text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
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, β¦, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+β¦+|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 β€ t β€ 2 β
10^4) β the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 β€ n β€ 10^5) β the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, β¦, p_{n} (1 β€ p_i β€ n, p_i are distinct) β 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, β¦, s_k β 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:
```
t = int(input())
for i in range(t):
n= int(input())
ar = [int(i) for i in input().split()]
seq= [ar[0]]
run = ar[0]
count = False
for j in range(n-1):
if (run>ar[j+1]):
run = ar[j+1]
if count==False:
seq.append(run)
sign=False
count=True
elif sign==False:
seq[-1]=run
else:
seq.append(run)
sign=False
elif (run<ar[j+1]):
run = ar[j+1]
if count== False:
seq.append(run)
sign=True
count=True
elif sign==True:
seq[-1]=run
else:
seq.append(run)
sign=True
else:
pass
print(len(seq))
print(' '.join(str(i) for i in seq))
```
Yes
| 92,800 |
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, β¦, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+β¦+|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 β€ t β€ 2 β
10^4) β the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 β€ n β€ 10^5) β the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, β¦, p_{n} (1 β€ p_i β€ n, p_i are distinct) β 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, β¦, s_k β 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
t = int(input())
for cnt in range(t):
n = map(int, input().split())
a = list(map(int, input().split()))
num = max(a)
b = list()
sum = 0
for i in a:
temp = len(b)
if(temp == 0):
b.append(i)
if(temp>=1):
if(i!=b[temp-1]):
b.append(i)
temp = len(b)
if(temp>2):
if(abs(b[temp-1]-b[temp-2]) +abs(b[temp-2]-b[temp-3]) <= abs(b[temp-1]-b[temp-3])):
b.pop(temp-2)
print(len(b))
print(*b,sep=" ")
```
Yes
| 92,801 |
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, β¦, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+β¦+|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 β€ t β€ 2 β
10^4) β the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 β€ n β€ 10^5) β the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, β¦, p_{n} (1 β€ p_i β€ n, p_i are distinct) β 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, β¦, s_k β 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:
```
#BECAUSE ONE TRUTH PREVAILS
import sys
input=sys.stdin.readline
from collections import defaultdict
for _ in range(int(input())):
N=int(input())
L=list(map(int,input().split()))
i=0
ans=[]
while i<N-1:
pp=L[i]
ans.append(pp)
if L[i]<L[i+1]:
while i<N-1 and L[i]<L[i+1]:
i+=1
ans.append(L[i])
else:
while i<N-1 and L[i]>L[i+1]:
i+=1
ans.append(L[i])
i+=1
if(ans[-1]!=L[N-1]):
ans.append(L[N-1])
print(len(ans))
print(*ans)
```
No
| 92,802 |
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, β¦, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+β¦+|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 β€ t β€ 2 β
10^4) β the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 β€ n β€ 10^5) β the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, β¦, p_{n} (1 β€ p_i β€ n, p_i are distinct) β 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, β¦, s_k β 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:
```
for i in range(int(input())):
a=int(input())
b=list(map(int,input().strip().split()))[:a]
j=0
while j<len(b)-2:
if b[j]==b[j+1]:
b.remove(b[j])
elif b[j]>b[j+1]>b[j+2]:
b.remove(b[j+1])
elif b[j]<b[j+1]<b[j+2]:
b.remove(b[j+1])
else:
j+=1
if j+2==len(b):
break
for k in b:
print(k,end=" ")
print(" ")
```
No
| 92,803 |
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, β¦, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+β¦+|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 β€ t β€ 2 β
10^4) β the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 β€ n β€ 10^5) β the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, β¦, p_{n} (1 β€ p_i β€ n, p_i are distinct) β 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, β¦, s_k β 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:
```
for _ in range(int(input())):
n = int(input())
l = list(map(int,input().split()))
l1 = l.copy()
if n==2:
print(n)
print(*l)
else:
for i in range(n-2):
if ((l[i]-l[i+1])>0 and (l[i+1]-l[i+2])>0) or ((l[i]-l[i+1])<0 and (l[i+1]-l[i+2])<0) or ((l[i]-l[i+1])+(l[i+1]-l[i+2])==0):
l1.remove(l[i+1])
print(*l1)
```
No
| 92,804 |
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, β¦, s_k of length at least 2 such that:
* |s_1-s_2|+|s_2-s_3|+β¦+|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 β€ t β€ 2 β
10^4) β the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer n (2 β€ n β€ 10^5) β the length of the permutation p.
The second line of each test case contains n integers p_1, p_2, β¦, p_{n} (1 β€ p_i β€ n, p_i are distinct) β 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, β¦, s_k β 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:
```
t = int(input())
for _ in range(t):
# n, x = map(int, input().split(' '))
# a = list(map(int, input().split(' ')))
n = int(input())
a = list(map(int, input().split(' ')))
ans = [a[0]]
if a[0] > a[1]:
temp = 'small'
else:
temp = 'big'
for i in range(1, n-1):
if (temp == 'big'):
if (a[i] < a[i+1]):
continue
else:
ans.append(a[i])
else:
if (a[i] > a[i+1]):
continue
else:
ans.append(a[i])
if a[i] > a[i+1]:
temp = 'small'
else:
temp = 'big'
ans.append(a[-1])
print(*ans)
```
No
| 92,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
def ok(a,b,n):
for i in range(n):
if(a[i]>b[i]):
return False
return True
MAX = 100001
parent = [0] * MAX
Rank = [0] * MAX
def find(x):
if parent[x] == x:
return x
else:
return find(parent[x])
def merge(r1, r2):
if (r1 != r2):
if (Rank[r1] > Rank[r2]):
parent[r2] = r1
Rank[r1] += Rank[r2]
else:
parent[r1] = r2
Rank[r2] += Rank[r1]
def minimumOperations(s1, s2):
for i in range(1, 26 + 1):
parent[i] = i
Rank[i] = 1
ans = []
for i in range(len(s1)):
if (s1[i] != s2[i]):
if (find(ord(s1[i]) - 96) !=
find(ord(s2[i]) - 96)):
x = find(ord(s1[i]) - 96)
y = find(ord(s2[i]) - 96)
merge(x, y)
ans.append([s1[i], s2[i]])
print(len(ans))
for _ in range(int(input())):
n = int(input())
s = [i for i in input()]
t = [i for i in input()]
if(ok(s,t,n)):
minimumOperations("".join(s),"".join(t))
else:
print(-1)
```
| 92,806 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
T=int(input())
for _ in range(T):
n=int(input())
a=input()
b=input()
flag=0
for i in range(n):
if (a[i]>b[i]):
flag=1
break
if (flag==1):
print(-1)
else:
G={}
for i in range(97,97+20):
G[chr(i)]=[]
for i in range(n):
if (a[i]!=b[i]):
G[a[i]].append(b[i])
G[b[i]].append(a[i])
v=0
vis=[-1 for i in range(20)]
for i in range(20):
if (vis[i]==-1):
A=[chr(97+i)]
T1=[]
while(len(A)!=0):
a=A.pop()
vis[ord(a)-97]=0
T1.append(a)
for i in G[a]:
if (vis[ord(i)-97]==-1):
vis[ord(i)-97]=0
A.append(i)
v=v+len(T1)-1
print(v)
```
| 92,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
def x():
n = int(input())
ans = 0
A = list(input())
B = list(input())
G = []
ver = set()
for i in range(n):
a = ord(A[i]) - ord('a')
b = ord(B[i]) - ord('a')
ver.add(a)
ver.add(b)
if a <= b:
G.append([a, b])
else:
print(-1)
return 0
G.sort()
res = []
tree_id = [i for i in range(20)]
for i in range(n):
a = G[i][0]
b = G[i][1]
if tree_id[a] != tree_id[b]:
res.append((a, b))
old_id = tree_id[b]
new_id = tree_id[a]
for j in range(20):
if tree_id[j] == old_id:
tree_id[j] = new_id
print(len(res))
for q in range(int(input())):
x()
```
| 92,808 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
def found(x):
global uf
temp = x
while uf[x] != x:
x = uf[x]
ans = x
while uf[temp] != temp:
parent = uf[temp]
uf[temp] = ans
temp = parent
return ans
inp = lambda : list(map(lambda x: ord(x) - 97, list(input())))
t = int(input())
for _ in range(t):
n = int(input())
a = inp()
b = inp()
res = 0
for i in range(n):
if a[i] > b[i]:
res = -1
break
if res == -1:
print(-1)
continue
edges = [[] for _ in range(20)]
for i in range(n):
if a[i] != b[i] and not b[i] in edges[a[i]]:
edges[a[i]].append(b[i])
uf = [i for i in range(20)]
for i in range(20):
if len(edges[i]) == 0:
continue
for j in edges[i]:
temp1 = found(i)
temp2 = found(j)
if temp1 != temp2:
uf[temp1] = temp2
res += 1
print(res)
```
| 92,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
import sys
import math
import heapq
import bisect
from collections import Counter
from collections import defaultdict
from io import BytesIO, IOBase
import string
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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
self.BUFSIZE = 8192
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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 = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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:
self.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")
def get_int():
return int(input())
def get_ints():
return list(map(int, input().split(' ')))
def get_int_grid(n):
return [get_ints() for _ in range(n)]
def get_str():
return input().split(' ')
def yes_no(b):
if b:
return "YES"
else:
return "NO"
def binary_search(good, left, right, delta=1, right_true=False):
"""
Performs binary search
----------
Parameters
----------
:param good: Function used to perform the binary search
:param left: Starting value of left limit
:param right: Starting value of the right limit
:param delta: Margin of error, defaults value of 1 for integer binary search
:param right_true: Boolean, for whether the right limit is the true invariant
:return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
alternatively returns False if no such value found
"""
limits = [left, right]
while limits[1] - limits[0] > delta:
if delta == 1:
mid = sum(limits) // 2
else:
mid = sum(limits) / 2
if good(mid):
limits[int(right_true)] = mid
else:
limits[int(~right_true)] = mid
if good(limits[int(right_true)]):
return limits[int(right_true)]
else:
return False
def prefix_sums(a, drop_zero=False):
p = [0]
for x in a:
p.append(p[-1] + x)
if drop_zero:
return p[1:]
else:
return p
def prefix_mins(a, drop_zero=False):
