1 >= 0 and luckstring[idx - 1] =='4'):
if(m & 1) == 1:
luckstring[idx] = '7'
m = 0
break
luckstring[idx] = '7'
if idx > 0:
start = idx - 1
else:
start = idx + 2
m -= 1
print("".join(luckstring))
``` | output | 1 | 35,255 | 20 | 70,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading zeroes. Let's call it d. The numeration starts with 1, starting from the most significant digit. Petya wants to perform the following operation k times: find the minimum x (1 ≤ x < n) such that dx = 4 and dx + 1 = 7, if x is odd, then to assign dx = dx + 1 = 4, otherwise to assign dx = dx + 1 = 7. Note that if no x was found, then the operation counts as completed and the array doesn't change at all.
You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of digits d, starting with d1. It is guaranteed that the first digit of the number does not equal zero.
Output
In the single line print the result without spaces — the number after the k operations are fulfilled.
Examples
Input
7 4
4727447
Output
4427477
Input
4 2
4478
Output
4478
Note
In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.
In the second sample: 4478 → 4778 → 4478. | instruction | 0 | 35,256 | 20 | 70,512 |
Tags: strings
Correct Solution:
```
n, k = map(int, input().split())
s = [i for i in input()]
flag = False
for i in range(len(s)-1):
if k == 0:
break
if s[i] == "4" and s[i+1] == "7":
if i % 2 == 0:
s[i] = "4"
s[i+1] = "4"
k -= 1
else:
if i != 0 and s[i-1] == "4":
k %= 2
if k == 1:
s[i] = "7"
k -= 1
else:
s[i] = "7"
s[i+1] = "7"
k -= 1
print(*s, sep="")
``` | output | 1 | 35,256 | 20 | 70,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading zeroes. Let's call it d. The numeration starts with 1, starting from the most significant digit. Petya wants to perform the following operation k times: find the minimum x (1 ≤ x < n) such that dx = 4 and dx + 1 = 7, if x is odd, then to assign dx = dx + 1 = 4, otherwise to assign dx = dx + 1 = 7. Note that if no x was found, then the operation counts as completed and the array doesn't change at all.
You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of digits d, starting with d1. It is guaranteed that the first digit of the number does not equal zero.
Output
In the single line print the result without spaces — the number after the k operations are fulfilled.
Examples
Input
7 4
4727447
Output
4427477
Input
4 2
4478
Output
4478
Note
In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.
In the second sample: 4478 → 4778 → 4478. | instruction | 0 | 35,257 | 20 | 70,514 |
Tags: strings
Correct Solution:
```
import sys
input=sys.stdin.readline
from math import *
n,m=map(int,input().split())
s=list(input().rstrip())
for i in range(n-1):
if m==0:
break
if i>0:
if s[i-1]=='4' and s[i]=='4' and s[i+1]=='7' and i%2==1:
if m%2==1:
s[i]='7'
break
if s[i]=='4' and s[i+1]=='7':
if i%2==0:
s[i]='4'
s[i+1]='4'
else:
s[i]='7'
m-=1
#print(s)
print("".join(s))
``` | output | 1 | 35,257 | 20 | 70,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading zeroes. Let's call it d. The numeration starts with 1, starting from the most significant digit. Petya wants to perform the following operation k times: find the minimum x (1 ≤ x < n) such that dx = 4 and dx + 1 = 7, if x is odd, then to assign dx = dx + 1 = 4, otherwise to assign dx = dx + 1 = 7. Note that if no x was found, then the operation counts as completed and the array doesn't change at all.
You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of digits d, starting with d1. It is guaranteed that the first digit of the number does not equal zero.
Output
In the single line print the result without spaces — the number after the k operations are fulfilled.
Examples
Input
7 4
4727447
Output
4427477
Input
4 2
4478
Output
4478
Note
In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.
In the second sample: 4478 → 4778 → 4478. | instruction | 0 | 35,258 | 20 | 70,516 |
Tags: strings
Correct Solution:
```
from math import *
from collections import deque
from copy import deepcopy
import sys
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def multi(): return map(int,input().split())
def strmulti(): return map(str, inp().split())
def lis(): return list(map(int, inp().split()))
def lcm(a,b): return (a*b)//gcd(a,b)
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def stringlis(): return list(map(str, inp().split()))
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def printlist(a) :
print(' '.join(str(a[i]) for i in range(len(a))))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#copied functions end
#start coding
n,k=multi()
s=inp()
a=list(s)
count=0
for i in range(1,len(a)):
if(i>1):
if(a[i-2]=='4' and (a[i-1]=='4' and a[i]=='7' and count<k)):
if((k-count)%2!=0):
if(i%2==0):
a[i-2]='4'
a[i-1]='7'
a[i]='7'
break
if(i%2==0):
break
if(a[i-1]=='4' and a[i]=='7' and count<k):
count+=1
if((i)%2!=0):
a[i-1]=a[i]='4'
else:
a[i-1]=a[i]='7'
continue
for i in a:
print(i,end="")
``` | output | 1 | 35,258 | 20 | 70,517 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.