text stringlengths 198 433k | conversation_id int64 0 109k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Tags: math, number theory
Correct Solution:
```
de = [2,3,5,7,11]
for i in range(int(input())):
a,b = map(int,input().split())
c = a-b
if c%2 == 1 and c>1:
c-=3
print('YES' if c%2 == 0 else 'NO')
```
| 92,700 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Tags: math, number theory
Correct Solution:
```
t = int(input())
for _ in range(t):
a, b = [int(i) for i in input().split()]
if a - 2 >= b:
print("YES")
else:
print("NO")
```
| 92,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
for i in range(int(input())):
a,b=list(map(int,input().split()))
print("NYOE S"[a-b>1::2])
```
Yes
| 92,702 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
elem = int(input())
for x in range(elem):
ar = []
for t in input().split(' '):
ar.append(int(t))
temp = ar[0] - ar[1]
if temp>1:
print('YES')
else:
print('NO')
```
Yes
| 92,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
def mp():
return map(int, input().split())
t = int(input())
for tt in range(t):
x, y = mp()
if x - y == 1:
print('NO')
else:
print('YES')
```
Yes
| 92,704 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
'''input
4
100 98
42 32
1000000000000000000 1
41 40
'''
for _ in range(int(input())):
x, y = [int(i) for i in input().split()]
d = x - y
if d > 1:
print("YES")
else:
print("NO")
```
Yes
| 92,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
n=int(input())
for i in range(n):
a,b=map(int,input().split())
if (abs(a-b)%2!=0)&(abs(a-b)%3!=0)&(abs(a-b)%5!=0)&(abs(a-b)%7!=0):
print('No')
else:
print('Yes')
```
No
| 92,706 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
t = int(input())
ans = [True] * t
for i in range(t):
x, y = map(int, input().split())
x -= y
if x == 1:
ans[i] = False
continue
d = 2
while d <= x:
if x % d:
d += 1
else:
x //= d
break
else:
ans[i] = False
for i in ans:
print('YES' if ans[i] else 'NO')
```
No
| 92,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
n, ans, z = int(input()), [], 0
for i in range(n):
x, y = map(int, input().split())
z = x - y
if z%2 == 0 or z%3 == 0 or z%5 == 0 or z%7 == 0 or z%11 == 0 or z%13 == 0 or z%17 == 0 or z%19 == 0 or z%23 == 0 or z%31 == 0:
ans.append(1)
else:
ans.append(0)
for i in range(n):
if ans[i] == 1:
print("YES")
else:
print("NO")
```
No
| 92,708 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y (it is guaranteed that x > y). You may choose any prime integer p and subtract it any number of times from x. Is it possible to make x equal to y?
Recall that a prime number is a positive integer that has exactly two positive divisors: 1 and this integer itself. The sequence of prime numbers starts with 2, 3, 5, 7, 11.
Your program should solve t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Then t lines follow, each describing a test case. Each line contains two integers x and y (1 ≤ y < x ≤ 10^{18}).
Output
For each test case, print YES if it is possible to choose a prime number p and subtract it any number of times from x so that x becomes equal to y. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
4
100 98
42 32
1000000000000000000 1
41 40
Output
YES
YES
YES
NO
Note
In the first test of the example you may choose p = 2 and subtract it once.
In the second test of the example you may choose p = 5 and subtract it twice. Note that you cannot choose p = 7, subtract it, then choose p = 3 and subtract it again.
In the third test of the example you may choose p = 3 and subtract it 333333333333333333 times.
Submitted Solution:
```
num = int(input())
for i in range(num):
l = input().split()
a = int(l[0])
b = int(l[1])
if a % b == 0:
print(0)
else:
print(int(b - (a % b)))
```
No
| 92,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
for tc in [0]*int(input()):
n = int(input())
a = list(map(int, input().split()))
i = 0
j = 0
while i < n - 1:
m = min(a[i:])
i = a.index(m)
a = a[:j] + [m] + a[j:i] + a[i + 1:]
if i == j:
i += 1
j = i
print(*a)
```
| 92,710 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
qq = int(input())
for q in range(qq):
n = int(input())
dat = list(map(int, input().split()))
already = [False] * 110
for j in range(0, n - 1):
#print("j:{0}".format(j))
ma = 99999
maindex = 99999
for i in range(0, n - 1):
if already[i]:
continue
if dat[i] < dat[i+1]:
continue
if dat[i + 1] < ma:
if dat[i+1] == n:
continue
maindex = i
ma = dat[i + 1]
if maindex == 99999:
break
else:
#print("maindex:{0}, ma:{1}".format(maindex, ma))
already[maindex] = True
dat[maindex], dat[maindex+1] = dat[maindex + 1], dat[maindex]
#print(dat)
dat = list(map(lambda x: str(x), dat))
print(" ".join(dat))
```
| 92,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
s = list(map(int, input().split()))
perform = [0 for i in range(n)]
rec = [0 for i in range(n)]
for i in range(n):
rec[s[i] - 1] = i
# print(rec)
op = n - 1
# lim_for_n = 0
for i in range(n):
if op == 0:
break
p = rec[i] - 1
temp = op
tempP = p
lim = p + 1
while tempP >= 0 and temp > 0:
if perform[tempP] == 1:
break
if s[tempP] > i + 1 and perform[p] == 0:
lim = tempP
tempP -= 1
temp -= 1
# print('for ', i + 1, lim, 'p', p)
# print('p', p)
while p >= lim and op > 0:
rec[i] = p
s[p], s[p + 1] = s[p + 1], s[p]
perform[p] = 1
# print(p + 1)
# print("s, ", s[p + 1])
rec[s[p + 1] - 1] = p + 1
p -= 1
op -= 1
# print('inter', s, i + 1)
for i in s:
print(i, end=' ')
print('')
```
| 92,712 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
q = int(input())
for i in range(q):
n = int(input())