p = [float('inf')]
for x in a:
p.append(min(p[-1], x))
if drop_zero:
return p[1:]
else:
return p
class DSU:
# Disjoint Set Union (Union-Find) Data Structure
def __init__(self, nodes):
# Parents
self.p = {i: i for i in nodes}
# Ranks
self.r = {i: 0 for i in nodes}
# Sizes
self.s = {i: 1 for i in nodes}
def get(self, u):
# Recursive Returns the identifier of the set that contains u, includes path compression
if u != self.p[u]:
self.p[u] = self.get(self.p[u])
return self.p[u]
def union(self, u, v):
# Unites the sets with identifiers u and v
u = self.get(u)
v = self.get(v)
if u != v:
if self.r[u] > self.r[v]:
u, v = v, u
self.p[u] = v
if self.r[u] == self.r[v]:
self.r[v] += 1
self.s[v] += self.s[u]
def solve():
n = get_int()
s1 = input().strip()
s2 = input().strip()
G = defaultdict(list)
for letter in list(string.ascii_lowercase)[:20]:
top_set = set(s2[i] for i in range(n) if s1[i] == letter)
if len(top_set) == 0:
pass
else:
for item in top_set:
if item < letter:
return -1
elif item > letter:
G[item].append(letter)
G[letter].append(item)
dsu = DSU(G.keys())
for x in G:
for y in G[x]:
dsu.union(x, y)
S = 0
for x in G:
if dsu.p[x] == x:
S += dsu.s[x] - 1
return S
t = get_int()
for _ in range(t):
print(solve())
```
| 92,810 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
t=int(input())
for j in range(t):
n=int(input())
a=input()
b=input()
# First constructing graph
gr=dict(list())
flag=0
for i in range(n):
if ord(a[i])>ord(b[i]):
flag=1
break
if gr.get(a[i],0)==0:
gr[a[i]]=[]
if gr.get(b[i],0)==0:
gr[b[i]]=[]
gr[a[i]].append(b[i])
gr[b[i]].append(a[i])
if flag==1:
print(-1)
else:
# print(gr)
ans=0
visited=[0 for i in range(20)]
def dfs(ii):
visited[ii-ord('a')]=1
# fgh=gr.get(chr(ii))
for j in gr.get(chr(ii),[]):
if visited[ord(j)-ord('a')]==0:
dfs(ord(j))
for i in range(ord('a'),ord('t')+1):
if visited[i-ord('a')]!=1:
dfs(i)
# print(chr(i))
ans+=1
print(20-ans)
```
| 92,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n, a, b, ans = int(input()), [char for char in input().strip()], [char for char in input().strip()], 0
for i in range(n):
if a[i] > b[i]:
ans = -1
if ans != -1:
for char in "abcdefghijklmnopqrst":
s = set()
for i in range(n):
if char == b[i] and a[i] < b[i]:
s.add(a[i]) # need to convert from
for i in range(n):
if a[i] in s:
a[i] = char
ans += len(s)
print(ans)
```
| 92,812 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(input())
b=list(input())
for i in range(n):
if a[i]>b[i]:
print(-1)
break
else:
s=sorted(list(set(a)|set(b)))
ans=0
for j in s:
f=False
x=set()
for k in range(n):
if a[k]==j and a[k]!=b[k]:
x.add(b[k])
f=True
x=sorted(list(x))
for k in range(n):
if a[k]==j and a[k]!=b[k]:
a[k]=x[0]
if f:ans+=1
print(ans)
```
| 92,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
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")
# --------------------------------------------------binary------------------------------------
for ik in range(int(input())):
n=int(input())
s1=input()
l=input()
tr2=defaultdict(set)
tr1=defaultdict(set)
f=0
for i in range(n):
if s1[i]>l[i]:
print(-1)
f=1
break
elif s1[i]==l[i]:
continue
else:
tr2[s1[i]].add(l[i])
tr1[l[i]].add(s1[i])
if f==1:
continue
ans=0
for i in sorted(tr2.keys(),reverse=True):
ans+=len(tr2[i])
for j in tr2[i]:
for k in sorted(tr2.keys()):
if i==k:
break
if j in tr2[k]:
tr2[k].remove(j)
tr2[k].add(i)
print(ans)
```
Yes
| 92,814 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
import os
import sys
debug = True
if debug and os.path.exists("input.in"):
input = open("input.in", "r").readline
else:
debug = False
input = sys.stdin.readline
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return s[:len(s) - 1] # Remove line char from end
def invr():
return (map(int, input().split()))
test_count = int(input())
for t in range(test_count):
if debug:
print("Test Case #", t + 1)
# Start code here
n = inp()
a = list(insr())
b = list(insr())
move = 0
possible = True
for i in range(0, n):
if b[i] < a[i]:
possible = False
break
if not possible:
print(-1)
else:
for i in range(0, 20):
ch = chr(ord("a") + i)
isAny = False
minimum = 'z'
for j in range(0, n):
if a[j] == ch and b[j] > a[j]:
minimum = min(minimum, b[j])
isAny = True
if isAny:
for j in range(0, n):
if a[j] == ch and b[j] > a[j]:
a[j] = minimum
move += 1
# print(minimum, a)
print(move)
```
Yes
| 92,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
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 Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
# ls = list(map(int, input().split()))
# n, k = map(int, input().split())
# n =int(input())
# e=list(map(int, input().split()))
# n =int(input())
from collections import Counter
for _ in range(int(input())):
n = int(input())
s=list(input())
t=input()
f=0
d={}
ls=[[]]*26
for i in range(n):
if ord(s[i])>ord(t[i]):
f=1
break
cnt=0
if f==1:
print(-1)
else:
for val in range(97, ord("t") + 1):
char = chr(val)
ls = []
y = "z"
for i in range(n):
if s[i] == char and t[i] > s[i]:
ls.append(i)
y = min(t[i], y)
if ls != []:
cnt += 1
for j in ls:
s[j] = y
print(cnt)
```
Yes
| 92,816 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
letters = 'abcdefghijklmnopqrst'
letters_index = {}
for let in letters:
letters_index[let] = letters.index(let)
#print(letters_index)
for _ in range(t):
l = int(input())
a = input().rstrip()
b = input().rstrip()
used = {}
an = ''
bn = ''
for j in range(len(a)):
if (a[j], b[j]) in used:
pass
else:
an += a[j]
bn += b[j]
used[(a[j], b[j])] = 1
a = an
b = bn
nogo = 0
for j in range(len(a)):
x = letters_index[b[j]] - letters_index[a[j]]
if x<0:
nogo = 1
break
if nogo==1:
print(-1)
else:
c = 0
while a!=b:
tbr = ('z', 999)
for j in range(len(a)):
x = letters_index[b[j]] - letters_index[a[j]]
if x!=0:
if a[j] < tbr[0]:
tbr = (a[j], x)
if x < tbr[1] and a[j]<=tbr[0]:
tbr = (a[j], x)
old_l = tbr[0]
new_l = letters_index[old_l] + tbr[1]
new_l = letters[new_l]
av = ''
bv = ''
for kk in range(len(a)):
if a[kk] != b[kk]:
if a[kk]==old_l:
av+= new_l
else:
av+= a[kk]
bv += b[kk]
a = av
b = bv
c+=1
print(c)
```
Yes
| 92,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
for nt in range(int(input())):
n = int(input())
s = input()
t = input()
new = []
alpha = []
for i in range(97,117):
alpha.append(chr(i))
flag = 0
for i in range(n):
if t[i]<s[i]:
flag = 1
break
if s[i]!=t[i]:
new.append([s[i],t[i]])
if flag:
print (-1)
continue
new.sort(key = lambda x: (x[0],x[1]))
ans = 0
move = {}
done = {}
used = [0]*len(new)
loc = {}
for i in range(len(new)):
loc[new[i][0],new[i][1]] = i
for i in range(len(new)):
if (new[i][0],new[i][1]) in done:
continue
flag = 0
for j in alpha:
if new[i][0]==j:
break
if (j,new[i][0]) in done and not used[loc[j,new[i][0]]] and (j,new[i][1]) in done and not used[loc[j,new[i][1]]]:
flag = 1
used[loc[j,new[i][0]]] = 1
used[loc[j,new[i][1]]] = 1
break
if not flag:
ans += 1
done[new[i][0],new[i][1]] = 1
print (ans)
```
No
| 92,818 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
for nt in range(int(input())):
n = int(input())
s = input()
t = input()
new = []
flag = 0
for i in range(n):
if t[i]<s[i]:
flag = 1
break
if s[i]!=t[i]:
new.append([s[i],t[i]])
if flag:
print (-1)
continue
new.sort(key = lambda x: (x[0],x[1]))
used = [0]*n
done = {}
first = {}
ans = 0
for i in range(len(new)):
if new[i][0] not in first:
first[new[i][0]] = 1
used[i] = 1
if new[i][1] not in done:
ans += 1
done[new[i][1]] = [i]
else:
flag = 0
for k in done[new[i][1]]:
if not used[k]:
used[k] = 1
flag = 1
break
if not flag:
ans += 1
done[new[i][1]].append(i)
# print (done,ans,used)
print (ans)
```
No
| 92,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
def solve(n,a,b):
for i in range(n):
if(a[i]>b[i]):
return -1
ans=0
for j in range(20):
cr=chr(97+j)
arr=[]
x='z'
for i in range(n):
if(a[i]==cr and b[i]>cr):
x=min(a[i],x)
arr.append(i)
if(len(arr)!=0):
ans+=1
for k in arr:
a[k]=x
return ans
for _ in range(int(input())):
n=int(input())
a=list(input())
b=list(input())
print(solve(n,a,b))
```
No
| 92,820 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
"""
Satwik_Tiwari ;) .
21st july , 2020 - tuesday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
#some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
ans = 1
while(b>0):
if(b%2==1):
ans*=a
a*=a
b//=2
return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def solve():
n = int(inp())
a = list(inp())
b = list(inp())
for i in range(n):
if(a[i] > b[i]):
print(-1)
return
cnt = [{} for i in range(20)]
for i in range(n):
if(a[i] in cnt[ord(b[i])-97]):
cnt[ord(b[i])-97][a[i]] +=1
else:
cnt[ord(b[i])-97][a[i]] = 1
# for i in range(20):
# print(cnt[i])
ans = 0
for i in range(1,20):
if(len(cnt[i]) == 0):
continue
else:
ans+=1
print(ans)
# testcase(1)
testcase(int(inp()))
```
No
| 92,821 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a_1, β¦, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).
The weight of a is defined as the maximum number of elements you can remove.
You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?
Input
The first line contains two integers n and q (1 β€ n, q β€ 3 β
10^5) β the length of the array and the number of queries.
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n) β elements of the array.
The i-th of the next q lines contains two integers x and y (x, y β₯ 0 and x+y < n).
Output
Print q lines, i-th line should contain a single integer β the answer to the i-th query.
Examples
Input
13 5
2 2 3 9 5 4 6 5 7 8 3 11 13
3 1
0 0
2 4
5 0
0 12
Output
5
11
6
1
0
Input
5 2
1 4 1 2 4
0 0
1 0
Output
2
0
Note
Explanation of the first query:
After making first x = 3 and last y = 1 elements impossible to remove, a becomes [Γ, Γ, Γ, 9, 5, 4, 6, 5, 7, 8, 3, 11, Γ] (we represent 14 as Γ for clarity).
Here is a strategy that removes 5 elements (the element removed is colored in red):
* [Γ, Γ, Γ, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, Γ]
* [Γ, Γ, Γ, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, Γ]
* [Γ, Γ, Γ, 9, 4, \color{red}{6}, 5, 7, 8, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 7, \color{red}{8}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, \color{red}{7}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 3, Γ] (final state)
It is impossible to remove more than 5 elements, hence the weight is 5.
