message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He has n balls, the i-th ball has size ti.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.
Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Examples
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output
YES
Note
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
Submitted Solution:
```
n = int(input())
t = list(set(map(int, input().split())))
t.sort()
l = len(t)
if l < 3:
print("NO")
exit()
f = t[0]
s = t[1]
tr = False
for i in range(2, len(t)):
if t[i]-f <= 2:
tr = True
break
else:
f = s
s = t[i]
if tr:
print("YES")
else:
print("NO")
``` | instruction | 0 | 97,465 | 14 | 194,930 |
Yes | output | 1 | 97,465 | 14 | 194,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He has n balls, the i-th ball has size ti.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.
Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Examples
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output
YES
Note
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
Submitted Solution:
```
#!/usr/bin/env python3
import sys
n = int(input())
t = sorted(list(set([int(x) for x in input().split()])))
for i in range(0, len(t) - 2):
if t[i+1] - t[i] == 1 and t[i+2] - t[i+1] == 1:
print('YES')
sys.exit(0)
print('NO')
``` | instruction | 0 | 97,466 | 14 | 194,932 |
Yes | output | 1 | 97,466 | 14 | 194,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He has n balls, the i-th ball has size ti.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.
Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Examples
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output
YES
Note
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
Submitted Solution:
```
_ = input()
balls = sorted(list(set(map(lambda x: int(x), input().split()))))
for i in range(len(balls) - 2):
a = balls[i]
b = balls[i + 2]
if b - a <= 2:
print('YES')
break
else:
print('NO')
``` | instruction | 0 | 97,467 | 14 | 194,934 |
Yes | output | 1 | 97,467 | 14 | 194,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He has n balls, the i-th ball has size ti.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.
Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Examples
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output
YES
Note
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
Submitted Solution:
```
if __name__=='__main__':
n = int(input())
b = list(map(int,input().split(' ')))
b.sort()
pos = False
for i in range(len(b)-2):
if len(set(b[i:i+3]))==3:
if b[i+2]-b[i]<=2 and b[i+2]-b[i+1]<=1:
pos = True
if pos: break
if pos: print('YES')
else: print('NO')
``` | instruction | 0 | 97,468 | 14 | 194,936 |
No | output | 1 | 97,468 | 14 | 194,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He has n balls, the i-th ball has size ti.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.
Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Examples
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output
YES
Note
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
Submitted Solution:
```
c=1; n=int(input()); l=sorted(map(int,input().split()))
for i in range(n-1):
if l[i+1]-l[i]==1:c+=1
else:c=1
if c==3:break
print(["NO","YES"][c==3])
``` | instruction | 0 | 97,469 | 14 | 194,938 |
No | output | 1 | 97,469 | 14 | 194,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He has n balls, the i-th ball has size ti.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.
Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Examples
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output
YES
Note
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
Submitted Solution:
```
from collections import defaultdict, deque, Counter, OrderedDict
def main():
n = int(input())
l = sorted([int(i) for i in input().split()])
check = False
for i in range(2,n):
a, b, c = l[i-2], l[i-1], l[i]
check |= (a != b != c and b-a < 3 and c-a < 3)
print("YES" if check else "NO")
if __name__ == "__main__":
"""sys.setrecursionlimit(400000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()"""
main()
``` | instruction | 0 | 97,470 | 14 | 194,940 |
No | output | 1 | 97,470 | 14 | 194,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He has n balls, the i-th ball has size ti.
Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:
* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).
Your task is to check whether Limak can choose three balls that satisfy conditions above.
Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.
Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).