A = list(map(int,input().split()))
B = []
k = 1
for i in range(n):
j = A.index(k)
if j != -1 and j not in B:
if len(B) == 0:
g = 0
else:
g = max(B) + 1
A.pop(j)
A.insert(g,k)
for f in range(g,j):
B.append(f)
if j == g:
B.append(j)
k += 1
print(*A)
```
| 92,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
from sys import stdin, stdout
from collections import defaultdict
# 4
# 3 4 1 2
# 3 1 2 4
# 1 3 2 4
def swap(A, d, x, y):
t = A[x]
A[x] = A[y]
A[y] = t
# print(d, A[x], A[y], x, y)
d[A[x]] = x
d[A[y]] = y
# print(d, A[x], A[y], x, y)
return A,d
def solve():
n = int(input())
A = [int(i) for i in stdin.readline().split()]
d = defaultdict()
st = set()
for i in range(len(A)):
d[A[i]] = i
for i in range(len(A) - 1):
while True:
if d[i+1] == 0: break
if A[d[i+1] - 1] > A[d[i+1]] and d[i+1] - 1 not in st:
st.add(d[i+1] - 1)
A,d = swap(A, d, d[i+1] - 1, d[i+1])
else: break
for i in A:
print(str(i) + " ", end="")
print()
return 0
t = int(input())
for i in range(t):
solve()
```
| 92,714 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
visited = set()
for i in range(len(arr)):
if i in visited:
continue
best_idx = i
for j in range(i + 1, len(arr)):
if j in visited:
break
if arr[j] < arr[best_idx]:
best_idx = j
while best_idx != i:
visited.add(best_idx - 1)
arr[best_idx], arr[best_idx - 1] = arr[best_idx - 1], arr[best_idx]
best_idx -= 1
print(*arr)
```
| 92,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
for i in range(int(input())):
n=int(input())
ar=list(map(int,input().split()))
if(n==1):
print(*ar)
elif(n==2):
ar.sort()
print(*ar)
else:
ans=[]
y=0
x=0
s=0
for j in range(1,n+1):
y=s
while(y<n):
if(ar[y]==j):
for k in range(y,x,-1):
if(ar[k-1]>ar[k]):
ar[k],ar[k-1]=ar[k-1],ar[k]
x=y
s=y
break
y+=1
print(*ar)
```
| 92,716 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Tags: greedy
Correct Solution:
```
def p_v(v):
for i in v:
print(i, end = " ")
return
def max_p(v, used):
ind_f = 0
ind_s = 1
mx_ind_s = -1
mx = 0
for i in range(len(v)- 1):
if v[ind_f] > v[ind_s] and used[ind_s] == 0:
if ind_s > mx:
mx = ind_s
mx_ind_s = ind_s
ind_f+=1
ind_s+=1
return mx_ind_s, used
m = int(input())
for i in range(m):
n = int(input())
v = a = list(map(int, input().split()))
used = [0 for i in range(n) ]
if len(v) == 1:
p_v(v)
else:
for j in range(n-1):
mx_ind, used = max_p(v, used)
if mx_ind > -1 and used[mx_ind] == 0:
used[mx_ind] = 1
t = v[mx_ind]
v[mx_ind] = v[mx_ind-1]
v[mx_ind-1] = t
else:
break
p_v(v)
```
| 92,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
import sys
input = sys.stdin.readline
q = int(input())
for _ in range(q):
w = int(input())
e = list(map(int, input().split()))
minpos = []
mn = w - 1
m = min(e)
nn = e.index(m)
while mn - nn > 0 and w > 0:
mn -= nn
if nn > 1:
minpos += [m] + e[:nn - 1]
e = [e[nn - 1]] + e[nn + 1:]
elif nn == 0:
minpos += [m]
e = e[nn + 1:]
w -= 1
else:
minpos += [m]
e = [e[nn - 1]] + e[nn + 1:]
w -= nn
if w > 0:
m = min(e)
nn = e.index(m)
if w > 0:
e = e[:nn - mn] + [e[nn]] + e[nn - mn:nn] + e[nn + 1:]
minpos += e
print(*minpos)
```
Yes
| 92,718 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
# sys.setrecursionlimit(10**6)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
def YES(c): return IF(c, "YES", "NO")
def Yes(c): return IF(c, "Yes", "No")
def main():
t = I()
rr = []
for _ in range(t):
n = I()
a = LI()
j = -1
for c in range(1,n+1):
nj = i = a.index(c)
while j < i and i >= c and a[i] < a[i-1]:
a[i],a[i-1] = a[i-1],a[i]
i -= 1
j = max(j, nj)
rr.append(JA(a, " "))
return JA(rr, "\n")
print(main())
```
Yes
| 92,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
q=int(input())
for j in range(q):
n=int(input())
ch=input()
l=ch.split()
for i in range(n):
l[i]=int(l[i])
if n==1:
print(l[0])
else:
s=set(range(n))
s.remove(0)
s.add(n)
if l[0]==1:
i=2
else:
i=1
h=l.index(1)
for k in range(h):
s.remove(k+1)
while h>0:
x=l[h]
l[h]=l[h-1]
l[h-1]=x
h-=1
i=2
while(s and i<n):
h=l.index(i)
if l[h]<l[h-1] and h in s:
x=l[h-1]
l[h-1]=l[h]
l[h]=x
s.remove(h)
else:
i+=1
for m in range(n):
print(l[m],end=' ')
print()
```
Yes
| 92,720 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
q = int(input())
for qi in range(q):
n = int(input())
a = list(map(int, input().split()))
used = [False] * n
for t in range(n):
for i in range(len(a) - 1, 0, -1):
if used[i]:
continue
if a[i] < a[i - 1]:
a[i], a[i - 1] = a[i - 1], a[i]
used[i] = True
print(' '.join(str(x) for x in a))
```
Yes
| 92,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
l = list(map(int,input().split()))
c = 1
i = 0
l2 = [i for i in range(1,n+1)]
w = l.index(c)
while i!=n-1:
if c == l.index(c) + 1:
if l == l2:
break
c+=1
w = l.index(c)
else:
if l == l2:
break
else:
dck = l[w]
l[w] = l[w-1]
l[w-1] = dck
w-=1
i+=1
print(l)
```
No
| 92,722 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
amount = int(input())
for i in range(amount):
n = int(input())
array = [int(s) for s in input().split()]
for j in range(len(array) - 1, 0, -1):
if array[j] < array[j - 1]:
array[j], array[j - 1] = array[j - 1], array[j]
print(" ".join([str(s) for s in array]))
```
No
| 92,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
seq = [int(x) for x in input().split()]
val = 1
op = [False]*(len(seq)-1)
op_perf = 0
for val in range(1, n+1):
swap = seq.index(val)
for i in range(swap-1,val-2,-1):
if not op[i]:
op[i] = True
swap = i
else:
break
seq.remove(val)
seq.insert(swap, val)
if all(op):
break
print(' '.join(map(str,seq)))
```
No
| 92,724 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).