Tags: binary search, constructive algorithms, data structures, greedy, two pointers
Correct Solution:
```
class Fenwick:
def __init__(self, size):
self.size = size
self.tree = [0] * (size + 1)
def add(self, idx, val):
idx = int(idx)
while idx <= self.size:
self.tree[idx] += val
idx += idx & (-idx)
def sum(self, idx):
ret = 0
idx = int(idx)
while idx > 0:
ret += self.tree[idx]
idx -= idx & (-idx)
return ret
n, q = map(int, input().split())
A = [int(x) for x in input().split()]
A = [A[i] - (i + 1) for i in range(n)]
query = [[] for _ in range(n + 1)]
for i in range(q):
x, y = map(int, input().split())
l, r = x, n - y - 1
query[r].append((l, i))
ft = Fenwick(n + 1)
# ans = [0 for _ in range(q + 3)]
# for r in range(n):
# ob = A[r]
# if ob <= 0:
# if ft.sum(1) >= -ob:
# low, high = 0, r
# while low + 1 < high:
# mid = low + high >> 1;
# if ft.sum(mid + 1) >= -ob:
# low = mid
# else: high = mid
# idx = low
# if ft.sum(high + 1) >= -ob:
# idx = max(idx, high)
# ft.add(1, 1)
# ft.add(idx + 2, -1)
# for qr in query[r]:
# ans[qr[1]] = ft.sum(qr[0] + 1)
#
# for _ in range(q):
# print(ans[_])
ans = [0 for _ in range(q + 3)]
for r in range(n):
ob = A[r]
if ob <= 0:
if ft.sum(1) >= -ob:
low, high = 0, r
while low + 1 < high:
mid = low + high >> 1
if ft.sum(mid + 1) >= -ob:
low = mid
else: high = mid
idx = high if ft.sum(high + 1) >= -ob else low
ft.add(1, 1)
ft.add(idx + 2, -1)
for qr in query[r]:
ans[qr[1]] = ft.sum(qr[0] + 1)
for _ in range(q):
print(ans[_])
```
| 92,822 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a_1, β¦, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).
The weight of a is defined as the maximum number of elements you can remove.
You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?
Input
The first line contains two integers n and q (1 β€ n, q β€ 3 β
10^5) β the length of the array and the number of queries.
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n) β elements of the array.
The i-th of the next q lines contains two integers x and y (x, y β₯ 0 and x+y < n).
Output
Print q lines, i-th line should contain a single integer β the answer to the i-th query.
Examples
Input
13 5
2 2 3 9 5 4 6 5 7 8 3 11 13
3 1
0 0
2 4
5 0
0 12
Output
5
11
6
1
0
Input
5 2
1 4 1 2 4
0 0
1 0
Output
2
0
Note
Explanation of the first query:
After making first x = 3 and last y = 1 elements impossible to remove, a becomes [Γ, Γ, Γ, 9, 5, 4, 6, 5, 7, 8, 3, 11, Γ] (we represent 14 as Γ for clarity).
Here is a strategy that removes 5 elements (the element removed is colored in red):
* [Γ, Γ, Γ, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, Γ]
* [Γ, Γ, Γ, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, Γ]
* [Γ, Γ, Γ, 9, 4, \color{red}{6}, 5, 7, 8, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 7, \color{red}{8}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, \color{red}{7}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 3, Γ] (final state)
It is impossible to remove more than 5 elements, hence the weight is 5.
Tags: binary search, constructive algorithms, data structures, greedy, two pointers
Correct Solution:
```
from sys import stdin
def bitadd(a,w,bit):
x = a
while x <= (len(bit)-1):
bit[x] += w
x += x & (-1 * x)
def bitsum(a,bit):
ret = 0
x = a
while x > 0:
ret += bit[x]
x -= x & (-1 * x)
return ret
class RangeBIT:
def __init__(self,N,indexed):
self.bit1 = [0] * (N+2)
self.bit2 = [0] * (N+2)
self.mode = indexed
def bitadd(self,a,w,bit):
x = a
while x <= (len(bit)-1):
bit[x] += w
x += x & (-1 * x)
def bitsum(self,a,bit):
ret = 0
x = a
while x > 0:
ret += bit[x]
x -= x & (-1 * x)
return ret
def add(self,l,r,w):
l = l + (1-self.mode)
r = r + (1-self.mode)
self.bitadd(l,-1*w*l,self.bit1)
self.bitadd(r,w*r,self.bit1)
self.bitadd(l,w,self.bit2)
self.bitadd(r,-1*w,self.bit2)
def sum(self,l,r):
l = l + (1-self.mode)
r = r + (1-self.mode)
ret = self.bitsum(r,self.bit1) + r * self.bitsum(r,self.bit2)
ret -= self.bitsum(l,self.bit1) + l * self.bitsum(l,self.bit2)
return ret
n,q = map(int,stdin.readline().split());a = list(map(int,stdin.readline().split()));qs = [ [] for i in range(n+1) ];ans = [None] * q;BIT = [0] * (n+1)
for loop in range(q):x,y = map(int,stdin.readline().split());l = x+1;r = n-y;qs[r].append((l,loop))
for r in range(1,n+1):
b = r-a[r-1]
if b >= 0:
L = 1;R = r+1
while R-L != 1:
M = (L+R)//2
if bitsum(M,BIT) >= b:L = M
else:R = M
if bitsum(L,BIT) >= b:bitadd(1,1,BIT);bitadd(L+1,-1,BIT)
for ql,qind in qs[r]:ans[qind] = bitsum(ql,BIT)
for i in ans:print (i)
```
| 92,823 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a_1, β¦, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).
The weight of a is defined as the maximum number of elements you can remove.
You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?
Input
The first line contains two integers n and q (1 β€ n, q β€ 3 β
10^5) β the length of the array and the number of queries.
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n) β elements of the array.
The i-th of the next q lines contains two integers x and y (x, y β₯ 0 and x+y < n).
Output
Print q lines, i-th line should contain a single integer β the answer to the i-th query.
Examples
Input
13 5
2 2 3 9 5 4 6 5 7 8 3 11 13
3 1
0 0
2 4
5 0
0 12
Output
5
11
6
1
0
Input
5 2
1 4 1 2 4
0 0
1 0
Output
2
0
Note
Explanation of the first query:
After making first x = 3 and last y = 1 elements impossible to remove, a becomes [Γ, Γ, Γ, 9, 5, 4, 6, 5, 7, 8, 3, 11, Γ] (we represent 14 as Γ for clarity).
Here is a strategy that removes 5 elements (the element removed is colored in red):
* [Γ, Γ, Γ, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, Γ]
* [Γ, Γ, Γ, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, Γ]
* [Γ, Γ, Γ, 9, 4, \color{red}{6}, 5, 7, 8, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 7, \color{red}{8}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, \color{red}{7}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 3, Γ] (final state)
It is impossible to remove more than 5 elements, hence the weight is 5.
Tags: binary search, constructive algorithms, data structures, greedy, two pointers
Correct Solution:
```
from sys import stdin
def bitadd(a,w,bit):
x = a
while x <= (len(bit)-1):
bit[x] += w
x += x & (-1 * x)
def bitsum(a,bit):
ret = 0
x = a
while x > 0:
ret += bit[x]
x -= x & (-1 * x)
return ret
class RangeBIT:
def __init__(self,N,indexed):
self.bit1 = [0] * (N+2)
self.bit2 = [0] * (N+2)
self.mode = indexed
def bitadd(self,a,w,bit):
x = a
while x <= (len(bit)-1):
bit[x] += w
x += x & (-1 * x)
def bitsum(self,a,bit):
ret = 0
x = a
while x > 0:
ret += bit[x]
x -= x & (-1 * x)
return ret
def add(self,l,r,w):
l = l + (1-self.mode)
r = r + (1-self.mode)
self.bitadd(l,-1*w*l,self.bit1)
self.bitadd(r,w*r,self.bit1)
self.bitadd(l,w,self.bit2)
self.bitadd(r,-1*w,self.bit2)
def sum(self,l,r):
l = l + (1-self.mode)
r = r + (1-self.mode)
ret = self.bitsum(r,self.bit1) + r * self.bitsum(r,self.bit2)
ret -= self.bitsum(l,self.bit1) + l * self.bitsum(l,self.bit2)
return ret
n,q = map(int,stdin.readline().split())
a = list(map(int,stdin.readline().split()))
qs = [ [] for i in range(n+1) ]
ans = [None] * q
for loop in range(q):
x,y = map(int,stdin.readline().split())
l = x+1
r = n-y
qs[r].append((l,loop))
BIT = [0] * (n+1)
for r in range(1,n+1):
b = r-a[r-1]
if b >= 0:
L = 1
R = r+1
while R-L != 1:
M = (L+R)//2
if bitsum(M,BIT) >= b:
L = M
else:
R = M
if bitsum(L,BIT) >= b:
bitadd(1,1,BIT)
bitadd(L+1,-1,BIT)
for ql,qind in qs[r]:
ans[qind] = bitsum(ql,BIT)
for i in ans:
print (i)
```
| 92,824 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a_1, β¦, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).
The weight of a is defined as the maximum number of elements you can remove.
You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?
Input
The first line contains two integers n and q (1 β€ n, q β€ 3 β
10^5) β the length of the array and the number of queries.
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ n) β elements of the array.
The i-th of the next q lines contains two integers x and y (x, y β₯ 0 and x+y < n).
Output
Print q lines, i-th line should contain a single integer β the answer to the i-th query.
Examples
Input
13 5
2 2 3 9 5 4 6 5 7 8 3 11 13
3 1
0 0
2 4
5 0
0 12
Output
5
11
6
1
0
Input
5 2
1 4 1 2 4
0 0
1 0
Output
2
0
Note
Explanation of the first query:
After making first x = 3 and last y = 1 elements impossible to remove, a becomes [Γ, Γ, Γ, 9, 5, 4, 6, 5, 7, 8, 3, 11, Γ] (we represent 14 as Γ for clarity).
Here is a strategy that removes 5 elements (the element removed is colored in red):
* [Γ, Γ, Γ, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, Γ]
* [Γ, Γ, Γ, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, Γ]
* [Γ, Γ, Γ, 9, 4, \color{red}{6}, 5, 7, 8, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 7, \color{red}{8}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, \color{red}{7}, 3, Γ]
* [Γ, Γ, Γ, 9, 4, 5, 3, Γ] (final state)
It is impossible to remove more than 5 elements, hence the weight is 5.