Examples
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output
YES
Note
In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes 18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
Submitted Solution:
```
i = int(input())
l = sorted(list(map(int, input().split())))
for x in range(0,i-2):
if l[x]+1 == l[x+1]:
if l[x] + 2 == l[x + 2]:
print('YES')
exit()
print('NO')
``` | instruction | 0 | 97,471 | 14 | 194,942 |
No | output | 1 | 97,471 | 14 | 194,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | instruction | 0 | 98,312 | 14 | 196,624 |
Tags: dp, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n - 1):
b.append(abs(a[i] - a[i + 1]))
c = []
s = 1
summ = 0
for i in range(n - 1):
summ += s * b[i]
s = -s
c.append(summ)
c.sort()
if c[0] < 0:
print(c[n - 2] - c[0])
else:
print(c[n - 2])
``` | output | 1 | 98,312 | 14 | 196,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | instruction | 0 | 98,313 | 14 | 196,626 |
Tags: dp, two pointers
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
da,p=[],1
for i in range(n-1):
da.append(p*abs(a[i]-a[i+1]));p*=-1
m1,m2,s1,s2=0,0,0,0
for x in da:
s1+=x
if s1<0: s1=0
s2-=x
if s2<0: s2=0
m1=max(m1,s1);m2=max(m2,s2)
print(max(m1,m2))
``` | output | 1 | 98,313 | 14 | 196,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | instruction | 0 | 98,316 | 14 | 196,632 |
Tags: dp, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
b = []
sign = 1
for i in range(len(a)-1):
b.append( abs(a[i]-a[i+1]) * sign )
sign *= -1
max_ = 0
max1, max2 = 0, 0
a1, a2 = 0, 0
for i in range(n-1):
if a1+b[i]>0:
a1 += b[i]
else:
a1 = 0
max1 = max(a1, max1)
if a2-b[i]>0:
a2 -= b[i]
else:
a2 = 0
max2 = max(a2, max2)
print(max(max1,max2))
``` | output | 1 | 98,316 | 14 | 196,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | instruction | 0 | 98,317 | 14 | 196,634 |
Tags: dp, two pointers
Correct Solution:
```
def maxSubArray( A):
if not A:
return 0
curSum = maxSum = A[0]
for num in A[1:]:
curSum = max(num, curSum + num)
maxSum = max(maxSum, curSum)
return maxSum
n = int(input())
L = list(map(int, input().split()))
D1 = []
D2 = []
for i in range(1, len(L)):
D1.append(abs(L[i] - L[i-1]) * (-1)**i)
D2.append(abs(L[i] - L[i-1]) * (-1)**(i+1))
print(max(maxSubArray(D1), maxSubArray(D2)))
``` | output | 1 | 98,317 | 14 | 196,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | instruction | 0 | 98,318 | 14 | 196,636 |
Tags: dp, two pointers
Correct Solution:
```
n = int(input())
num = list(map(int, input().split()))
diff = [abs(j-i) for i, j in zip(num, num[1:])]
s1 = 0
s2 = 0
m1 = 0
m2 = 0
for i, x in enumerate(diff):
if s1 < 0:
s1 = 0
if s2 < 0:
s2 = 0
if i % 2 == 0:
s1 += x
s2 -= x
else:
s1 -= x
s2 += x
if s1 > m1:
m1 = s1
if s2 > m2:
m2 = s2
print(max(m1, m2))
``` | output | 1 | 98,318 | 14 | 196,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | instruction | 0 | 98,319 | 14 | 196,638 |
Tags: dp, two pointers
Correct Solution:
```
def max_subarray(A):
max_ending_here = max_so_far = A[0]
for x in A[1:]:
max_ending_here = max(x, max_ending_here + x)
max_so_far = max(max_so_far, max_ending_here)
return max_so_far
n = int(input())
a = [int(v) for v in input().split()]
b = [abs(a[i]-a[i+1])*(-1)**i for i in range(n-1)]
print(max(max_subarray(b), max_subarray([-v for v in b])))
``` | output | 1 | 98,319 | 14 | 196,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,758 | 14 | 197,516 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
from sys import *
n=int(stdin.readline())
a=[int(x) for x in stdin.readline().split()]
b=[int(x) for x in stdin.readline().split()]
count=0
c=[]
for i in range(n):
c.append(a[i]-b[i])
c=sorted(c)
i=0
j=n-1
while(i<j):
if(c[i]+c[j]>0):
count+=j-i
j=j-1
else:
i=i+1
print(count)
``` | output | 1 | 98,758 | 14 | 197,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,759 | 14 | 197,518 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = [a[i]-b[i] for i in range(n)]
c.sort()
ind = n - 1
wyn = 0
for i in range(n):
while True:
if ind > i and c[i] + c[ind] > 0:
ind -= 1
else:
break
if ind <= i:
ind = min(n-1,i)
wyn += (n-1-ind)
print(wyn)
``` | output | 1 | 98,759 | 14 | 197,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,760 | 14 | 197,520 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