You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.
Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.
You can see the definition of the lexicographical order in the notes section.
You have to answer q independent test cases.
For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:
1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2];
2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3];
3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3].
4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3];
Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.
The second line of the test case contains n distinct integers from 1 to n — the given permutation.
Output
For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.
Example
Input
4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1
Output
1 5 2 4 3
1 2 3 4
1
1 4 3 2
Note
Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:
* p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j),
* p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j).
Submitted Solution:
```
for __ in range(int(input())):
n = int(input())
ar = list(map(int, input().split()))
x = 1
i = ar.index(x)
num = 0
while x < n and i < n:
if x != 1:
while n > i >= ar.index(x):
i += 1
x += 1
if i < n:
i = ar.index(x)
for j in range(i, num, -1):
if ar[j] < ar[j - 1]:
ar[j], ar[j - 1] = ar[j - 1], ar[j]
num = i
x += 1
print(*ar)
```
No
| 92,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
d = {}
for i in a:
s = d.get(i, 0)
if s == 0:
d[i] = 1
else:
d[i] = d[i] + 1
c = 0
for i in d.keys():
c += d[i] // 2
print(c // 2)
```
| 92,726 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
n=int(input())
arr=list(map(int,input().split()))
s=set(arr)
rem=0
c=0
for x in s:
if(arr.count(x)>=4):
c=c+(arr.count(x)//4)
rem=rem+((arr.count(x)%4)//2)
else:
rem=rem+(arr.count(x)//2)
c=c+(rem//2)
print(c)
```
| 92,727 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from copy import copy
n=int(input())
f={}
d=0
p=[]
deleted=[]
def refresh():
global p,d
if len(p)==2:
d+=1
f[p[0]]=0
f[p[1]]=0
p=[]
for x in input().split():
try:
f[x]+=1
except:
f[x]=1
for x in f.keys():
if f[x]>=4:
j=copy(f[x])
f[x]=j%4
d+=j//4
if f[x]>=2:
p.append(x)
refresh()
print(d)
```
| 92,728 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
n = int(input())
a = [0] * 110
sticks = list(map(int, input().split()))
for i in range(len(sticks)):
a[sticks[i]] += 1
counter = 0
for i in range(len(a)):
counter += a[i] // 2
print(counter // 2)
```
| 92,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
def arr_inp():
return [int(x) for x in input().split()]
from collections import *
n, a = int(input()), arr_inp()
c = Counter(a)
# print(c)
print(int(sum(list(map(lambda x:x//2, c.values())))//2))
```
| 92,730 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
s = set(l)
c = []
for i in s:
t = l.count(i)
if t%2==0:
c.append(t)
else:
c.append(t-1)
s = sum(c)
print(s//4)
```
| 92,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
"""
Oh, Grantors of Dark Disgrace,
Do Not Wake Me Again.
"""
from collections import Counter
ii = lambda: int(input())
mi = lambda: map(int, input().split())
li = lambda: list(mi())
si = lambda: input()
n = ii()
l = li()
cc = Counter(l)
e = [(i-i%2) for i in cc.values()]
print(sum(e)//4)
```
| 92,732 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Tags: implementation
Correct Solution:
```
from collections import defaultdict
n=int(input())
c=list(map(int,input().split()))
d=defaultdict(int)
for i in range(len(c)):
d[c[i]]+=1
count=0
for val in d.values():
count+=val//2
if count%2==0:
print(count//2)
else:
print((count-1)//2)
```
| 92,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
d={}
for x in map(int,[*open(0)][1].split()):
d[x]=d.get(x,0)+1
r=0
for x in d.values():r+=x//2
print(r//2)
```
Yes
| 92,734 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
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")
from collections import defaultdict
from math import ceil,floor,sqrt,log2,gcd
from heapq import heappush,heappop
from bisect import bisect_left,bisect
import sys
abc='abcdefghijklmnopqrstuvwxyz'
ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
n=int(input())
arr=list(map(int,input().split()))
d=defaultdict(int)
for i in arr:
d[i]+=1
ans=0
for i in d:
ans+=d[i]//2
print(ans//2)
```
Yes
| 92,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
if __name__ == '__main__':
Y = lambda: list(map(int, input().split()))
N = lambda: int(input())
n = N()
a = Y()
d = dict()
for i in range(n):
d[a[i]] = d.get(a[i], 0) + 1
for v in d.keys():
d[v] = d[v]//2
print(sum(d.values())//2)
```
Yes
| 92,736 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
import sys
f = sys.stdin
#f = open("input.txt", "r")
n = int(f.readline().strip())
a = [int(i) for i in f.readline().strip().split()]
a.sort()
k = list(set(a))
counts = []
for i in k:
counts.append(a.count(i))
counts.sort(reverse=True)
cnt = 0
i = 0
while i < len(counts):
if counts[i] >= 4:
cnt += counts[i]//4
counts[i] -= (counts[i]//4)*4
i += 1
while 0 in counts:
counts.remove(0)
counts.sort(reverse=True)
i = 0
while i < len(counts)-1:
if counts[i] >= 2 and counts[i+1] >= 2:
cnt += min(counts[i], counts[i+1])//2
i += 2
else:
i += 1
print(cnt)
```
Yes
| 92,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
z,w,m=input,int,sorted
n=w(z())
l1=list(map(int,input().split()))
l2=set(l1)
l={}
for i in l2:
l[i]=0
for i in l1:
l[i]+=1
l=sorted(l.values())
l=l[::-1]
c=0
ans=0
for i in l:
ans+=(i+c)//4
c=i%4
if c>=2:
c=c//2
else:
c=0
print(ans)
```
No
| 92,738 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
n = int(input())
l = list(map(int , input().split()))
i = 0
arr=[]
while i<len(l):
x = l.count(l[i])
if x>=4:
arr.append(1)
arr.append(1)
l.remove(l[i])
l.remove(l[i])
l.remove(l[i])
l.remove(l[i])
i = i-1
elif x>=2:
arr.append(1)
l.remove(l[i])
l.remove(l[i])
i = i-1
i = i + 1
ans = len(arr)//2
print(ans)
```
No
| 92,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
freq={}
for x in a:
if x in freq:
freq[x]+=1
else:
freq[x]=1
count=0
for x in freq:
if freq[x]>1:
count+=freq[x]
print(count//4)
```
No
| 92,740 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has n sticks whose lengths equal a1, a2, ... an. Nicholas does not want to break the sticks or glue them together. To make a h × w-sized frame, he needs two sticks whose lengths equal h and two sticks whose lengths equal w. Specifically, to make a square frame (when h = w), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of sticks. The second line contains n space-separated integers. The i-th integer equals the length of the i-th stick ai (1 ≤ ai ≤ 100).