Tags: binary search, constructive algorithms, data structures, greedy, two pointers
Correct Solution:
```
from sys import stdin
def bitadd(a,w,bit):
x = a
while x <= (len(bit)-1):bit[x] += w;x += x & (-1 * x)
def bitsum(a,bit):
ret = 0;x = a
while x > 0:ret += bit[x];x -= x & (-1 * x)
return ret
class RangeBIT:
def __init__(self,N,indexed):self.bit1 = [0] * (N+2);self.bit2 = [0] * (N+2);self.mode = indexed
def bitadd(self,a,w,bit):
x = a
while x <= (len(bit)-1):bit[x] += w;x += x & (-1 * x)
def bitsum(self,a,bit):
ret = 0;x = a
while x > 0:ret += bit[x];x -= x & (-1 * x)
return ret
def add(self,l,r,w):l = l + (1-self.mode);r = r + (1-self.mode);self.bitadd(l,-1*w*l,self.bit1);self.bitadd(r,w*r,self.bit1);self.bitadd(l,w,self.bit2);self.bitadd(r,-1*w,self.bit2)
def sum(self,l,r):l = l + (1-self.mode);r = r + (1-self.mode);ret = self.bitsum(r,self.bit1) + r * self.bitsum(r,self.bit2);ret -= self.bitsum(l,self.bit1) + l * self.bitsum(l,self.bit2);return ret
n,q = map(int,stdin.readline().split());a = list(map(int,stdin.readline().split()));qs = [ [] for i in range(n+1) ];ans = [None] * q;BIT = [0] * (n+1)
for loop in range(q):x,y = map(int,stdin.readline().split());l = x+1;r = n-y;qs[r].append((l,loop))
for r in range(1,n+1):
b = r-a[r-1]
if b >= 0:
L = 1;R = r+1
while R-L != 1:
M = (L+R)//2
if bitsum(M,BIT) >= b:L = M
else:R = M
if bitsum(L,BIT) >= b:bitadd(1,1,BIT);bitadd(L+1,-1,BIT)
for ql,qind in qs[r]:ans[qind] = bitsum(ql,BIT)
for i in ans:print (i)
```
| 92,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
for _ in range(int(input())):
input();a = list(map(int, input().split()))
s = sum(a)
if s==0:print('NO')
else:
print('YES')
if s>0: print(*sorted(a, reverse=True))
else: print(*sorted(a))
```
| 92,826 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
t = int(input())
q = []
for i in range(t):
n = int(input())
l = list(map(int, list(input().split())))
if sum(l) == 0:
q.append(["NO"])
else:
l.sort()
q.append(["YES"])
if sum(l) > 0:
q.append(l[::-1])
else:
q.append(l)
for i in q:
for j in i:
print(j, end=" ")
print()
```
| 92,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if sum(a) == 0:
print('NO')
elif sum(a) > 0:
print('YES')
print(*sorted(a, reverse=True))
else:
print('YES')
print(*sorted(a))
```
| 92,828 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(sorted(map(int,input().split())))
def judge(a):
cum = 0
for i in range(n):
cum += a[i]
if cum == 0:
flag = True
for k in range(i+1,n):
if a[k] != a[i]:
cum += a[k] - a[i]
tmp = a[i]
a[i] = a[k]
a[k] = tmp
flag = False
break
if flag:
return False
return True
if judge(a):
print("YES")
print(*a)
elif judge(a[::-1]):
print("YES")
print(*a[::-1])
else:
print("NO")
```
| 92,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
def function():
n=int(input())
a=list(map(int,input().split()))
temp=[]
b=[]
flag=0
if sum(a)==0:
print("NO")
else:
sum1=0
sum2=0
for i in a:
if i>0:
sum1=sum1+i
temp.append(i)
for i in a:
if i<0:
sum2=sum2+i
temp.append(i)
if abs(sum1)<abs(sum2):
temp.reverse()
# print(temp)
for i in a:
if i==0:
temp.append(i)
s_um=0
for value in temp:
s_um=s_um+value
if s_um==0:
print("NO")
flag=1
return
if flag==0:
print("YES")
print(*temp)
return
if __name__=="__main__":
tests=int(input())
for test in range(tests):
function()
```
| 92,830 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
T = int( input() )
for t in range(T):
n = int( input() )
st = input().split()
A = []
s = 0
for i in range(n):
A.append( int( st[i] ) )
s += A[i]
if s == 0:
print("NO")
continue
else:
print("YES")
if s > 0:
A = sorted(A, reverse=True)
else:
A = sorted(A)
if A[0] == 0:
for i in range(n):
if A[i] != 0:
A[0], A[i] = A[i], A[0]
break
for i in range(n):
print(A[i], end = " ")
print()
```
| 92,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
for test in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort()
if sum(a)==0:
print("NO")
elif sum(a)>0:
print("YES")
print(*a[::-1])
else:
print("YES")
print(*a)
```
| 92,832 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Tags: math, sortings
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
arr = []
if sum(a) == 0:
print("NO")
elif sum(a) > 0:
a.sort(reverse=True)
print("YES")
print(*a)
elif sum(a) < 0:
a.sort()
print("YES")
print(*a)
```
| 92,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def main():
n = int(input())
alst = list(map(int, input().split()))
total = sum(alst)
if total == 0:
print("NO")
return
print("YES")
alst.sort()
if total > 0:
print(*alst[::-1])
else:
print(*alst)
for _ in range(int(input())):
main()
```
Yes
| 92,834 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
if sum(arr) == 0:
print('NO')
elif sum(arr) > 0:
print('YES')
arr = sorted(arr, reverse=True)
print(' '.join(list(map(str, arr))))
elif sum(arr) < 0:
print('YES')
arr = sorted(arr)
print(' '.join(list(map(str, arr))))
```
Yes
| 92,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
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")
##########################################################
import math
import bisect
# for _ in range(int(input())):
from collections import Counter
# sys.setrecursionlimit(10**6)
# dp=[[-1 for i in range(n+5)]for j in range(cap+5)]
# arr= list(map(int, input().split()))
# n,m= map(int, input().split())
# arr= list(map(int, input().split()))
# for _ in range(int(input())):
import bisect
# n=int(input())
for _ in range(int(input())):
n=int(input())
#n,m=map(int, input().split())
arr= list(map(int, input().split()))
#arr=[0]*50
#ls=sorted(arr)
#ls=ls[::-1]
if sum(arr)==0:
print("NO")
else:
ne=[]
p=[]
z=[]
for i in range(n):
if arr[i]>0:
p.append(arr[i])
elif arr[i]==0:
z.append(0)
else:
ne.append(arr[i])
if sum(p)>abs(sum(ne)):
ans=p+ne+z
print("YES")
print(*ans)
elif sum(p)<abs(sum(ne)):
ans=ne+p+z
print("YES")
print(*ans)
else:
print("NO")
```
Yes
| 92,836 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
if(sum(a)==0):
print("NO")
else:
l1=[]
l2=[]
print("YES")
pos=0
neg=0
z=[]
for i in range(n):
if(a[i]>0):
l1.append(a[i])
pos=pos+a[i]
elif(a[i]<0):
l2.append(a[i])
neg=neg+a[i]
else:
z.append(a[i])
if(pos>abs(neg)):
print(*l1,*l2,*z)
else:
print(*l2,*l1,*z)
```
Yes
| 92,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Submitted Solution:
```
# ========== //\\ //|| ||====//||
# || // \\ || || // ||
# || //====\\ || || // ||
# || // \\ || || // ||
# ========== // \\ ======== ||//====||
# code
def solve():
n = int(input())
a = list(map(int, input().split()))
if sum(a) == 0:
print('NO')
return
a.sort(reverse = True)
b = []
c = []
s = 0
for i in a:
s += i
if s == 0:
s -= i
c.append(i)
else:
b.append(i)
print('YES')
print(*(b + c))
return
def main():
t = 1
t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
```
No
| 92,838 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Submitted Solution:
```
# import sys; input = sys.stdin.buffer.readline
# sys.setrecursionlimit(10**7)
from collections import defaultdict
mod = 10 ** 9 + 7; INF = float("inf")
def getlist():
return list(map(int, input().split()))
def main():
T = int(input())
for _ in range(T):
N = int(input())
A = getlist()
S = sum(A)
if S == 0:
print("NO")
elif S < 0:
A.sort()
print("YES")
print(*A)
else:
print(A.sort(reverse=True))
print("YES")
print(*A)
if __name__ == '__main__':
main()
```
No
| 92,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Submitted Solution:
```
def main():
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
if(sum(a)==0):
print('NO')
else:
print('YES')
l=[]
i,s=0,0
while(i<n):
for j in range(len(a)):
if(s+a[j]!=0):
l.append(a[j])
s+=a[j]
del a[j]
break
i+=1
print(*l)
main()
```
No
| 92,840 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n integers a_1,a_2,...,a_n.
You have to create an array of n integers b_1,b_2,...,b_n such that:
* The array b is a rearrangement of the array a, that is, it contains the same values and each value appears the same number of times in the two arrays. In other words, the multisets \\{a_1,a_2,...,a_n\} and \\{b_1,b_2,...,b_n\} are equal.
For example, if a=[1,-1,0,1], then b=[-1,1,1,0] and b=[0,1,-1,1] are rearrangements of a, but b=[1,-1,-1,0] and b=[1,0,2,-3] are not rearrangements of a.
* For all k=1,2,...,n the sum of the first k elements of b is nonzero. Formally, for all k=1,2,...,n, it must hold $$$b_1+b_2+β
β
β
+b_knot=0 .$$$
If an array b_1,b_2,..., b_n with the required properties does not exist, you have to print NO.
Input
Each test contains multiple test cases. The first line contains an integer t (1β€ t β€ 1000) β the number of test cases. The description of the test cases follows.
The first line of each testcase contains one integer n (1β€ nβ€ 50) β the length of the array a.
The second line of each testcase contains n integers a_1,a_2,..., a_n (-50β€ a_iβ€ 50) β the elements of a.
Output
For each testcase, if there is not an array b_1,b_2,...,b_n with the required properties, print a single line with the word NO.
Otherwise print a line with the word YES, followed by a line with the n integers b_1,b_2,...,b_n.
If there is more than one array b_1,b_2,...,b_n satisfying the required properties, you can print any of them.
Example
Input
4
4
1 -2 3 -4
3
0 0 0
5
1 -1 1 -1 1
6
40 -31 -9 0 13 -40
Output
YES
1 -2 3 -4
NO
YES
1 1 -1 1 -1
YES
-40 13 40 0 -9 -31
Note
Explanation of the first testcase: An array with the desired properties is b=[1,-2,3,-4]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is -1.
* The sum of the first three elements of b is 2.
* The sum of the first four elements of b is -2.
Explanation of the second testcase: Since all values in a are 0, any rearrangement b of a will have all elements equal to 0 and therefore it clearly cannot satisfy the second property described in the statement (for example because b_1=0). Hence in this case the answer is NO.
Explanation of the third testcase: An array with the desired properties is b=[1, 1, -1, 1, -1]. For this array, it holds:
* The first element of b is 1.
* The sum of the first two elements of b is 2.
* The sum of the first three elements of b is 1.
* The sum of the first four elements of b is 2.
* The sum of the first five elements of b is 1.
Explanation of the fourth testcase: An array with the desired properties is b=[-40,13,40,0,-9,-31]. For this array, it holds:
* The first element of b is -40.
* The sum of the first two elements of b is -27.
* The sum of the first three elements of b is 13.
* The sum of the first four elements of b is 13.
* The sum of the first five elements of b is 4.