part = []
for i in range(n):
part.append(a[i] - b[i])
part.sort()
ans = 0
for i in range(n):
if part[i] > 0:
ans += n - i - 1
else:
L = 0
R = n
while L < R - 1:
mid = (L + R) // 2
if part[mid] > abs(part[i]):
R = mid
else:
L = mid
ans += n - R
print(ans)
``` | output | 1 | 98,760 | 14 | 197,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,761 | 14 | 197,522 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
from bisect import bisect_left
n = int(input())
x = [*map(int, input().split())]
y = [xi - yi for xi, yi in zip(x, map(int, input().split()))]
#print(y)
y.sort()
ans = 0
for i in range(n):
k = y[i]
a = bisect_left(y, -k+1)
ans += n-a
print((ans - len([i for i in y if i > 0]))// 2)
``` | output | 1 | 98,761 | 14 | 197,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,762 | 14 | 197,524 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
n=int(input())
ans=i=0
j=n-1
diff=[]
p=[]
j=n-1
list1=list(map(int,input().split()))
list2=list(map(int,input().split()))
zip_object = zip(list1,list2)
for list1_i, list2_i in zip_object:
diff.append(list1_i-list2_i)
#print(diff)
diff.sort()
#print(diff)
while(i<j):
if diff[i]+diff[j]>0:
ans+=j-i
j=j-1
else:
i+=1
print(ans)
``` | output | 1 | 98,762 | 14 | 197,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,763 | 14 | 197,526 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
import collections
import bisect
def solve(n, teacherInterests, studentInterests):
diffs = []
for t, s in zip(teacherInterests, studentInterests):
diffs.append(t-s)
diffs.sort()
ans = 0
# print(diffs)
for i, difference in enumerate(diffs):
if difference <= 0:
continue
pos = bisect.bisect_left(diffs, -difference + 1)
ans += i - pos
return ans
n = int(input().strip())
teacherInterests = list(map(int, input().strip().split()))
studentInterests = list(map(int, input().strip().split()))
print(solve(n, teacherInterests, studentInterests))
``` | output | 1 | 98,763 | 14 | 197,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,764 | 14 | 197,528 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
def binarySearch(A,start,end,x):
if start>end:
return start
mid=(start+end)//2
if A[mid]==x:
return mid
elif A[mid]>x:
return binarySearch(A,start,mid-1,x)
else:
return binarySearch(A,mid+1,end,x)
n=int(input())
A=[int(i) for i in input().split()]
B=[int(i) for i in input().split()]
pos=[]
neg=[]
for i in range(n):
tmp=A[i]-B[i]
if tmp>0:
pos.append(tmp)
else:
neg.append(tmp)
pos.sort()
neg.sort()
toplam=0
for i in range(len(neg)):
tmp=-neg[i]+1
s=binarySearch(pos,0,len(pos)-1,tmp)
if s<len(pos) and pos[s]==tmp:
while(pos[s]==tmp and s>=0):
s-=1
s+=1
toplam+=len(pos)-s
toplam+=(len(pos)*(len(pos)-1))//2
print(toplam)
``` | output | 1 | 98,764 | 14 | 197,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0 | instruction | 0 | 98,765 | 14 | 197,530 |
Tags: binary search, data structures, sortings, two pointers
Correct Solution:
```
def binary_search(arr, st, ed, comp):
ans = -1
while st <= ed:
mid = (st + ed) // 2
if arr[mid] + comp > 0:
ans = mid
ed = mid-1
else:
st = mid+1
return ans
n = int(input())
a = input().split()
b = input().split()
nums = []
for i in range(n):
a[i] = int(a[i])
b[i] = int(b[i])
nums.append(a[i] - b[i])
nums.sort()
num_pairs = 0
for i in range(n):
large_start_idx = binary_search(nums, i+1, n-1, nums[i])
if large_start_idx == -1:
continue
num_pairs += n-large_start_idx
print(num_pairs)
``` | output | 1 | 98,765 | 14 | 197,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,864 | 14 | 197,728 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
# SHRi GANESHA author: Kunal Verma #
import os
import sys
from bisect import bisect_right
from collections import Counter, defaultdict, deque
from heapq import *
from io import BytesIO, IOBase
from math import gcd, inf, sqrt, ceil
def lcm(a, b):
return (a * b) // gcd(a, b)
'''
mod = 10 ** 9 + 7
fac = [1]
for i in range(1, 2 * 10 ** 5 + 1):
fac.append((fac[-1] * i) % mod)
fac_in = [pow(fac[-1], mod - 2, mod)]
for i in range(2 * 10 ** 5, 0, -1):
fac_in.append((fac_in[-1] * i) % mod)
fac_in.reverse()
def comb(a, b):
if a < b:
return 0
return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''
#MAXN = 10000004