Output
Print the single number — the maximum number of frames Nicholas can make for his future canvases.
Examples
Input
5
2 4 3 2 3
Output
1
Input
13
2 2 4 4 4 4 6 6 6 7 7 9 9
Output
3
Input
4
3 3 3 5
Output
0
Submitted Solution:
```
n = int(input())
l = list(input().split())
d = {}
for gar in l:
if gar in d:
d[gar] = d[gar] + 1
else:
d[gar] = 1
pari = 0
for gar in d:
pari = pari + d[gar] // 2
print(pari // 2)
```
No
| 92,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
matrix=[[],[],[],[]]
for i in range(1,4):
x=list(input())
x.insert(0,'')
matrix[i]=x
if(matrix[1][1] == matrix[3][3] and matrix[1][3] == matrix[3][1] and matrix[1][2] == matrix[3][2] and matrix[2][1] == matrix[2][3]):
print("YES")
else:
print("NO")
```
| 92,742 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
l=[]
for jk in range(0,3):
line=input()
l.append(line)
chl1=l[2][::-1]
chl2=l[1][::-1]
chl3=l[0][::-1]
if chl1==l[0] and chl2==l[1] and chl3==l[2]:
print("YES")
else:
print("NO")
```
| 92,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
mat = []
for i in range(3): mat.append(list(input()))
mat1 = [x[::-1] for x in mat]
if mat1[::-1] == mat:print('YES')
else: print('NO')
```
| 92,744 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
i = input
s = i() + i() + i()
print("Yes" if s==s[::-1] else "No")
```
| 92,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
m = []
d = {(0, 0): (2, 2), (0, 1): (2, 1), (0, 2): (2, 0), (2, 2): (0, 0), (2, 1): (0, 1), (2, 0): (0, 2), (1, 0): (1, 2),
(1, 2): (1, 0), (1, 1): (1, 1)}
for k in range(3):
m.append(input())
temp_dict = {}
for i in range(3):
for j in range(3):
if m[i][j] == 'X' and d[i, j] in temp_dict:
del temp_dict[d[i, j]]
elif m[i][j] == 'X' and (i, j) != (1, 1):
temp_dict[i, j] = ''
if temp_dict:
print('NO')
else:
print('YES')
```
| 92,746 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
password = [input() for i in range(3)]
sym = True
if password[0] != password[2][::-1]:
sym = False
if password[1][0] != password[1][2]:
sym = False
if sym:
print('YES')
else:
print('NO')
```
| 92,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
a=int(3)
b=[]
for i in range(int(a)):
b.append(list(input()))
if b[0][0]==b[2][2] and b[0][1]==b[2][1] and b[0][2]==b[2][0] and b[1][0]==b[1][2]:
print("YES")
else:
print("NO")
```
| 92,748 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Tags: implementation
Correct Solution:
```
s=""
for i in range(3):
s+=input()
print(["NO","YES"][s==s[::-1]])
```
| 92,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
mat=[]
for _ in range(3):
s=input()
l=[]
l.append(s[0])
l.append(s[1])
l.append(s[2])
mat.append(l)
#print(mat)
if(mat[0][1]==mat[2][1] and mat[1][0]==mat[1][2] and mat[0][0]==mat[2][2] and mat[0][2]==mat[2][0]):
print("YES")
else:
print('NO')
```
Yes
| 92,750 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
t = [input().strip() for i in range(3)]
print("YES" if "".join(t) == "".join([i[::-1] for i in t[::-1]]) else "NO")
```
Yes
| 92,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
a = input()
b = input()
c = input()
i = 0
if a[0]==c[2]:
i = i+1
if a[1]==c[1]:
i = i+1
if a[2]==c[0]:
i = i+1
if b[0]==b[2]:
i = i+1
if i==4 :
print('YES')
else :
print('NO')
```
Yes
| 92,752 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
l = []
for i in range(3):
l.append(input())
if ((l[0][0] == l[2][2]) and (l[0][1] == l[2][1]) and (l[0][2] == l[2][0]) and (l[1][0] == l[1][2])):
print("YES")
else:
print("NO")
```
Yes
| 92,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
# 12A Super Agent
key = []
for i in range(3):
key.append(input())
if key[0][0] == key[2][2] and key[0][1] == key[2][1] and key[0][2] == key[2][0]:
print("YES")
else:
print("NO")
# XX.
# ...
# .XX
```
No
| 92,754 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
list=[]
for i in range(0,3):
reg=input()
list.append(reg)
if list[2][1]==list[0][1] and list[0][1]==list[2][1] and list[0][0]==list[2][2] and list[0][2]==list[2][0]:
print ("YES")
else:
print ("NO")
```
No
| 92,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
import sys
n = [i.strip().rstrip() for i in sys.stdin.readlines()]
if n[0][0] == n[0][2] and n[0][2] == n[2][0] and n[0][1] == n[2][1] and n[1][0] == n[1][2]:
print("YES")
else:
print("NO")
```
No
| 92,756 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base.
Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3 × 3 with digits from 1 to 9.
Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Output
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
Examples
Input
XX.
...
.XX
Output
YES
Input
X.X
X..
...
Output
NO
Note
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
Submitted Solution:
```
k=[]
n=3
for i in range(3):
k.append(input())
m=True
for i in range(2):
for j in range(3):
if i!=1:
if (k[i][j]=="X" and k[n-i-1][n-j-1]=="X") or (k[i][j]=="." and k[n-i-1][n-j-1]=="."):
continue
else:
print("NO")
m=False
break
else:
if (k[i][j]=="X" and k[i][n-j-1]=="X") or (k[i][j]=="." and k[i][n-j-1]=="."):
continue
else:
print("NO")
m=False
break
if m:
print("YES")
```
No
| 92,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
S = input().rstrip()
A = [0]
for s in S:
a = 1 if s == "(" else -1
A.append(A[-1]+a)
isOK = True
ans = 0
last0 = 0
for i, a in enumerate(A):
if a == 0:
if not isOK:
ans += i - last0
last0 = i
isOK = True
if a < 0:
isOK = False
if A[-1] != 0:
print(-1)
else:
print(ans)
```
| 92,758 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
n = int(input())
a = input()
if a.count('(') != a.count(')'):
print("-1")
else :
c=0
d=0
for i in a :
if i == '(' :
c+=1
else:
c-=1
if c<0:
d+=2
print(d)
```
| 92,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
import sys
#import math
#from collections import deque
#import heapq
input=sys.stdin.readline
n=int(input())
s=input()
l=list()
open=list()
close=list()
for i in range(n):
if(s[i]=='('):
open.append(i)
else:
close.append(i)
if(len(open)!=len(close)):
print(-1)
else:
ans=0
for i in range(len(open)):
if(open[i]>close[i]):
ans+=2
print(ans)
```
| 92,760 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
t=int(input())
s=input()
if s.count('(') != s.count(')'):
print(-1)
else:
ans=[]
negval=0
val=0
for i in range(len(s)):
if s[i] == '(':
val+=1
if val == 0:
ans.append(-1)
else:
ans.append(val)
else:
val-=1
ans.append(val)
a=0
for i in ans:
if i < 0:
a+=1
print(a)
```
| 92,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
#import io,os
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
import sys
input = sys.stdin.readline
from collections import defaultdict
def main():
#for _ in range(int(input())):
n = int(input())
s = input()
cnt = defaultdict(int)
for i in range(n):
cnt[s[i]] += 1
if cnt["("] != cnt[")"]:
print(-1)
exit()
ans = 0
bal = 0
bad = False
last = 0
for i in range(n):
if s[i] == "(":
bal += 1
else:
bal -= 1
if bal == 0:
if bad:
ans += i - last + 1
last = i + 1
bad = False
else:
last = i + 1
bad = False
elif bal < 0:
bad = True
print(min(n, ans))
main()
```
| 92,762 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
p=int(input())
q=input()
x=q.count(')')
y=q.count('(')
c=0
l=0
r=0
z=0
sum=0
if x!=y :
print(-1)
c=1
if c==0 :
for i in range(p) :
if q[i]=='(' :
l+=1
else :
r+=1
if r>l :
z=1
if l==r and z==0 :
# print("xxxxxxxxxx",l,r)
l=0
r=0
if l==r and z==1 :
# print("yyyyyyyyy",l,r)
sum+=l+r
l=0
r=0
z=0
print(sum)
```
| 92,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
n = int(input())
s = input()
stack = []
stack2 = []
pair = 0
if s.count(')') != s.count('('):
print(-1)
else:
for i in s:
if i == '(':
stack2.append(i)
if i == ')':
if stack2:
del stack2[-1]
else:
stack.append(i)
elif stack and i == '(':
del stack[-1]
del stack2[-1]
pair += 2
print(pair)
```
| 92,764 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Tags: greedy
Correct Solution:
```
n=int(input())
s=input()
a=0
b=0
for i in range(0,len(s)):
if s[i]==')':
a+=1
else:
b+=1
if a!=b:
print(-1)
else:
count=0
stack=[]
i=0
while i<len(s):
a=0
b=0
if s[i]=='(':
stack.append(s[i])
else:
if stack==[]:
for j in range(i,len(s)):
if s[j]==')':
a+=1
else:
b+=1
if a==b:
k=j
break
count=count+k+1-i
i=k
else:
stack.pop(-1)
i+=1
print(count)
```
| 92,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
from bisect import bisect_left as bl, bisect_right as br, insort
import sys
import heapq
from math import *
from collections import defaultdict as dd, deque
def data(): return sys.stdin.readline().strip()
def mdata(): return map(int, data().split())
#sys.setrecursionlimit(100000)
n=int(data())
s=data()
l=[]
o=c=0
k=0
ans=0
for i in range(n):
if s[i]=='(':
o+=1
l.append('(')
else:
c+=1
if len(l)>0:
if l[-1]=='(':
l.pop()
else:
l.append(')')
else:
l.append(')')
if o==c and len(l)!=0:
ans+=o+c
l=[]
o=c=0
if len(l)==0:
o=c=0
if o!=c:
print(-1)
else:
print(ans)
```
Yes
| 92,766 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
def check(s):
st=[]
for i in s:
if i=="(":
st.append("(")
else:
if len(st)==0:
return False
st.pop()
else:
return True
n=int(input())
s=list(input())
if s.count("(")!=s.count(")"):
print(-1)
exit()
if check(s):
print(0)
exit()
c=0
x=0
y=0
for i in range(n):
if s[i]=="(":
if x==0 and y==0:
j=i
x+=1
else:
if x==0 and y==0:
j=i
y+=1
if x==y:
if not check(s[j:i+1]):
c+=(x+y)
x=0
y=0
print(c)
```
Yes
| 92,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
n = int(input())
z = input()
pravych = 0
lavych = 0
for i in range(n):
if z[i] == ')':
pravych += 1
else:
lavych += 1
def najdi_usek_viac_lavych(index):
pp = 0
ll = 0
for i in range(index,n):
if z[i] == ')':
pp = pp+1