* The sum of the first six elements of b is -27.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
if set(l)=={0}:
print('NO')
else:
pos,neg=[],[]
s=0
for x in range(n):
if l[x] >=0:
pos.append(l[x])
s+=l[x]
else:
neg.append(l[x])
for x in pos:
if -x not in pos:
print('YES')
pos=pos[::-1]
for x in range(len(neg)):
if neg[x]==-(s) and s!=0:
pos.append(neg[x])
pos[x],pos[x-1]=pos[x-1],pos[x]
else:pos.append(neg[x])
print(*pos)
break
else:
print('NO')
```
No
| 92,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
from sys import stdin
def main():
input = lambda: stdin.readline()[:-1]
T = int(input())
for _ in [0] * T:
N = int(input())
A = input()
ans = "".join(sorted(A))
print(ans)
main()
```
| 92,842 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
from sys import stdout,stdin,maxsize
from collections import defaultdict,deque
import math
import bisect
t=int(stdin.readline())
for _ in range(t):
n=int(stdin.readline())
#n,m=map(int,stdin.readline().split())
#l=list(map(int,stdin.readline().split()))
s=input()
print(''.join(sorted(s)))
```
| 92,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
y122=int(input())
for _ in range(y122):
s=input()
s=input()
s=sorted(s)
print("".join(s))
```
| 92,844 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
s=str(input())
countt=0
countr=0
st=""
for i in range(n):
if(s[i]=="t"):
countt+=1
elif(s[i]=="r"):
countr+=1
else:
st+=s[i]
k=countr*"r"+st+countt*"t"
print(k)
```
| 92,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
from collections import Counter
for y in range(int(input())):
n=int(input())
s=input()
dic=Counter(s)
#print(dic)
for i in dic:
if i!='t' and i!='r' and i!='y' and i!='g' and i!='u' and i!='b':
print(i*dic[i],end="")
print('r'*dic['r']+'t'*dic['t']+'g'*dic['g']+'y'*dic['y']+'b'*dic['b']+'u'*dic['u'])
```
| 92,846 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
t = int(input())
for j in range(0,t):
s = int(input())
inp = []
inp_str = input()
for i in inp_str:
inp.append(i)
l2 = ['r','y','g','u','b']
matched = False
m = '*'
for i in l2:
for k in inp:
if i==k:
matched = True
m = i
break
if matched==True:
break
count = 0
if matched==False:
print(inp_str)
else:
for i in range(0,s):
if inp[i]==m:
count+=1
inp[i] = '.'
for i in range(0,count):
print(m,end='')
for i in inp:
if i!='.':
print(i,end='')
print()
```
| 92,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
import io
import os
import sys
import math
import heapq
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
s = input().rstrip()
s = sorted(s)
print(''.join(s))
```
| 92,848 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Tags: constructive algorithms, sortings
Correct Solution:
```
inpt = input().split()
inpt = list(map(int,inpt))
for i in range(inpt[0]):
string = input()
string = input()
ans=''
print(ans.join(sorted(string)))
```
| 92,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
import sys
import math
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
def FACT(n, mod):
s = 1
facts = [1]
for i in range(1,n+1):
s*=i
s%=mod
facts.append(s)
return facts[n]
def C(n, k, mod):
return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
t = II()
for q in range(t):
n = II()
s = list(SI())
count = 0
for i in range(n):
if s[i] == 't':
s[i] = ""
count+=1
s+=['t']*count
print("".join(s))
```
Yes
| 92,850 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s=str(input())
k=sorted(s)
print("".join(k))
```
Yes
| 92,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
# import sys
# input = lambda: sys.stdin.readline().rstrip()
import sys
import math
input = sys.stdin.readline
from collections import deque
from queue import LifoQueue
def binary_search(a, n):
L = 0
R = len(a)-1
while L<=R:
mid = L+(R-L)//2
if a[mid]==n:
return mid
elif a[mid]<n:
L = mid+1
else:
R = mid-1
return -1
for _ in range(int(input())):
n = int(input())
s = input()
for i in range(n):
if s[i]=='b':
s = 'b'+s[:i]+s[i+1:]
print(s)
```
Yes
| 92,852 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=input()
n=a.count('b')
x=a.replace('b','')
ans='b'*n
ans+=x
print(ans)
```
Yes
| 92,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
s = input()
new = s[::-1]
if new == s:
new = s[-1]
new += s[:-1]
print(new)
```
No
| 92,854 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
inpt = input().split()
inpt = list(map(int,inpt))
for i in range(inpt[0]):
string = input()
string = input()
print(string.replace('trygub','gubyrt'))
```
No
| 92,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
f = input()
s = "trygub"
l=[]
l1=[]
def iS(str1,str2,l,l1):
m = len(str1)
n = len(str2)
j = 0
i = 0
while j<m and i<n:
if str1[j] == str2[i]:
l.append(i)
l1.append(str2[i])
j = j+1
i = i + 1
return j==m
def ls(s):
str1 = ""
for ele in s:
str1 += ele
return str1
if iS(s, f,l,l1):
p=list(f)
j=len(l)
p.reverse()
print(ls(p))
else:
print(f)
```
No
| 92,856 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".
You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.
In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.
We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.
Input
The first line contains a single integer t (1β€ tβ€ 100) β the number of test cases.
The first line of each test case contains a single integer n (1β€ nβ€ 200) β the length of a.
The next line contains the string a of length n, consisting of lowercase English letters.
Output
For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.
If there exist multiple possible strings b, you can print any.
Example
Input
3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh
Output
bugyrtnotna
bestcoordinator
bruhtrywatchinggura
Note
In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.
In the second test case, we did not change the order of characters because it is not needed.
In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = input()
print(str(sorted(a, reverse = True)))
```
No
| 92,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S β \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j β [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is β_{i β S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 β€ n β€ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 β€ b_i β€ 10^5).
Output
Print one integer β the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Tags: flows, math
Correct Solution:
```
from typing import NamedTuple, Optional, List, cast
class MFGraph:
class Edge(NamedTuple):
src: int
dst: int
cap: int
flow: int
class _Edge:
def __init__(self, dst: int, cap: int) -> None:
self.dst = dst
self.cap = cap
self.rev: Optional[MFGraph._Edge] = None
def __init__(self, n: int) -> None:
self._n = n
self._g: List[List[MFGraph._Edge]] = [[] for _ in range(n)]
self._edges: List[MFGraph._Edge] = []
def add_edge(self, src: int, dst: int, cap: int) -> int:
assert 0 <= src < self._n
assert 0 <= dst < self._n
assert 0 <= cap
m = len(self._edges)
e = MFGraph._Edge(dst, cap)
re = MFGraph._Edge(src, 0)
e.rev = re
re.rev = e
self._g[src].append(e)
self._g[dst].append(re)
self._edges.append(e)
return m
def get_edge(self, i: int) -> Edge:
assert 0 <= i < len(self._edges)
e = self._edges[i]
re = cast(MFGraph._Edge, e.rev)
return MFGraph.Edge(
re.dst,
e.dst,
e.cap + re.cap,
re.cap
)
def edges(self) -> List[Edge]:
return [self.get_edge(i) for i in range(len(self._edges))]
def change_edge(self, i: int, new_cap: int, new_flow: int) -> None:
assert 0 <= i < len(self._edges)
assert 0 <= new_flow <= new_cap
e = self._edges[i]
e.cap = new_cap - new_flow
assert e.rev is not None
e.rev.cap = new_flow
def flow(self, s: int, t: int, flow_limit: Optional[int] = None) -> int:
assert 0 <= s < self._n
assert 0 <= t < self._n
assert s != t
if flow_limit is None:
flow_limit = cast(int, sum(e.cap for e in self._g[s]))
current_edge = [0] * self._n
level = [0] * self._n
def fill(arr: List[int], value: int) -> None:
for i in range(len(arr)):
arr[i] = value
def bfs() -> bool:
fill(level, self._n)
queue = []
q_front = 0
queue.append(s)
level[s] = 0
while q_front < len(queue):
v = queue[q_front]
q_front += 1
next_level = level[v] + 1
for e in self._g[v]:
if e.cap == 0 or level[e.dst] <= next_level:
continue
level[e.dst] = next_level
if e.dst == t:
return True
queue.append(e.dst)
return False
def dfs(lim: int) -> int:
stack = []
edge_stack: List[MFGraph._Edge] = []
stack.append(t)
while stack:
v = stack[-1]
if v == s:
flow = min(lim, min(e.cap for e in edge_stack))
for e in edge_stack:
e.cap -= flow
assert e.rev is not None
e.rev.cap += flow
return flow
next_level = level[v] - 1
while current_edge[v] < len(self._g[v]):
e = self._g[v][current_edge[v]]
re = cast(MFGraph._Edge, e.rev)
if level[e.dst] != next_level or re.cap == 0:
current_edge[v] += 1
continue
stack.append(e.dst)
edge_stack.append(re)
break
else:
stack.pop()
if edge_stack:
edge_stack.pop()
level[v] = self._n
return 0
flow = 0
while flow < flow_limit:
if not bfs():
break
fill(current_edge, 0)
while flow < flow_limit:
f = dfs(flow_limit - flow)
flow += f
if f == 0:
break
return flow
def min_cut(self, s: int) -> List[bool]:
visited = [False] * self._n
stack = [s]
visited[s] = True
while stack:
v = stack.pop()
for e in self._g[v]:
if e.cap > 0 and not visited[e.dst]:
visited[e.dst] = True
stack.append(e.dst)
return visited
INF=10**20
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
k=max(b)
if k<=0:
print(0)
exit()
g=MFGraph(n+2)
for i in range(n):
b[i]=k-b[i]
for i in range(n):
g.add_edge(n,i,b[i])
g.add_edge(i,n+1,k)
for i in range(n):
q=[0]*101
for j in range(i-1,-1,-1):
if a[i]%a[j]==0 and q[a[j]]==0:
q[a[j]]=1
g.add_edge(j,i,INF)
print(k*n-g.flow(n,n+1))
```
| 92,858 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S β \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j β [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is β_{i β S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 β€ n β€ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 β€ b_i β€ 10^5).
Output
Print one integer β the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Tags: flows, math
Correct Solution:
```
# Author: yumtam
# Created at: 2021-03-02 20:19
def main():
n = int(input())
A = [int(t) for t in input().split()]
cost = [int(t) for t in input().split()]
g = Flow(n+2)
s, t = n, n+1
last = [-1] * 101
for i, x in enumerate(A):
for d in range(1, 101):
if x % d == 0 and last[d] >= 0:
g.add_edge(i, last[d], float('inf'))
last[x] = i
if cost[i] >= 0:
g.add_edge(s, i, cost[i])
else:
g.add_edge(i, t, -cost[i])
min_cut = g.calc(s, t)
ans = sum(max(c, 0) for c in cost) - min_cut
print(ans)
class Flow:
def __init__(self, n):
self.n = n
self.g = [dict() for _ in range(n)]
def add_edge(self, u, v, w):
self.g[u][v] = w
self.g[v][u] = 0
def bfs(self, s, t):
q = [s]
vis = [0] * self.n
vis[s] = 1
prev = [-1] * self.n
found = False
for ver in q:
for nei, w in self.g[ver].items():
if not vis[nei] and w > 0:
vis[nei] = 1
prev[nei] = ver
q.append(nei)
if nei == t:
found = True
break
if found:
break
if not vis[t]:
return 0
flow = float('inf')
ver = t
while ver != s:
p = prev[ver]
flow = min(flow, self.g[p][ver])
ver = p
ver = t
while ver != s:
p = prev[ver]
self.g[p][ver] -= flow
self.g[ver][p] += flow
ver = p
return flow
def calc(self, s, t):
res = 0
while True:
flow = self.bfs(s, t)
res += flow
if not flow:
return res
main()
```
| 92,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S β \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j β [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is β_{i β S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 β€ n β€ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 β€ b_i β€ 10^5).