# spf = [0 for i in range(MAXN)]
# adj = [[] for i in range(MAXN)]
def sieve():
global spf, adj, MAXN
spf[1] = 1
for i in range(2, MAXN):
spf[i] = i
for i in range(2, MAXN):
if i * i > MAXN:
break
if (spf[i] == i):
for j in range(i * i, MAXN, i):
if (spf[j] == j):
spf[j] = i
def getdistinctFactorization(n):
global adj, spf, MAXN
for i in range(1, n + 1):
index = 1
x = i
if (x != 1):
adj[i].append(spf[x])
x = x // spf[x]
while (x != 1):
if (adj[i][index - 1] != spf[x]):
adj[i].append(spf[x])
index += 1
x = x // spf[x]
def printDivisors(n):
i = 2
z = [1, n]
while i <= sqrt(n):
if (n % i == 0):
if (n / i == i):
z.append(i)
else:
z.append(i)
z.append(n // i)
i = i + 1
return z
def create(n, x, f):
pq = len(bin(n)[2:])
if f == 0:
tt = min
else:
tt = max
dp = [[inf] * n for _ in range(pq)]
dp[0] = x
for i in range(1, pq):
for j in range(n - (1 << i) + 1):
dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
return dp
def enquiry(l, r, dp, f):
if l > r:
return inf if not f else -inf
if f == 1:
tt = max
else:
tt = min
pq1 = len(bin(r - l + 1)[2:]) - 1
return tt(dp[pq1][l], dp[pq1][r - (1 << pq1) + 1])
def SieveOfEratosthenes(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
x = []
for i in range(2, n + 1):
if prime[i]:
x.append(i)
return x
def main():
n,m=map(int,input().split())
z=[[0,0]for i in range(n+1) ]
xx=[]
for i in range(n):
p=int(input())
if p>0:
z[p][1]+=1
else:
z[abs(p)][0]+=1
xx.append(p)
s=0
for i in range(1,n+1):
s+=z[i][1]
an=[]
# print(z)
for i in range(1,n+1):
# print(n - z[i][0] - s + z[i][1],s)
if n-z[i][0]-s+z[i][1]==m:
an.append(i)
z=Counter(an)
m=[0]*n
# print(an)
for j in range(n):
if xx[j]>0:
if z[xx[j]]==len(an):
m[j]=1
elif z[xx[j]]==0:
m[j]=0
else:
m[j]=0.7
else:
if z[abs(xx[j])]==len(an):
m[j]=0
elif z[abs(xx[j])]>0:
m[j]=0.7
else:
m[j]=1
#print(an)
for i in range(n):
if m[i]==1:
m[i]="Truth"
elif m[i]==0:
m[i]= "Lie"
else:
m[i]="Not defined"
print(*m,sep='\n')
# Fast IO 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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 98,864 | 14 | 197,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,865 | 14 | 197,730 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
a,b=map(int,input().split())
r=[0]*a
c1=[0]*(a+1)
c2=[0]*(a+1)
f=0
for i in range(a):
n=int(input())
r[i]=n
if n>0:c1[n]+=1
else:
f+=1
c2[-n]+=1
possible=[False]*(a+1)
# print(c1,c2)
np=0#number of suspects
for i in range(1,a+1):
if c1[i]+f-c2[i]==b:
#f-c2[i] is truth
possible[i]=True
np+=1
# print(possible)
for i in range(a):
if r[i]>0:#he said +
if possible[r[i]] and np==1:#unique
print("Truth")
elif not possible[r[i]]:#Lie
print("Lie")
else:
print("Not defined")
else:#claims r[i] is not
r[i]*=-1
if possible[r[i]] and np==1:print("Lie")
elif not possible[r[i]]:print("Truth")
else:print("Not defined")
# print (possible)
``` | output | 1 | 98,865 | 14 | 197,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,866 | 14 | 197,732 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
import sys
input=lambda:sys.stdin.readline().strip()
print=lambda s:sys.stdout.write(str(s)+"\n")
a,b=map(int,input().split())
r=[0]*a
c1=[0]*(a+1)
c2=[0]*(a+1)
f=0
for i in range(a):
n=int(input())
r[i]=n
if n>0:c1[n]+=1
else:
f+=1
c2[-n]+=1
possible=[False]*(a+1)
# print(c1,c2)
np=0#number of suspects
for i in range(1,a+1):
if c1[i]+f-c2[i]==b:
#f-c2[i] is truth
possible[i]=True
np+=1
# print(possible)
for i in range(a):
if r[i]>0:#he said +
if possible[r[i]] and np==1:#unique
print("Truth")
elif not possible[r[i]]:#Lie
print("Lie")
else:
print("Not defined")
else:#claims r[i] is not
r[i]*=-1
if possible[r[i]] and np==1:print("Lie")
elif not possible[r[i]]:print("Truth")
else:print("Not defined")
# print (possible)
``` | output | 1 | 98,866 | 14 | 197,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,867 | 14 | 197,734 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
from collections import Counter
c = Counter()
n, m = map(int, input().split())
sumdc = 0
mark = []
suspect = set()
for _ in range(n):
k = int(input())
mark.append(k)
if k < 0: sumdc += 1
c = Counter(mark)
k = 0
for i in range(1, n + 1):
if c[i] + sumdc - c[-1 * i] == m:
suspect.add(i)
k += 1
if k==0:print("Not defined\n"*n)
elif k==1:
j=suspect.pop()
for x in mark:
if x==j or(x<0 and abs(x)!=j):print("Truth")