else:
ll = ll+1
if ll > pp:
return i-index
def prehadzuj(z,n,spolu):
p = 0
l = 0
i = 0
while(i<n):
if z[i] == ')':
p = p+1
else:
l = l+1
if l < p:
out = najdi_usek_viac_lavych(i+1)
spolu = spolu + out + 2
i = i+2+out
p=0
l=0
else:
i = i+1
return spolu
if n%2 == 1:
print(-1)
elif pravych != lavych:
print(-1)
else:
print(prehadzuj(z,n,0))
```
Yes
| 92,768 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
n=int(input())
s=input()
op,cl=0,0
ind=0
flag=0
for i in range(len(s)):
if(op<0):flag=1
else:flag=0
if(s[i]=='('):op+=1
else:op-=1
if(op==-1 and s[i]==')'):
ind=i
#print(i)
if(flag and op==0):
#print(i,ind)
cl+=((i-ind)+1)
if(op!=0):
print(-1)
else:
print(cl)
```
Yes
| 92,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
n = int(input())
a = input()
if a.count('(') != a.count(')'):
print("-1")
else :
b = []
c = 0
for i in range(n):
if len(b) == 0:
b.append(a[i])
else:
if b[len(b)-1] == '(' and a[i] == ')' and len(b)%2 != 0 :
b.pop()
elif b[len(b)-1] == '(' and a[i] == ')' and len(b)%2 == 0 :
b.append(a[i])
elif b[len(b)-1] == '(' and a[i] == '(' :
b.append(a[i])
elif b[len(b)-1] == ')' and a[i] == '(' :
b.append(a[i])
elif b[len(b)-1] == ')' and a[i] == ')' :
b.append(a[i])
print(len(b))
```
No
| 92,770 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
# Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout
# -------------- INPUT FUNCTIONS ------------------
def get_ints_in_variables(): return map(
int, sys.stdin.readline().strip().split())
def get_int(): return int(sys.stdin.readline())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
# -------- SOME CUSTOMIZED FUNCTIONS-----------
def myceil(x, y): return (x + y - 1) // y
# -------------- SOLUTION FUNCTION ------------------
def Solution(s, n):
# Write Your Code Here
opn = 0
close = 0
for c in s:
if c == "(":
opn += 1
else:
close += 1
if n % 2 or opn != close:
print(-1)
return
i = 0
ans = 0
while i < n:
opn = 0
clse = 0
l = i
firstOpen = True
if s[i] == "(":
while i < n and s[i] == "(":
i += 1
opn += 1
else:
firstOpen = False
while i < n and s[i] == ")":
i += 1
clse += 1
flg = True
if firstOpen:
while i < n and clse != opn:
if s[i] == "(":
opn += 1
flg = False
else:
clse += 1
i += 1
else:
while i < n and clse != opn:
if s[i] == ")":
clse += 1
flg = False
else:
opn += 1
i += 1
if (not flg) or (not firstOpen):
ans += (i-l)
print(ans)
def main():
# Take input Here and Call solution function
n = get_int()
Solution(get_string(), n)
# calling main Function
if __name__ == '__main__':
main()
```
No
| 92,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
if __name__ == '__main__':
n=int(input())
brackets=input()
wrongbra=0
tl,tr=0,0
l,r=0,0
wrongNow=False
for i,bra in enumerate(brackets):
if bra==")":
r+=1
tr+=1
else:
l+=1
tl+=1
if l<r:
wrongNow=True
if wrongNow:
if l==r:
wrongNow=False
wrongbra+=(l+r)
l,r=0,0
else:
if l==r:
l,r=0,0
print(wrongbra)
```
No
| 92,772 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracketed sequence is called correct (regular) if by inserting "+" and "1" you can get a well-formed mathematical expression from it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not.
The teacher gave Dmitry's class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima's turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word "correct" in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It's easy to see that reorder operation doesn't change the number of opening and closing brackets. For example for "))((" he can choose the substring ")(" and do reorder ")()(" (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it's impossible.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of Dima's sequence.
The second line contains string of length n, consisting of characters "(" and ")" only.
Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or "-1" if it is impossible to do so.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with "()()", the whole sequence will be "()()())(". And then reorder the segment from the seventh to eighth character, replacing it with "()". In the end the sequence will be "()()()()", while the total time spent is 4 + 2 = 6 nanoseconds.
Submitted Solution:
```
n=int(input())
s=list(input())
if s.count("(")!=s.count(")"):
print(-1)
exit()
st=[]
for i in s:
if i=="(":
st.append("(")
else:
if len(st)==0:
break
st.pop()
else:
print(0)
exit()
c=0
for i in range(0,n,2):
if s[i]!="(" or s[i+1]!=")":
c+=2
print(c)
```
No
| 92,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
t = int(input())
for ii in range(t):
t = input()
if t == '0' * len(t) or t == '1' * len(t):
print(t)
else:
print('01' * len(t))
```
| 92,774 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
T = int(input())
for _ in range(T) :
t = input()
isEqual = True
for i in range (len(t)-1):
if t[i] != t[i+1]:
isEqual = False
break
if isEqual:
print (t)
else:
output = ""
for i in range (len(t)-1):
if t[i] == t[i+1]:
output += t[i]
if t[i] == "0":
output += "1"
else:
output += "0"
else:
output += t[i]
output += t[len(t)-1]
print (output)
# region fastio
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 = lambda s: self.buffer.write(s.encode()) if self.writable else None