Output
Print one integer β the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Tags: flows, math
Correct Solution:
```
from collections import deque
class Dinic:
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
# 1ζΉε
def add_edge(self, fr, to, cap):
forward = [to, cap, None]
forward[2] = backward = [fr, 0, forward]
self.G[fr].append(forward)
self.G[to].append(backward)
# δΈ‘ζΉε
def add_multi_edge(self, v1, v2, cap1, cap2):
edge1 = [v2, cap1, None]
edge1[2] = edge2 = [v1, cap2, edge1]
self.G[v1].append(edge1)
self.G[v2].append(edge2)
def bfs(self, s, t):
self.level = level = [None]*self.N
deq = deque([s])
level[s] = 0
G = self.G
while deq:
v = deq.popleft()
lv = level[v] + 1
for w, cap, _ in G[v]:
if cap and level[w] is None:
level[w] = lv
deq.append(w)
return level[t] is not None
def dfs(self, v, t, f):
if v == t:
return f
level = self.level
for e in self.it[v]:
w, cap, rev = e
if cap and level[v] < level[w]:
d = self.dfs(w, t, min(f, cap))
if d:
e[1] -= d
rev[1] += d
return d
return 0
def flow(self, s, t):
flow = 0
INF = 10**9 + 7
G = self.G
while self.bfs(s, t):
*self.it, = map(iter, self.G)
f = INF
while f:
f = self.dfs(s, t, INF)
flow += f
return flow
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
# s:n+1,g:0
f = Dinic(n+2)
INF = 10**15
for i in range(n):
if b[i] > 0:
f.add_edge(n+1,i+1,b[i])
elif b[i] < 0:
f.add_edge(i+1,0,-b[i])
for i in range(n):
used = [False]*101
for j in reversed(range(i)):
if a[i] % a[j] == 0 and not used[a[j]]:
f.add_edge(i+1,j+1,INF)
used[a[j]] = True
ans = sum(b[i]*(b[i] > 0) for i in range(n)) - f.flow(n+1,0)
print(ans)
```
| 92,860 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S β \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j β [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is β_{i β S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 β€ n β€ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 β€ b_i β€ 10^5).
Output
Print one integer β the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Submitted Solution:
```
print('quypnpnpnp')
```
No
| 92,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S β \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j β [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is β_{i β S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 β€ n β€ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 β€ b_i β€ 10^5).
Output
Print one integer β the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Submitted Solution:
```
n = int(input())
count = 0
for x in range (1,n-2):
for y in range (1,n-2):
if ((x % y) == 0):
count += 1
print (count % 998244353)
```
No
| 92,862 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S β \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j β [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is β_{i β S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 β€ n β€ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 β€ b_i β€ 10^5).
Output
Print one integer β the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Submitted Solution:
```
# Author: yumtam
# Created at: 2021-03-02 20:19
def main():
n = int(input())
A = [int(t) for t in input().split()]
cost = [int(t) for t in input().split()]
g = Flow(n+2)
s, t = n, n+1
last = [-1] * 101
for i, x in enumerate(A):
for d in range(1, 101):
if x % d == 0 and last[d] >= 0:
g.add_edge(i, last[d], 10**5)
last[x] = i
if cost[i] >= 0:
g.add_edge(s, i, cost[i])
else:
g.add_edge(i, t, -cost[i])
min_cut = g.calc(s, t)
ans = sum(max(c, 0) for c in cost) - min_cut
print(ans)
class Flow:
def __init__(self, n):
self.n = n
self.g = [dict() for _ in range(n)]
def add_edge(self, u, v, w):
self.g[u][v] = w
self.g[v][u] = 0
def bfs(self, s, t):
q = [s]
vis = [0] * self.n
vis[s] = 1
prev = [-1] * self.n
for ver in q:
for nei, w in self.g[ver].items():
if not vis[nei] and w > 0:
vis[nei] = 1
prev[nei] = ver
q.append(nei)
if not vis[t]:
return 0
flow = float('inf')
ver = t
while ver != s:
p = prev[ver]
flow = min(flow, self.g[p][ver])
ver = p
ver = t
while ver != s:
p = prev[ver]
self.g[p][ver] -= flow
self.g[ver][p] += flow
ver = p
return flow
def calc(self, s, t):
res = 0
while True:
flow = self.bfs(s, t)
res += flow
if not flow:
return res
main()
```
No
| 92,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S β \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j β [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is β_{i β S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 β€ n β€ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 β€ b_i β€ 10^5).
Output
Print one integer β the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key, lru_cache
import sys
input = sys.stdin.readline
M = mod = 10 ** 9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
n = val()
a = li()
b = li()
dp = [set() for i in range(n)]
for i in range(n):
dp[i].add(i)
for j in range(i - 1, -1, -1):
if j in dp[i]:continue
if a[i] % a[j] == 0:
dp[i] |= dp[j]
# print(dp)
l = []
for i in dp:
l.append(sum(b[j] for j in i))
currset = set()
for i in range(n):
if l[i] > 0:
currset |= dp[i]
print(sum(b[i] for i in currset))
```
No
| 92,864 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
k=0
s=0
while s<n:
if len(l)==0:
k=-1
break
h=max(l)
s=s+h
l.remove(h)
k=k+1
print(k)
```
| 92,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Wed Apr 15 15:11:16 2020
@author: DELL
"""
k=int(input())
l=list(map(int,input().split()))
l.sort()
c=0
u=0
o=False
for i in range(len(l)-1,-1,-1):
i=l[i]
if c>=k:
o=True
break
c+=i
u+=1
if o or c>=k:
print(u)
else:
print(-1)
```
| 92,866 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
l.sort(reverse=True)
s=0
a=0
flag=0
if n > 0:
for i in range(len(l)):
s+=l[i]
a+=1
if n<=s:
flag=1
break
if flag == 1:
print(a)
else:
print(-1)
else:
print(0)
```
| 92,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
n = int(input())
a = map(int, input().split())
if n is 0:
print('0')
exit()
c = 0
for i, v in enumerate(reversed(sorted(a))):
c += v
if c >= n:
print(i + 1)
exit()
print('-1')
```
| 92,868 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
n=int(input())
l1=list(map(int,input().split()))
l1=sorted(l1,reverse=True)
a=sum(l1)
if a<n:
print("-1")
else:
i=0
b=0
while b<n:
b+=l1[i]
i+=1
print(i)
```
| 92,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
n=int(input())
num=[int(x) for x in input().split()]
list.sort(num,reverse=True)
a=0
b=0
for i in range(len(num)):
a=a+int(num[i])
if a>=n:
b=i+1
break
if n==0:
print(0)
else:
if b==0:
print(-1)
else:
print(b)
```
| 92,870 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
k= int(input())
a = list(map(int, input().split()))
if sum(a)<k:
print(-1)
exit()
if k<=0:
print(0)
exit()
else:
a.sort(reverse = True)
sum = 0
i = 0
while(sum<k):
sum +=a[i]
i+=1
print(i)
```
| 92,871 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Tags: greedy, implementation, sortings
Correct Solution:
```
def business(n, ls):
s = 0
count = 0
if(n == 0):
return 0
s1 = sum(ls)
while(s < n):
m = max(ls)
s = s + m
i = ls.index(max(ls))
ls[i] = 0
count = count + 1
if(s == s1 and s < n):
return -1
return count
n = int(input())
ls = [int(i) for i in input().split()]
print(business(n, ls))
```
| 92,872 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
#ashu@gate22
k=int(input())
l=list(map(int,input().split()))
l.sort(reverse=True)
if k==0:
print("0")
else:
s=0
no=0
while (s<k and no<len(l)):
s+=l[no]
no+=1
if s<k:
print("-1")
else:
print(no)
```
Yes
| 92,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
k = int(input())
a = input().split()
for i in range(len(a)):
a[i] = int(a[i])
a.sort(reverse=True)
m = 0
c = 0
for i in a:
if m >= k:
break
m += int(i)
c += 1
if m < k:
c = -1
print(c)
```
Yes
| 92,874 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
k=int(input())
b=sorted(int(i) for i in input().split())
b.reverse()
sum=0
if k==0:
ans=0
else:
ans=-1
for i in range(len(b)):
sum+=b[i]
if sum>=k:
ans=i+1
break
print(ans)
```
Yes
| 92,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
k=int(input());l=sorted(list(map(int,input().split())));count=0
while k>0 and count<12:k-=l[-1];l.remove(l[-1]);count+=1
if k<=0:print(count)
else:print("-1")
```
Yes
| 92,876 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
R = lambda:list(map(int,input().split()))
k = int(input())
a = R()
a.sort(reverse=True)
z = 0
i = 0
while (z < k and i < 12):
z += a[i]
i += 1
if i == 12:
print(-1)
else:
print(i)
```
No
| 92,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
k = int(input())
a = sorted(map(int,input().split()))
r = 0
g = 0
i = 0
while g < k:
if i == 11:
print("-1")
exit()
g += a[i]
r += 1
print(r)
```
No
| 92,878 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
k = int(input())
li = [int(x) for x in input().split()]
f = False
if k==0:
print('0')
f = True
ct = 0
sum = 0
li.sort(reverse=True)
flag = False
for i in range(11,-1,-1):
if i<12 and i>-1:
sum += li[i]
else:
f = False
break
ct += 1
if sum>=k:
f = True
print(ct)
break
if not f:
print('-1')
```
No
| 92,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" β thought Petya. He know for a fact that if he fulfills the parents' task in the i-th (1 β€ i β€ 12) month of the year, then the flower will grow by ai centimeters, and if he doesn't water the flower in the i-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by k centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by k centimeters.
Input
The first line contains exactly one integer k (0 β€ k β€ 100). The next line contains twelve space-separated integers: the i-th (1 β€ i β€ 12) number in the line represents ai (0 β€ ai β€ 100).
Output
Print the only integer β the minimum number of months when Petya has to water the flower so that the flower grows no less than by k centimeters. If the flower can't grow by k centimeters in a year, print -1.
Examples
Input
5
1 1 1 1 2 2 3 2 2 1 1 1
Output
2
Input
0
0 0 0 0 0 0 0 1 1 2 3 0
Output
0
Input
11
1 1 4 1 1 5 1 1 4 1 1 1
Output
3
Note
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (k = 0). So, it is possible for Petya not to water the flower at all.