else: print("Lie")
else:
suspect.update({-x for x in suspect})
for x in mark:
if x in suspect:print("Not defined")
elif x<0:print("Truth")
else:print("Lie")
``` | output | 1 | 98,867 | 14 | 197,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,868 | 14 | 197,736 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
n, m = map(int, input().split())
t = [int(input()) for i in range(n)]
s, p = 0, [0] * (n + 1)
for i in t:
if i < 0:
m -= 1
p[-i] -= 1
else: p[i] += 1
q = {i for i in range(1, n + 1) if p[i] == m}
if len(q) == 0: print('Not defined\n' * n)
elif len(q) == 1:
j = q.pop()
print('\n'.join(['Truth' if i == j or (i < 0 and i + j) else 'Lie' for i in t]))
else:
q.update({-i for i in q})
print('\n'.join(['Not defined' if i in q else ('Truth' if i < 0 else 'Lie') for i in t]))
``` | output | 1 | 98,868 | 14 | 197,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,869 | 14 | 197,738 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
n,m=list(map(int,input().split()))
a=[[0,0] for i in range(n)]
b=[]
e=0
f=0
for i in range(n):
c=input()
d=int(c[1:])
if c[0]=='+':
a[d-1][0]+=1
e+=1
else:
a[d-1][1]+=1
f+=1
b.append([c[0],d])
g=[a[i][0]+f-a[i][1] for i in range(n)]
h=g.count(m)
for i in range(n):
d=b[i][1]
if b[i][0]=='+':
if g[d-1]==m:
if h>1:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
else:
if h>1 or h==1 and g[d-1]!=m:
if g[d-1]==m:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
``` | output | 1 | 98,869 | 14 | 197,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,870 | 14 | 197,740 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
from sys import stdin, stdout
n, m = map(int, stdin.readline().split())
cntf = [0 for i in range(n + 1)]
cnts = [0 for i in range(n + 1)]
challengers = []
for i in range(n):
s = stdin.readline().strip()
challengers.append(s)
if s[0] == '+':
cntf[int(s[1:])] += 1
else:
cnts[int(s[1:])] += 1
first = sum(cntf)
second = sum(cnts)
ans = set()
for i in range(1, n + 1):
if cntf[i] + second - cnts[i] == m:
ans.add(i)
for i in range(n):
s = challengers[i]
if s[0] == '+':
if int(s[1:]) in ans:
if len(ans) > 1:
stdout.write('Not defined\n')
else:
stdout.write('Truth\n')
else:
stdout.write('Lie\n')
else:
if int(s[1:]) in ans:
if len(ans) > 1:
stdout.write('Not defined\n')
else:
stdout.write('Lie\n')
else:
stdout.write('Truth\n')
``` | output | 1 | 98,870 | 14 | 197,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one. | instruction | 0 | 98,871 | 14 | 197,742 |
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
from collections import Counter
n, m = map(int, input().split())
pank = 0
buben = []
bober = set()
for _ in range(n):
k = int(input())
buben.append(k)
if k < 0:
pank += 1
cnt = Counter(buben)
k = 0
for i in range(1, n + 1):
if cnt[i] + pank - cnt[-1 * i] == m:
bober.add(i)
k += 1
if k == 0:
print('Not defined\n' * n)
elif k == 1:
j = bober.pop()
for x in buben:
if x == j or (x < 0 and abs(x) != j):
print('Truth')
else:
print('Lie')
else:
bober.update({-x for x in bober})
for x in buben:
if x in bober:
print('Not defined')
elif x < 0:
print('Truth')
else:
print('Lie')
``` | output | 1 | 98,871 | 14 | 197,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
"""
Brandt Smith, Lemuel Gorion and Peter Haddad
codeforces.com
Problem 156B
"""
import sys
n, m = map(int, input().split(' '))
inp = []
guess = [0] * (n + 1)
for i in range(n):
temp = int(input())
inp.append(temp)
if temp < 0:
m -= 1
guess[-temp] -= 1
else:
guess[temp] += 1
dic = {temp for temp in range(1, n + 1) if guess[temp] == m}
if len(dic) == 0:
for i in range(n):
print('Not defined')
elif len(dic) == 1:
temp = dic.pop()
for i in inp:
if i == temp or (i < 0 and i + temp):
print('Truth')
else:
print('Lie')
else:
temp = dic.update({-i for i in dic})
for i in inp:
if i in dic:
print('Not defined')
else:
if i < 0:
print('Truth')
else:
print('Lie')
``` | instruction | 0 | 98,872 | 14 | 197,744 |
Yes | output | 1 | 98,872 | 14 | 197,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
from collections import defaultdict
n, m = map(int, input().split())
a = []
for i in range(n):
a.append(int(input()))
d = defaultdict(int)
pos, neg = 0, 0
for x in a:
d[x] += 1
if x > 0:
pos += 1
else:
neg += 1
possible = [False] * n