def read(self):
if self.buffer.tell():
return self.buffer.read().decode("ascii")
return os.read(self._fd, os.fstat(self._fd).st_size).decode("ascii")
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().decode("ascii")
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
def print(*args, sep=" ", end="\n", file=sys.stdout, flush=False):
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(end)
if flush:
file.flush()
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
sys.setrecursionlimit(10000)
if __name__ == "__main__":
main()
```
| 92,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
from collections import Counter
t=int(input())
for _ in range(t):
s=input()
dic=Counter(s)
if '0' not in dic:
print(s)
elif '1' not in dic:
print(s)
else:
front=s[0]
ans=""
if(front=='1'):
ans = "10"
for i in range(1,len(s)):
ans += "10"
print(ans)
else:
ans="01"
for i in range(1,len(s)):
ans += "01"
print(ans)
```
| 92,776 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
cases = int(input())
for i in range(cases):
t = list(input())
if len(set(t)) == 1:
print("".join(t))
else:
t = [int(q) for q in t]
n = []
for j in range(len(t)-1):
if t[j] == t[j+1]:
n.append(t[j])
n.append(1-t[j])
else:
n.append(t[j])
n.append(t[len(t)-1])
#n.append(1-t[len(t)-1])
st = ""
for j in n:
st += str(j)
print(st)
```
| 92,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
for _ in range(int(input())):
t=input()
z = t.count('0')
o = t.count('1')
import re
if z == 0 or o == 0:
print(t)
continue
if z > o:
while "00" in t:
t = re.sub("00", "010", t)
while "11" in t:
t = re.sub("11", "101", t)
print(t)
else:
while "00" in t:
t = re.sub("00", "010", t)
while "11" in t:
t = re.sub("11", "101", t)
print(t)
```
| 92,778 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
from sys import stdin, stdout
from math import *
from heapq import *
from collections import *
def main():
ntest=int(stdin.readline())
for testcase in range(ntest):
t=stdin.readline().strip()
s=[]
if len(set(t))==2:
s=['01']*len(t)
else:
s=[t]
stdout.write("%s\n"%("".join(s)))
return 0
if __name__ == "__main__":
main()
```
| 92,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
def main():
for _ in range(int(input())):
t=list(input())
l=[t[0]]
if(t.count('0')>0 and t.count('1')>0):
for i in range(1,len(t)):
if(l[len(l)-1]=='0' and t[i]=='0'):
l.append('1')
l.append('0')
elif(l[len(l)-1]=='1' and t[i]=='1'):
l.append('0')
l.append('1')
elif(l[len(l)-1]=='0' and t[i]=='1'):
l.append('1')
else:
l.append('0')
print("".join(l))
else:
print("".join(t))
main()
```
| 92,780 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Tags: constructive algorithms, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(s[:len(s) - 1])
def invr():
return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
t = inp()
for _ in range(t):
s = input().rstrip()
has = True
l = s[0]
for i in range(1,len(s)):
if l != s[i]:
has = False
break
if has:
print(s)
else:
ans = []
for i in range(len(s)):
ans.append('1')
ans.append('0')
print(''.join(ans))
```
| 92,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
T = int(input())
for times in range(T):
s = input()
if(s.find('1')!=-1 and s.find('0') != -1):
print("10" * len(s))
else:
print(s)
```
Yes
| 92,782 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
from collections import Counter
T = int(input())
ss = []
for _ in range(T):
ss.append(input())
for s in ss:
counter = Counter(s)
if counter["1"] > 0 and counter["0"] > 0:
if s[0] == "1":
print("".join(["1", "0"] * len(s)))
else:
print("".join(["0", "1"] * len(s)))
else:
print(s)
```
Yes
| 92,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
t= int(input())
def inv(ch):
return '1' if ch=='0' else '0'
for i in range(t):
s=input()
if len(s)==2:
print(s)
else:
stri=''
temp= True
for i in range(len(s)-1):
if(s[i]!=s[i+1]):
temp=False
if temp==True:
print(s+s)
else:
for i in range(len(s)-1):
stri+= s[i]
if s[i]==s[i+1]:
stri+= inv(s[i])
c=len(s)-1
stri+=s[c]
if(len(stri)%2==1):
stri+= inv(s[c])
print(stri)
```
Yes
| 92,784 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
for _ in range(int(input())):
t = input()
flag = t.count('1') and t.count('0')
if flag:
print(len(t)*'10')
else:
print(t)
```
Yes
| 92,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
for i in range(int(input())):
a=input()
print(a*2)
```
No
| 92,786 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
t=int(input())
while t:
t-=1
string=input()
ans=""
for i in range(len(string)-1):
if string[i]+string[i+1]=="00":
ans+=string[i]+"1"
elif string[i]+string[i+1]=="11":
ans+=string[i]+"0"
else:
ans+=string[i]
ans+=string[-1]
print(ans)
```
No
| 92,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
t=int(input())
for i in range(t):
s=input()
arr=[]
if(len(s)==1 or len(s)==2):
print(s)
else:
a=s[0]
b="0"
if(a=="0"):
b="1"
for j in range(len(s)):
arr.append(a)
arr.append(b)
print(*arr,sep="")
```
No
| 92,788 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.
You are given string t consisting only of 0's and 1's and you need to find such string s that:
1. String s consists only of 0's and 1's;
2. The length of s doesn't exceed 2 ⋅ |t|;
3. String t is a subsequence of string s;
4. String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".
Input
The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
Input
4
00
01
111
110
Output
00
01
11111
1010
Note
In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
Submitted Solution:
```
for _ in range(int(input())):
t=input()
n=len(t)
n1=t.count('1')
n0=n-n1
if n1==0 or n0==0:
print(t)
else:
k=max(n1,n0)
if k==n//2:
k+=1
print('10'*k)
```
No
| 92,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
t=int(input())
while t>0 :
n=int(input())
l=list(map(int,input().split()))
ans=[0,abs(l[1]-l[0])]
a=[l[0]]
for i in range(2,n) :
if ans[i-1]+abs(l[i]-l[i-1]) > ans[i-2]+abs(l[i]-l[i-2]) :
ans.append(ans[i-1]+abs(l[i]-l[i-1]))
a.append(l[i-1])
else :
ans.append(ans[i-2]+abs(l[i]-l[i-2]))
a.append(l[-1])
print(len(a))
for i in a :
print(i,end=" ")
print()
#print(ans)
#print(ans[n-1])
t-=1
```
| 92,790 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
for T in range(int(input())) :
n = int(input())
arr = list(map(int ,input().split()))
inc = False
dec = False
res =[arr[0]]
for i in range(1,n):
if arr[i] < arr[i-1] :
if dec :
res.pop()
res.append(arr[i])
dec = True
inc = False
elif arr[i] > arr[i-1]:
if inc :
res.pop()
res.append(arr[i])
dec = False
inc = True
print(len(res))
print(*res)
```
| 92,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
ar=list(map(int,input().split()))
ans=[ar[0]]
for i in range(1,n-1):
if ar[i]>ar[i-1] and ar[i]>ar[i+1]:
ans.append(ar[i])
elif ar[i]<ar[i-1] and ar[i]<ar[i+1]:
ans.append(ar[i])
ans.append(ar[-1])
print(len(ans))
for i in ans:
print(i,end= ' ')
print('')
```
| 92,792 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
l = list(map(int,input().split()))
k = [] # new list after removing
c = ''
if l[0] < l[1]:
c = 'i' # increasing
else:
c = 'd' # decreasing
# x = 1 # start of sequence
# count = 0
# for i in range(0,n-1):
# if (l[i] <= l[i+1] and c == 'd') or (l[i] >= l[i+1] and c == 'i'): # change in increse/decrease
# # sequence: 1 -> i
# k.append(str(x)) # 1st element
# k.append(str(i)) # last element
# x = i # reset first element to last element of previous sequence
# count += 1
#
# if count == 0:
# print(2)
# print(l[0],l[-1])
k.append(l[0])
k.append(l[1])
for i in range(1,n-1):
if l[i] <= l[i+1]:
if c == 'i':
k[len(k)-1] = l[i+1]
else:
k.append(l[i+1])
c = 'i'
elif l[i] >= l[i+1]:
if c == 'd':
k[len(k)-1] = l[i+1]
else:
k.append(l[i+1])
c = 'd'
print(len(k))
print(' '.join(map(str,k)))
# 3 7 2 4 5 6 1
# 4 5 2 1 1 5 = 18
# 3 7 2 x x 6 1
```
| 92,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
import sys
max_int = 1000000001 # 10^9+1
min_int = -max_int
t = int(input())
for _t in range(t):
n = int(sys.stdin.readline())
p = list(map(int, sys.stdin.readline().split()))
d = prev_d = 0
out = [p[0]]
for i in range(1, n):
if p[i] == p[i - 1]:
continue
if p[i] > p[i - 1]:
d = 1
else:
d = -1
if not prev_d:
prev_d = d
if d and d != prev_d:
out.append(p[i - 1])
prev_d = d
out.append(p[-1])
print(len(out))
print(' '.join(map(str, out)))
```
| 92,794 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from itertools import (
accumulate, combinations, combinations_with_replacement, groupby,
permutations, product)
from math import (acos, asin, atan, ceil, cos, degrees, factorial, gcd, hypot,
log2, pi, radians, sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits
def inp():
return(int(input()))
def inlist():
return(list(map(int, input().split())))
def instr():
s = input()
return(list(s[:len(s)]))
def invr():
return(map(int, input().split()))
t = inp()
for _ in range(t):
n = inp()
a = inlist()
res = []
res.append(a[0])
flip = 0
if a[1] < a[0]:
flip = 1
for i in range(1, n):
if flip == 0 and a[i] < a[i-1]:
res.append(a[i-1])
flip = 1
elif flip == 1 and a[i] >= a[i-1]:
res.append(a[i-1])
flip = 0
res.append(a[n-1])
print(len(res))
print(*res)
```
| 92,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
import sys
from sys import stdin
input = sys.stdin.readline
for _ in range(int(input())):
n=int(input())
arr=[int(j) for j in input().split()]
res=[]
res.append(arr[0])
count=0
for i in range(1,n-1):
if (arr[i-1]<arr[i] and arr[i]>arr[i+1]) or (arr[i-1]>arr[i] and arr[i]<arr[i+1]) :
res.append(arr[i])
res.append(arr[-1])
print(len(res))
print(*res)
```
| 92,796 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Tags: greedy, two pointers
Correct Solution:
```
t = int(input())
for _ in range(t):
n1 = int(input())
a= list(map(int,input().split()))
g = []
inc = -1
c=0
n=0
# f = a[i-1]
for i in range(1,n1):
if a[i]>a[i-1]:
if n>0:
g.append(f)
# g.append(a[i-1])
n=0
c=1
inc=0
f=a[i-1]
elif inc==0:
c+=1
elif c==0:
inc=0
c=1
f = a[i-1]
else:
if c>0:
g.append(f)
# g.append(a[i-1])
inc=1
c=0
n=1
inc=1
f=a[i-1]
elif inc==1:
n+=1
elif n==0:
inc=1
n=1
f=a[i-1]
if n==0 and c>=1:
g.append(f)
g.append(a[i])
if c==0 and n>=1:
g.append(f)
g.append(a[i])
print(len(g))
print(*g, sep=" ")
```
| 92,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Submitted Solution:
```
def main():
t = int(input())
for i in range(t):
solve()
def solve():
n = int(input())
arr = list(map(int, input().split(" ")))
if n == 2:
print(n)
print(*arr)
return
prev = arr.pop(0)
numbers = [prev]
amount = 1
while len(arr) > 1:
current = arr.pop(0)
if (current > numbers[amount - 1] and current > arr[0]) or (current < numbers[amount - 1] and current < arr[0]):
amount += 1
prev = current
numbers.append(current)
numbers.append(arr.pop(0))
amount += 1
print(amount)
print(*numbers)
main()
```
Yes
| 92,798 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2.
* Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.
A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.
Input
The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.
The sum of n across the test cases doesn't exceed 10^5.
Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.
Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2
Note
In the first test case, there are 4 subsequences of length at least 2:
* [3,2] which gives us |3-2|=1.
* [3,1] which gives us |3-1|=2.
* [2,1] which gives us |2-1|=1.
* [3,2,1] which gives us |3-2|+|2-1|=2.
So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
def YES(c): return IF(c, "YES", "NO")
def main():
t = I()
rr = []
for _ in range(t):
n = I()
a = LI()
t = set()
for i in range(1,n-1):
if a[i] == a[i-1]:
t.add(i)
elif a[i-1] < a[i] < a[i+1]:
t.add(i)
elif a[i-1] > a[i] > a[i+1]:
t.add(i)
if a[-1] == a[-2]:
t.add(n-1)
r = [a[i] for i in range(n) if i not in t]
rr.append(len(r))
rr.append(JA(r, " "))
return JA(rr, "\n")
print(main())
```
Yes
| 92,799 |
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