Submitted Solution:
```
k = int(input())
arr = [int(y) for y in input().split()]
arr.sort()
month_count = 0
for i in range(12):
k -= arr[::-1][i]
month_count += 1
if k <= 0:
break
print(month_count)
```
No
| 92,880 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
from sys import stdin,stdout
from heapq import heapify,heappush,heappop,heappushpop
from collections import defaultdict as dd, deque as dq,Counter as C
from bisect import bisect_left as bl ,bisect_right as br
from itertools import combinations as cmb,permutations as pmb
from math import factorial as f ,ceil,gcd,sqrt,log,inf
mi = lambda : map(int,input().split())
ii = lambda: int(input())
li = lambda : list(map(int,input().split()))
mati = lambda r : [ li() for _ in range(r)]
lcm = lambda a,b : (a*b)//gcd(a,b)
MOD=10**9+7
def soe(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return(prime)
def solve():
n,m=mi()
arr=[x for x in input().strip()]
ans=True
while(ans and m):
a2=False
s=["0" for x in range(n)]
for x in range(n):
if arr[x]=="0":
c=0
if x>0 and arr[x-1]=="1":
c+=1
if x<n-1 and arr[x+1]=="1":
c+=1
if c==1:
s[x]="1"
a2=True
else:
s[x]="1"
arr=s[:]
m-=1
ans=ans and a2
print("".join(arr))
for _ in range(ii()):
solve()
```
| 92,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt
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)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
mod = int(1e9) + 7
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def totalPrimeFactors(n):
count = 0
if (n % 2) == 0:
count += 1
while (n % 2) == 0:
n //= 2
i = 3
while i * i <= n:
if (n % i) == 0:
count += 1
while (n % i) == 0:
n //= i
i += 2
if n > 2:
count += 1
return count
# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
# spf[1] = 1
# for i in range(2, MAXN):
# spf[i] = i
# for i in range(4, MAXN, 2):
# spf[i] = 2
#
# for i in range(3, mt.ceil(mt.sqrt(MAXN))):
# if (spf[i] == i):
# for j in range(i * i, MAXN, i):
# if (spf[j] == j):
# spf[j] = i
#
#
# def getFactorization(x):
# ret = 0
# while (x != 1):
# k = spf[x]
# ret += 1
# # ret.add(spf[x])
# while x % k == 0:
# x //= k
#
# return ret
# Driver code
# precalculating Smallest Prime Factor
# sieve()
def main():
for _ in range(int(input())):
n,m = map(int, input().split())
# = int(input())
# = list(map(int, input().split()))
S = input()
s=[]
for i in S:
if i=='0' or i=='1':
s.append(int(i))
while m>0:
f=0
t=[]
for i in range(n):
if s[i]==0:
if i==0:
if s[1]==1:
t.append(1)
f=1
else:
t.append(0)
elif i==n-1:
if s[n-2]==1:
t.append(1)
f=1
else:
t.append(0)
else:
if s[i-1]+s[i+1]==1:
t.append(1)
f=1
else:
t.append(0)
else:
t.append(1)
if f==0:
break
s=[]
#print(t)
for i in t:
s.append(i)
m-=1
for i in s:
print(i, end='')
print('')
return
if __name__ == "__main__":
main()
```
| 92,882 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
import sys
from os import path
if(path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
# sys.stdin.readline().rstrip()
#n, m = map(int, input().split())
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
a = list(map(int, list(input())))
for j in range(min(m,n)):
for i in range(n):
prev = 0 if i - 1 < 0 else a[i - 1]
if prev == 2: prev = 0
next = 0 if i + 1 >= n else a[i + 1]
if next == 2: next = 0
if a[i] == 0 and (next + prev) == 1:
a[i] = 2
for i in range(n):
if a[i] == 2: a[i] = 1
print(''.join(map(str, a)))
```
| 92,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
for _ in range(int(input())):
n, m = map(int, input().split())
s = input()
s = '0' + s
l = [-1e10] * (n+2)
r = [1e10] * (n+2)
for i in range(1, n+1):
l[i] = l[i-1]
if s[i] == '1':
l[i] = i
for i in range(n, 0, -1):
r[i] = r[i+1]
if s[i] == '1':
r[i] = i
res = ''
for i in range(1, n+1):
if s[i] == '1':
res += '1'
else:
if s[i] == '0' and (l[i] + 1 < i or r[i] -1 > i) and (l[i] + m >= i or r[i] - m <= i) and (l[i] + r[i] != 2*i):
res += '1'
else:
res += '0'
print(res)
```
| 92,884 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n,m = [int(d) for d in input().split()]
s = list(input())
idx = []
for i in range(len(s)):
if (s[i] == '1'):
idx.append(i)
if (idx != []):
if (idx[0] > 0):
l = idx[0]
c = 0
while (c < min(l,m)):
s[l-c-1] = '1'
c = c + 1
for i in range(len(idx)-1):
l = idx[i]
r = idx[i+1]
c = 0
while (c < min((r-l-1)//2, m)):
s[l+c+1] = '1'
s[r-c-1] = '1'
c = c + 1
if (idx[-1] < n-1):
c = 0
l = idx[-1]
while (c < min(m,n-1-l)):
s[l+c+1] = '1'
c = c + 1
print("".join(s))
```
| 92,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
import io
import os
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from heapq import heappush, heappop, heapify
from math import gcd, inf
def solve(N, M, A):
for _ in range(M):
B = [0] * N
for i in range(N):
if A[i]:
B[i] = 1
else:
nbr = (0 <= i - 1 and A[i - 1]) + (i + 1 < N and A[i + 1])
if nbr == 1:
B[i] = 1
if A == B:
break
A = B
return "".join(map(str, A))
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
TC = int(input())
for tc in range(1, TC + 1):
(N, M) = [int(x) for x in input().split()]
A = [int(x) for x in input().decode().rstrip()]
ans = solve(N, M, A)
print(ans)
```
| 92,886 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
for __ in range(int(input())):
n, m = map(int, input().split())
s = list(input())
zeros = s.count('0')
ones = n - zeros
j = 0
while j < m and (zeros != n or ones != n):
indices = []
flag = True
for i in range(n):
if s[i] == '0':
cnt = 0
if i > 0 and s[i-1] == '1':
cnt += 1
if i < n-1 and s[i+1] == '1':
cnt += 1
if cnt == 1:
indices.append(i)
for i in indices:
flag = False
s[i] = '1'
if flag:
break
j += 1
print("".join(s))
```
| 92,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Tags: implementation
Correct Solution:
```
for _ in range(int(input())):
s = [int(i)for i in input().split()][1]
x = input()
a = -1
ans = []
ans_str = ''
try:
if '1' in x:
for i in range(len(x)):
if x[i] == '0' and a == -1:
a = i
elif a != -1 and x[i] == '1':
if a != i-1 or a == 0:
ans.append([a, i-1])
a = -1
if i == len(x)-1 and x[i] == '0':
ans.append([a, i])
if ans:
a=0
for i in ans:
q = i[1] + 1 - i[0]
if i[0] == 0:
ans_str += '1'*q if q <= s \
else '0'*(q-s) + '1'*s
a = i[1]+1
elif i[1] == len(x) -1:
ans_str += x[a:i[0]]
ans_str += '1'*q if q <= s \
else '1'*s + '0'*(q-s)
a = i[1]+1
else:
ans_str += x[a:i[0]]
if q%2==0:
ans_str += '1'*q if q/2 <= s \
else '1'*s + '0'*(q-2*s) + '1'*s
else:
ans_str += '1'*((q-1)//2) + '0' + '1'*((q-1)//2) if q/2 <= s \
else '1'*s + '0'*(q-2*s) + '1'*s
a = i[1]+1
if a != len(x):
ans_str += x[a:]
except:
ans_str = ''
print(ans_str if ans_str else x)
```
| 92,888 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect, insort
from time import perf_counter
from fractions import Fraction
import copy
from copy import deepcopy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(*ele,end="\n")
for _ in range(L()[0]):
n,m=L()
A=["0"]+list(input())+["0"]
cnt=0
while(cnt<m):
B=[A[i] for i in range(n+2)]
for i in range(1,n+1):
if A[i]=="0" and A[i-1]!=A[i+1]:
B[i]="1"
if B==A:
break
A=B[::]
cnt+=1
print("".join(B[1:n+1]))
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
```
Yes
| 92,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
t = int(input())
for j in range(t):
n,m=map(int,input().split())
a= input()
x=[]
d=len(a)
a=int(a)
for i in range(d):
x.append(a%(10))
a=a//10
x=x[::-1]
l=x
arr=l[:]
for i in range(min(m,n)):
for k in range(n):
if x[k] == 0:
if k >= 1 and k < n - 1:
if x[k-1]^x[k+1]==1:
arr[k]=1
elif k == 0:
if x[k+1]==1:
arr[k]=1
else:
if x[k-1]==1:
arr[k]=1
x=arr[:]
print(*arr,sep="")
```
Yes
| 92,890 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
'''
|\_/|
| @ @ Woof!
| <> _
| _/\------____ ((| |))
| `--' |
____|_ ___| |___.'
/_/_____/____/_______|
I am here to gaurd this code, woof!