for i in range(1, n + 1):
t = d[i] + neg - d[-i]
if t == m:
possible[i - 1] = True
cnt = sum(possible)
for i in range(n):
if cnt == 0:
print('Lie')
continue
if a[i] > 0:
if possible[a[i] - 1]:
print('Truth' if cnt == 1 else 'Not defined')
else:
print('Lie')
else:
if not possible[-a[i] - 1]:
print('Truth')
else:
print('Lie' if cnt == 1 else 'Not defined')
``` | instruction | 0 | 98,873 | 14 | 197,746 |
Yes | output | 1 | 98,873 | 14 | 197,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
"""
Brandt Smith, Lemuel Gorion and Peter Haddad
codeforces.com
Problem 156B
"""
import sys
n, m = map(int, input().split(' '))
inp = []
guess = [0] * (n + 1)
for i in range(n):
temp = int(input())
inp.append(temp)
if temp < 0:
m -= 1
guess[-temp] -= 1
else:
guess[temp] += 1
dic = {temp for temp in range(1, n + 1) if guess[temp] == m}
if len(dic) == 0:
for i in range(n):
print('Not defined')
elif len(dic) == 1:
temp = dic.pop()
for i in inp:
if i == temp or (i < 0 and i + temp):
print('Truth')
else:
print('Lie')
else:
temp = dic.update({-i for i in dic})
for i in inp:
if i in dic:
print('Not defined')
else:
if i < 0:
print('Truth')
else:
print('Lie')
# Made By Mostafa_Khaled
``` | instruction | 0 | 98,874 | 14 | 197,748 |
Yes | output | 1 | 98,874 | 14 | 197,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
n,m=list(map(int,input().split()))
a=[[0,0] for i in range(n)]
b=[]
e=0
f=0
for i in range(n):
c=input()
d=int(c[1:])
if c[0]=='+':
a[d-1][0]+=1
e+=1
else:
a[d-1][1]+=1
f+=1
b.append([c[0],d])
g=[a[int(b[i][1])-1][0]+f-a[int(b[i][1])-1][1] for i in range(n)]
h=g.count(m)
for i in range(n):
d=b[i][1]
if b[i][0]=='+':
if g[i]==m:
if h>1:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
else:
if h>1 or h==1 and g[i]!=m:
if g[i]==m:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
``` | instruction | 0 | 98,875 | 14 | 197,750 |
No | output | 1 | 98,875 | 14 | 197,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
a,b=map(int,input().split())
r=[0]*a
c1=[0]*(a+1)
c2=[0]*(a+1)
f=0
for i in range(a):
n=int(input())
r[i]=n
if n>0:c1[n]+=1
else:
f+=1
c2[n]+=1
possible=[False]*(a+1)
np=0#number of suspects
for i in range(1,a+1):
if c1[i]+f-c2[i]==b:
#f-c2[i] is truth
possible[i]=True
np+=1
for i in range(a):
if r[i]>0:#he said +
if possible[r[i]] and np==1:#unique
print("Truth")
elif not possible[r[i]]:#Lie
print("Lie")
else:
print("Not defined")
else:#claims r[i] is not
r[i]*=-1
if possible[r[i]] and np==1:print("Lie")
elif not possible[r[i]]:print("Truth")
else:print("Not defined")
# print (possible)
``` | instruction | 0 | 98,876 | 14 | 197,752 |
No | output | 1 | 98,876 | 14 | 197,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
n,m=map(int,input().split())
c=[0]*(n+1)
dc=[0]*(n+1)
sumdc=0
sus=0
mark=[]
suspect=[]
for _ in range(n):
k=int(input())
mark.append(k)
if k>0:
c[k]+=1
else:
dc[abs(k)]+=1
sumdc+=1
for i in range(1,n+1): #假定每一个I 为罪犯
if c[i]+sumdc-dc[i]==m:
suspect.append(i)
if len(suspect)==1:
for x in mark:
print ("Truth") if x==suspect[0] else print("Lie")
else:
for x in mark:
if x>0:
print("Not defined") if suspect.count(x)>0 else print("Lie")
else:
if suspect.count(abs(x))>0:
print("Not defined")
elif suspect.count(abs(x))==0:
print("Truth")
``` | instruction | 0 | 98,877 | 14 | 197,754 |
No | output | 1 | 98,877 | 14 | 197,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
n,m=map(int,input().split())
c=[0]*(n+1)
dc=[0]*(n+1)
sumdc=0
sus=0
mark=[]
suspect=[]
for _ in range(n):
k=int(input())
mark.append(k)
if k>0:
c[k]+=1
else:
dc[abs(k)]+=1
sumdc+=1
for i in range(1,n+1): #假定每一个I 为罪犯
if c[i]+sumdc-dc[i]==m:
suspect.append(i)
#print(suspect)
if len(suspect)==1:
for x in mark:
if x>0:
print ("Truth") if x==suspect[0] else print("Lie")
else:
print ("Lie") if abs(x)==suspect[0] else print("True")
else:
for x in mark:
if x>0:
print("Not defined") if suspect.count(x)>0 else print("Lie")
else:
if suspect.count(abs(x))>0:
print("Not defined")
elif suspect.count(abs(x))==0:
print("Truth")
``` | instruction | 0 | 98,878 | 14 | 197,756 |
No | output | 1 | 98,878 | 14 | 197,757 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image> | instruction | 0 | 98,879 | 14 | 197,758 |
"Correct Solution:
```
n, k, t = map(int, input().split())
l = (1 << (n + 1)) - 1
p = [tuple(map(int, input().split())) for i in range(k)]
p.sort(key = lambda i: i[2])
a, b = -1, sum(i[2] for i in p)
def g(j, s, u, v):
global k, p, t
k += 1
if k > t: return
for i, (h, w, r) in enumerate(p[j:], j + 1):
if r > s: break
if ((u >> h) & 1) and ((v >> w) & 1):
g(i, s - r, u - (1 << h), v - (1 << w))
def f(s):
global k, l
k = 0
g(0, s, l, l)
return k
while b - a > 1:
c = (a + b) // 2
if f(c) < t: a = c
else: b = c
print(b)
``` | output | 1 | 98,879 | 14 | 197,759 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image> | instruction | 0 | 98,880 | 14 | 197,760 |
"Correct Solution:
```
def g(t, d):
j = len(d) - 1
i = len(t) - 1
while j >= 0:
while i >= 0 and t[i] > d[j]: i -= 1
t.insert(i + 1, d[j])
j -= 1
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, a, b = [0], [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k, t
r = [(p[j][1], p[j][2], s + p[j][0], j + 1) for j in range(i, k) if a[p[j][1]] == b[p[j][2]] == 0]
g(q, [s[2] for s in r])
if len(q) > t: q = q[:t]
for h, w, s, j in r:
if len(q) == t and s >= q[t - 1]: break
a[h], b[w] = 1, 1
f(j, s)
a[h], b[w] = 0, 0
f(0, 0)
print(q[t - 1])
``` | output | 1 | 98,880 | 14 | 197,761 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image> | instruction | 0 | 98,881 | 14 | 197,762 |
"Correct Solution:
```
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, a, b = [], [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k
q.append(s)
while i < k:
h, w = p[i][1], p[i][2]
if a[h] == b[w] == 0:
a[h], b[w] = 1, 1
f(i + 1, s + p[i][0])
a[h], b[w] = 0, 0
i += 1
f(0, 0)
q.sort()
print(q[t - 1])
``` | output | 1 | 98,881 | 14 | 197,763 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image> | instruction | 0 | 98,882 | 14 | 197,764 |
"Correct Solution:
```
I=lambda:list(map(int,input().split()))
n,k,T=I()
t=[I()for _ in '0'*k]
def b(h,w,r,a):
if h>n:a+=[r]
else:
b(h+1,w,r,a)
for f,s,v in t:
if f==h and s in w:b(h+1,w-set([s]),r+v,a)
return a
print(sorted(b(1,set(range(1,n+1)), 0,[]))[T-1])
``` | output | 1 | 98,882 | 14 | 197,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
Submitted Solution:
```
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, a, b = [0], [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k, t
r = [(p[j][1], p[j][2], s + p[j][0], j + 1) for j in range(i, k) if a[p[j][1]] == b[p[j][2]] == 0]
print(r)
q += [s[2] for s in r]
if len(q) > t:
q.sort()
q = q[:t]
for h, w, s, j in r:
if len(q) == t and s >= q[t - 1]: break
a[h], b[w] = 1, 1
f(j, s)
a[h], b[w] = 0, 0
f(0, 0)
q.sort()
print(q[t - 1])
``` | instruction | 0 | 98,883 | 14 | 197,766 |
No | output | 1 | 98,883 | 14 | 197,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
Submitted Solution:
```
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, g, a, b = [0], 10 ** 10, [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k, t, g
r = [(p[j][1], p[j][2], s + p[j][0]) for j in range(i, k) if a[p[j][1]] == b[p[j][2]] == 0]
q += [s[2] for s in r]
i += 1
if len(q) > t:
q.sort()
g = q[t - 1] - 1
for h, w, s in r:
if s > g: break
a[h], b[w] = 1, 1
f(i, s)
a[h], b[w] = 0, 0
f(0, 0)
print(g + 1)
``` | instruction | 0 | 98,884 | 14 | 197,768 |
No | output | 1 | 98,884 | 14 | 197,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
Submitted Solution:
```
n, k, t = map(int, input().split())
l = (1 << (n + 1)) - 1
p = [tuple(map(int, input().split())) for i in range(k)]
p.sort(key = lambda i: i[2])
a, b = 0, sum(i[2] for i in p)
def g(j, s, u, v):
global k, p, t
k += 1
if k > t: return
for i, (h, w, r) in enumerate(p[j:], j + 1):
if r > s: break
if ((u >> h) & 1) and ((v >> w) & 1):
g(i, s - r, u - (1 << h), v - (1 << w))
def f(s):
global k, l
k = 0
g(0, s, l, l)
return k
while b - a > 1:
c = (a + b) // 2
if f(c) < t: a = c
else: b = c
print(b)
``` | instruction | 0 | 98,885 | 14 | 197,770 |
No | output | 1 | 98,885 | 14 | 197,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,134 | 14 | 198,268 |
Tags: dfs and similar, math
Correct Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
# from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x))
mod = int(1e9)+7
inf = float("inf")
n, = gil()
nxt = [0] + gil()
vis = [0]*(n+1)
c = set()
def cycle(p):
tt = 0
start = p