'''
from sys import stdin, stdout
from math import ceil, floor, sqrt, log, log2, log10
from collections import Counter
int_max = float('inf')
int_min = -(float('inf'))
input = stdin.readline
def solve():
pass
t = int(input())
# t = 1
for _ in range(t):
# n = int(input())
# s = input()
n, m = map(int, input().split())
# arr = list(map(int, input().split()))
arr = list(input())
arr.pop(-1)
arr = list(map(int, arr))
output = arr[:]
z = arr.count(0)
previousZ = z
k = 0
while k < m:
# print(arr)
for i in range(n):
if i == 0:
if arr[i] == 0 and arr[i + 1] == 1:
output[i] = 1
z -= 1
elif i == n - 1:
if arr[i] == 0 and arr[i - 1] == 1:
output[i] = 1
z -= 1
else:
if arr[i] == 0 and arr[i-1] ^ arr[i + 1] == 1:
output[i] = 1
z -= 1
if previousZ == z:
break
else:
arr = output[:]
previousZ = z
k += 1
print(*output, sep="")
```
Yes
| 92,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
from collections import deque
def read_int():
return int(input())
def read_ints():
return map(int, input().split(' '))
t = read_int()
for case_num in range(t):
n, m = read_ints()
s = input()
dq = deque()
alive_day = [-1] * n
for i in range(n):
if s[i] == '1':
dq.append(i)
alive_day[i] = 0
while len(dq) > 0:
i = dq.popleft()
if alive_day[i] == -1:
continue
if alive_day[i] == m:
break
if i >= 1:
if alive_day[i - 1] == -1:
alive_day[i - 1] = alive_day[i] + 1
dq.append(i - 1)
elif alive_day[i - 1] == alive_day[i] + 1:
alive_day[i - 1] = -1
if i + 1 < n:
if alive_day[i + 1] == -1:
alive_day[i + 1] = alive_day[i] + 1
dq.append(i + 1)
elif alive_day[i + 1] == alive_day[i] + 1:
alive_day[i + 1] = -1
print(''.join(map(lambda x: '0' if x == -1 else '1', alive_day)))
```
Yes
| 92,892 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
t = int(input())
while(t > 0):
n, m = map(int,input().split(" "))
arr = list(map(int,input()))
i = 0
if arr[0] == 0 and arr[1] == 1:
arr[0] = 1
if arr[n -1] == 0 and arr[n - 2] == 1:
arr[n- 1] = 1
k = 0
changed = []
while(k<n):
changed.append(arr[k])
k = k + 1
while(i < m):
j = 1
while(j < len(arr)-1):
if arr[j] == 0:
if (((arr[j - 1] == 1) and (arr[j + 1] == 0)) or ((arr[j - 1] == 0) and (arr[j + 1] == 1))):
changed[j] =1
j = j + 1
i = i + 1
k = 0
while(k<n):
arr[k] = changed[k]
k = k + 1
print(arr)
t = t - 1
```
No
| 92,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
#DaRk DeveLopeR
import sys
#taking input as string
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
import math
from bisect import *
from heapq import *
from collections import defaultdict as dd
from collections import OrderedDict as odict
from collections import Counter as cc
from collections import deque
from itertools import groupby
sys.setrecursionlimit(20*20*20*20+10) #this is must for dfs
def solve():
n,m=takeivr()
string=list(map(str,input().strip()))
index=[]
for i in range(n):
if string[i]=='1':
index.append(i)
if len(index)==0:
print("".join(map(str,string)))
return
index.append(sys.maxsize)
k=1
for i in range(n):
if index[k]==i:
k+=1
else:
if (min(i-index[k-1],index[k]-i)<=m and (i-index[k-1])!=(index[k]-i)):
string[i]='1'
print("".join(map(str,string)))
def main():
global tt
if not ONLINE_JUDGE:
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
t = 1
t = takein()
#t = 1
for tt in range(1,t + 1):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed :",time.time() - start_time,"seconds")
sys.stdout.close()
#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def takein():
return (int(sys.stdin.readline().rstrip("\r\n")))
# input the string
def takesr():
return (sys.stdin.readline().rstrip("\r\n"))
# input int array
def takeiar():
return (list(map(int, sys.stdin.readline().rstrip("\r\n").split())))
# input string array
def takesar():
return (list(map(str, sys.stdin.readline().rstrip("\r\n").split())))
# innut values for the diffrent variables
def takeivr():
return (map(int, sys.stdin.readline().rstrip("\r\n").split()))
def takesvr():
return (map(str, sys.stdin.readline().rstrip("\r\n").split()))
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def ispalindrome(s):
return s==s[::-1]
def invert(bit_s):
# convert binary string
# into integer
temp = int(bit_s, 2)
# applying Ex-or operator
# b/w 10 and 31
inverse_s = temp ^ (2 ** (len(bit_s) + 1) - 1)
# convert the integer result
# into binary result and then
# slicing of the '0b1'
# binary indicator
rslt = bin(inverse_s)[3 : ]
return str(rslt)
def counter(a):
q = [0] * max(a)
for i in range(len(a)):
q[a[i] - 1] = q[a[i] - 1] + 1
return(q)
def counter_elements(a):
q = dict()
for i in range(len(a)):
if a[i] not in q:
q[a[i]] = 0
q[a[i]] = q[a[i]] + 1
return(q)
def string_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return(q)
def factorial(n,m = 1000000007):
q = 1
for i in range(n):
q = (q * (i + 1)) % m
return(q)
def factors(n):
q = []
for i in range(1,int(n ** 0.5) + 1):
if n % i == 0: q.append(i); q.append(n // i)
return(list(sorted(list(set(q)))))
def prime_factors(n):
q = []
while n % 2 == 0: q.append(2); n = n // 2
for i in range(3,int(n ** 0.5) + 1,2):
while n % i == 0: q.append(i); n = n // i
if n > 2: q.append(n)
return(list(sorted(q)))
def transpose(a):
n,m = len(a),len(a[0])
b = [[0] * n for i in range(m)]
for i in range(m):
for j in range(n):
b[i][j] = a[j][i]
return(b)
def power_two(x):
return (x and (not(x & (x - 1))))
def ceil(a, b):
return -(-a // b)
def seive(n):
a = [1]
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p ** 2,n + 1, p):
prime[i] = False
p = p + 1
for p in range(2,n + 1):
if prime[p]:
a.append(p)
return(a)
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
input = sys.stdin.readline
main()
```
No
| 92,894 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
t = int(input())
def should_be_ones(s):
idxs = []
for i in range(1, len(s) - 1):
if sum([int(s[i-1]), int(s[i + 1])]) == 1:
idxs.append(i)
if s[0] == '0' and s[1] == '1':
idxs.append(0)
if s[-1] == 0 and s[-2] == '1':
idxs.append(len(s) - 1)
return idxs
for _ in range(t):
n, m = list(map(int, input().split()))
s = list(input())
idxs = should_be_ones(s)
while m > 0 and len(idxs) > 0:
for idx in idxs:
s[idx] = '1'
m -= 1
idxs = should_be_ones(s)
print(''.join(s))
```
No
| 92,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.
Evolution of the array in William's cellular automaton occurs iteratively in the following way:
* If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor.
* William is a humane person so all alive elements stay alive.
Check the note section for examples of the evolution.
You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 β€ t β€ 10^3). Description of the test cases follows.
The first line of each test case contains two integers n and m (2 β€ n β€ 10^3, 1 β€ m β€ 10^9), which are the total number of cells in the array and the number of iterations.
The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
In each test case output a string of length n, made up of characters "0" and "1" β the state of the array after m iterations of evolution.
Example
Input
4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000
Output
11111001111
1110111101
10101
000
Note
Sequence of iterations of evolution for the first test case
* 01000000001 β initial state
* 11100000011 β first iteration of evolution
* 11110000111 β second iteration of evolution
* 11111001111 β third iteration of evolution
Sequence of iterations of evolution for the second test case
* 0110100101 β initial state
* 1110111101 β first iteration of evolution
* 1110111101 β second iteration of evolution
Submitted Solution:
```
from sys import stdin
for _ in range(int(stdin.readline())):
n,m=map(int,stdin.readline().split())
s=list(input())
l=[]
j=0
for i in s:
if i=='1':
l.append(j)
j+=1
#print(l)
k=0
while k<len(l):
for i in range(l[k],l[k]+m+1):
if i>=n :
break
if k+1<len(l) and i+1==l[k+1]:
pass
else:
s[i]='1'
for i in range(l[k],l[k]-m-1,-1):
if i<0:
break
if k-1>=0 and i-1==l[k-1]:
pass
else:
s[i]='1'
k+=1
print(''.join(s))
```
No
| 92,896 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has n pebbles. He arranges them in a equal rows, each row has b pebbles (a > 1). Note that the Beaver must use all the pebbles he has, i. e. n = aΒ·b.
<image> 10 pebbles are arranged in two rows, each row has 5 pebbles
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, b pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of a and b) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers c1, ..., ck, where:
* c1 = n
* ci + 1 is the number of pebbles that the Beaver ends up with after the i-th move, that is, the number of pebbles in a row after some arrangement of ci pebbles (1 β€ i < k). Note that ci > ci + 1.
* ck = 1
The result of the game is the sum of numbers ci. You are given n. Find the maximum possible result of the game.
Input
The single line of the input contains a single integer n β the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
* 2 β€ n β€ 50
The input limitations for getting 100 points are:
* 2 β€ n β€ 109
Output
Print a single number β the maximum possible result of the game.
Examples
Input
10
Output
16
Input
8
Output
15
Note
Consider the first example (c1 = 10). The possible options for the game development are:
* Arrange the pebbles in 10 rows, one pebble per row. Then c2 = 1, and the game ends after the first move with the result of 11.
* Arrange the pebbles in 5 rows, two pebbles per row. Then c2 = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) β 2 rows, one pebble per row. c3 = 1, and the game ends with the result of 13.
* Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to c2 = 5, c3 = 1, and the game ends with the result of 16 β the maximum possible result.
Tags: number theory
Correct Solution:
```
def flag(x):
# First x in the below expression
# is for the case when x is 0
return (x and (not(x & (x - 1))) )
#print(pow(2,22))
import math
def big(n):
z=int(math.sqrt(n))
for i in range(2,z+1):
if n%i==0:
return n//i
return 1
def solve(n):
if flag(n):
return n+(n-1)
else:
ans=0
while(n%2==0):
ans+=n
n=n//2
ans+=n
#print(n)
while(n!=1):
#print(big(n))
z=big(n)
n=z
#print(z)
ans+=z
return ans
n=int(input())
print(solve(n))
```
| 92,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has n pebbles. He arranges them in a equal rows, each row has b pebbles (a > 1). Note that the Beaver must use all the pebbles he has, i. e. n = aΒ·b.
<image> 10 pebbles are arranged in two rows, each row has 5 pebbles
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, b pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of a and b) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers c1, ..., ck, where:
* c1 = n
* ci + 1 is the number of pebbles that the Beaver ends up with after the i-th move, that is, the number of pebbles in a row after some arrangement of ci pebbles (1 β€ i < k). Note that ci > ci + 1.
* ck = 1
The result of the game is the sum of numbers ci. You are given n. Find the maximum possible result of the game.
Input
The single line of the input contains a single integer n β the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
* 2 β€ n β€ 50
The input limitations for getting 100 points are:
* 2 β€ n β€ 109
Output
Print a single number β the maximum possible result of the game.
Examples
Input
10
Output
16
Input
8
Output
15
Note
Consider the first example (c1 = 10). The possible options for the game development are:
* Arrange the pebbles in 10 rows, one pebble per row. Then c2 = 1, and the game ends after the first move with the result of 11.
* Arrange the pebbles in 5 rows, two pebbles per row. Then c2 = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) β 2 rows, one pebble per row. c3 = 1, and the game ends with the result of 13.
* Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to c2 = 5, c3 = 1, and the game ends with the result of 16 β the maximum possible result.
Tags: number theory
Correct Solution:
```
import sys,math,string,bisect
input=sys.stdin.readline
from collections import deque
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
n=I()
c=n
while(n!=1):
p=2
k=n
while(p<n**0.5+1):
if(n%p==0):
c+=n//p
n=n//p
break
p+=1
if(n==k):
c+=1
n=1
print(c)
```
| 92,898 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has n pebbles. He arranges them in a equal rows, each row has b pebbles (a > 1). Note that the Beaver must use all the pebbles he has, i. e. n = aΒ·b.
<image> 10 pebbles are arranged in two rows, each row has 5 pebbles
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, b pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of a and b) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers c1, ..., ck, where:
* c1 = n
* ci + 1 is the number of pebbles that the Beaver ends up with after the i-th move, that is, the number of pebbles in a row after some arrangement of ci pebbles (1 β€ i < k). Note that ci > ci + 1.
* ck = 1
The result of the game is the sum of numbers ci. You are given n. Find the maximum possible result of the game.
Input
The single line of the input contains a single integer n β the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
* 2 β€ n β€ 50
The input limitations for getting 100 points are:
* 2 β€ n β€ 109
Output
Print a single number β the maximum possible result of the game.
Examples
Input
10
Output
16
Input
8
Output
15
Note
Consider the first example (c1 = 10). The possible options for the game development are:
* Arrange the pebbles in 10 rows, one pebble per row. Then c2 = 1, and the game ends after the first move with the result of 11.
* Arrange the pebbles in 5 rows, two pebbles per row. Then c2 = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) β 2 rows, one pebble per row. c3 = 1, and the game ends with the result of 13.
* Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to c2 = 5, c3 = 1, and the game ends with the result of 16 β the maximum possible result.
Tags: number theory
Correct Solution:
```
def prime(n):
if n == 2 or n == 3:
return True
elif n%2 == 0:
return False
else:
for i in range (3,int(n**0.5)+1,2):
if n%i == 0:
return False
return True
def factors(n):
l = []
for i in range (2,int(n**0.5)+1):
if n%i == 0:
l.append(i)
if i != n**0.5:
l.append(n//i)
return l
def solution(n):
if prime(n):
return n+1
else:
ans = n
l = factors(n)
x = min(l)
y = n//x
ans = ans + y
n = y
while not prime(n):
l = factors(n)
x = min(l)
y = n//x
ans = ans + y
n = y
ans = ans + 1
return ans
n = int(input())
print(solution(n))
```
| 92,899 |
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