while vis[p] == 0:
tt += 1
vis[p] = 1
p = nxt[p]
if p != start:
print(-1)
exit()
return tt
for p in range(1, n+1):
if vis[p] : continue
tt = cycle(p)
if tt&1 :
c.add(tt)
else:
c.add(tt//2)
if 1 in c:c.remove(1)
c = list(c)
# print(c)
if len(c) <= 1:
print(c[0] if c else 1)
exit()
pr = {}
def getPrime(x):
ppr = {}
for i in range(2, int(sqrt(x))+1):
po = 0
while x%i == 0:
x //= i
po += 1
if po:ppr[i] = po
if x > 1:
ppr[x] = 1
return ppr
for v in c:
for pi, po in getPrime(v).items():
pr[pi] = max(pr.get(pi, 0), po)
ans = 1
for pi, po in pr.items():
ans *= pi**po
print(ans)
``` | output | 1 | 99,134 | 14 | 198,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,135 | 14 | 198,270 |
Tags: dfs and similar, math
Correct Solution:
```
from math import gcd
n = int(input())
crush = list(map(int,input().split()))
crush = [0] + crush
vis,dis,cyc,flag = [0]*(n+1), [-1]*(n+1), [], 1
def dfs(u,d):
dis[u] = d
vis[u] = 1
global flag
v = crush[u]
if vis[v]==0:
dfs(v,d+1)
else:
if dis[v]==1:
cyc.append(d)
else:
flag = 0
dis[u] = -1
for i in range(1,n+1):
if not flag:
break
if vis[i]==0:
dfs(i,1)
if not flag or len(cyc)==0:
print(-1)
else:
def lcm(a,b):
return a*b//gcd(a,b)
L = cyc[0]
if not L%2:
L //= 2
for i in range(1,len(cyc)):
s = cyc[i]
if not s%2:
s //= 2
L = lcm(L,s)
print(L)
# C:\Users\Usuario\HOME2\Programacion\ACM
``` | output | 1 | 99,135 | 14 | 198,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,136 | 14 | 198,272 |
Tags: dfs and similar, math
Correct Solution:
```
def gcd(a, b):
while b:
a, b = b, a % b
return a
def lcm(a, b):
return a * b // gcd(a, b)
n = int(input())
a = list(map(int, input().split()))
if sorted(a) != [i + 1 for i in range(n)]:
print(-1)
else:
ans = 1
used = [0 for i in range(n)]
for i in range(n):
if used[i] == 0:
j = i
am = 0
while used[j] == 0:
am += 1
used[j] = 1
j = a[j] - 1
if am % 2:
ans = lcm(ans, am)
else:
ans = lcm(ans, am // 2)
print(ans)
``` | output | 1 | 99,136 | 14 | 198,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,137 | 14 | 198,274 |
Tags: dfs and similar, math
Correct Solution:
```
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
n=int(input())
l=list(map(lambda x:int(x)-1,input().split()))
use=[]
valid=1
for i in range(n):
t=i
for j in range(n+5):
t=l[t]
if t==i:
if (j+1)%2==0: use.append((j+1)//2)
else: use.append(j+1)
break
else:
valid=0
if not valid: print("-1")
else:
# get lcm
ans=1
for i in use:
t=ans
while ans%i:
ans+=t
print(ans)
``` | output | 1 | 99,137 | 14 | 198,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,138 | 14 | 198,276 |
Tags: dfs and similar, math
Correct Solution:
```
from math import gcd
n = int(input())
arr = map(int, input().split())
arr = list(map(lambda x: x-1, arr))
res = 1
for i in range(n):
p, k = 0, i
for j in range(n):
k = arr[k]
if k == i:
p = j
break
if k != i:
print(-1)
exit()
p += 1
if p % 2 == 0: p //= 2
res = res * p // gcd(res, p)
print(res)
``` | output | 1 | 99,138 | 14 | 198,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,139 | 14 | 198,278 |
Tags: dfs and similar, math
Correct Solution:
```
input()
crush = [0] + [int(x) for x in input().split()]
visited = set()
circle_sizes = []
def gcd(a, b):
return a if b == 0 else gcd(b, a%b)
def lcm(a, b):
return a * b // gcd(a, b)
def solve():
for i in range(len(crush)):
if i not in visited:
start, cur, count = i, i, 0
while cur not in visited:
visited.add(cur)
count += 1
cur = crush[cur]
if cur != start:
return -1
circle_sizes.append(count if count % 2 else count // 2)
if len(circle_sizes) == 1:
return circle_sizes[0]
ans = lcm(circle_sizes[0], circle_sizes[1])
for size in circle_sizes[2:]:
ans = lcm(ans, size)
return ans
print(solve())
``` | output | 1 | 99,139 | 14 | 198,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,140 | 14 | 198,280 |
Tags: dfs and similar, math
Correct Solution:
```
from fractions import gcd
n = int(input())
crush = list(map(int,input().split()))
was = [0] * n
ans = 1
for i in range(n):
if was[i]:
continue
num = i
j = 0
while not was[num]:
was[num] = 1
num = crush[num] - 1
j += 1
if i != num:
ans = -1
break
else:
if j % 2 == 0:
j //= 2
ans = ans * j // gcd(ans,j)
print(ans)
``` | output | 1 | 99,140 | 14 | 198,281 |
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