text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
'''
Created on 2016-4-28
@author: chronocorax
'''
from heapq import heapify, heappush, heappop
S = input()
k = int(input())
n = len(S)
H = [([S[i]], i + 1) for i in range(n)]
if k > n * (n + 1) // 2:
print('No such line.')
exit()
heapify(H)
for i in range(k):
s, nxt = heappop(H)
if nxt < n and i + 1 < k:
s.append(S[nxt])
heappush(H, (s, nxt + 1))
print(''.join(s))
```
| 5,900 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
import heapq
string = input()
k = int(input())
if k > (len(string) + 1) * len(string) / 2:
print("No such line.")
else:
substring = []
for i in range(len(string)):
heapq.heappush(substring,(string[i],i + 1))
for i in range(k):
a,b = heapq.heappop(substring)
if b < len(string):
heapq.heappush(substring,(a + string[b],b + 1))
print(a)
```
Yes
| 5,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
import heapq
s = input()
k = int(input())
N = len(s)
if k>N*(N+1)//2:
print('No such line.')
exit()
A = [(c,j) for j,c in enumerate(list(s))]
heapq.heapify(A)
for i in range(k):
if not A:
break
min_item, index = heapq.heappop(A)
if index + 1 < N:
heapq.heappush(A, (min_item + s[index+1], index+1))
print(min_item)
```
Yes
| 5,902 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
from heapq import *
string = input()
k = int(input())
m = k
t = len(string)
u = t * (t + 1) / 2
if k > u:
print('No such line.')
exit()
substring = [[string[i],i] for i in range(t)]
heapify(substring)
while m > 0:
m -= 1
a = heappop(substring)
if a[1] < t - 1:
b = a[0] + string[a[1]+1]
heappush(substring,[b, a[1] + 1])
print(a[0])
```
Yes
| 5,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
from heapq import heapify, heappush, heappop
n = input()
k = int(input())
i = 0
h = [(n[i], i + 1) for i in range(len(n))]
heapify(h)
while True:
i += 1
if (not h):
print("No such line.")
exit(0)
s, nxt = heappop(h)
if (i == k):
print(s)
exit(0)
if (nxt >= len(n)):
continue
s = s + n[nxt]
heappush(h, (s, nxt + 1))
```
Yes
| 5,904 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
from heapq import *
l=input()
k=int(input())
n=len(l)
if k>n*(n+1)/2:
print("No such line.")
quit()
ss=[(l[i],i) for i in range(n)]
while k:
k-=1
t=heappop(ss)
if k==0:
print(t[0])
else:
if t[1]<n-1:
heappush(ss,(t[0]+l[t[1]+1],t[1]+1))
```
No
| 5,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
# -*-encoding=gbk-*-
ch = input()
K = int(input() )
if len(ch)*(len(ch) + 1)/2 < K :
print ("No such line.")
else:
nn = len(ch)
pos = [i for i in range(nn)]
print(pos)
ich = [ord(i) for i in ch]
print(ich)
ans = ""
while True:
num = [0] * 128
for i in pos :
num[ ich[i] ] += nn - i
print(num[ich[i]],i)
addChar = 0
for c in range(ord('a'), ord('z')+1):
if num[c] >= K:
addChar = c
break
K -= num[c]
pos = [i for i in pos if ich[i]==addChar]
ans += chr(addChar)
if len(pos) >= K:
break
K -= len(pos)
pos = [i+1 for i in pos if i+1 < nn]
print (ans)
```
No
| 5,906 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
string = input()
n = len(string)
s = int(input())
l = []
q = 0
if n > 20:
print('a')
q = 1
for i in range (0,n):
if q == 1:
exit
for o in range (i,n):
k = string[i:o+1]
while len(k)<n:
k += "`"
kk = ""
for p in range(0,n):
kk +=str(ord(k[p])-30)
l.append(int(kk))
l.sort()
out = str(l[s-1])
out += "66"
outt = ""
for i in range(0,n):
if q==1:
exit
outt += chr(int(out[2*i:2*i+2])+30)
if int(out[2*i+2:2*i+4]) == 66:
break
if q !=1:
print(outt)
```
No
| 5,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Submitted Solution:
```
'''
Created on 2016-4-28
@author: chronocorax
'''
from heapq import heapify, heappush, heappop
S = input()
k = int(input())
n = len(S)
H = [([S[i]], i + 1) for i in range(n)]
if k > n * (n + 1) // 2:
print('No such line.')
exit()
heapify(H)
for _ in range(k):
s, nxt = heappop(H)
res = s
if nxt < n:
s.append(S[nxt])
heappush(H, (s, nxt + 1))
print(''.join(res))
```
No
| 5,908 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
for t in range(input()):
n,k=in_arr()
s=raw_input().strip()
d=defaultdict(Counter)
for i in range(n):
d[i%k][s[i]]+=1
c=n/k
ans=0
for i in range(k/2):
mn=1000000000
for j in [chr(s) for s in range(97,97+26)]:
sm=d[i][j]+d[k-i-1][j]
#print i,j,sm
if (2*c)-sm<mn:
mn=(2*c)-sm
ans+=mn
if k%2:
mn=1000000000
for j in [chr(s) for s in range(97,97+26)]:
mn=min(mn,(c)-d[(k/2)][j])
ans+=mn
pr_num(ans)
```
| 5,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
from collections import Counter
for lp in range(int(input())):
n,k=map(int,input().split())
s=input()
q,t=0,1
for j in range((k+1)//2):
l=[]
for i in range(j,n,k):l.append(s[i])
if k%2==0 or j!=(k+1)//2-1:
for i in range(k-j-1,n,k):l.append(s[i])
else:t=0
q+=(n//k)*(1+t)-Counter(l).most_common(1)[0][1]
print(q)
```
| 5,910 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
from sys import stdin
from collections import defaultdict, Counter
from math import ceil
t = int(input())
for _ in range(t):
n, k = map(int, stdin.readline().split())
S = stdin.readline().strip()
count = 0
sub_k = []
for i in range(n//k):
sub_k.append(S[k*i:k*i + k])
for i in range(int(ceil(k/2))):
c = Counter()
for j in range(n//k):
if i != k - 1 - i:
c[sub_k[j][i]] += 1
c[sub_k[j][k - 1 - i]] += 1
else:
c[sub_k[j][i]] += 1
M = max(c.values())
if i != k - 1 - i:
count += 2*(n//k) - M
else:
count += n//k - M
print (count)
```
| 5,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
#input = raw_input
from collections import Counter
def answer():
n, k = map(int, input().split(" "))
s = input()
bunkatu_s_list = [s[i:i+k] for i in range(0, n, k)]
bunkatu_n = n // k
half_s_list = []
out = 0
if k % 2 == 0:
half_k = k // 2
for i in range(half_k):
i1 = i
i2 = k - (i+1)
#tmp1 = [s[i] for i in range(i1, n, k)]
tmp1 = s[i1::k]
tmp2 = s[i2::k]
#tmp2 = [s[i] for i in range(i2, n, k)]
tmp = len(tmp1) + len(tmp2) - Counter(tmp1+tmp2).most_common(1)[0][1]
out += tmp
aidas = None
else:
half_k = (k-1) // 2
for i in range(half_k):
i1 = i
i2 = k - (i+1)
#tmp1 = [s[i] for i in range(i1, n, k)]
#tmp2 = [s[i] for i in range(i2, n, k)]
tmp1 = s[i1::k]
tmp2 = s[i2::k]
tmp = len(tmp1) + len(tmp2) - Counter(tmp1+tmp2).most_common(1)[0][1]
out += tmp
aidas = [bunkatu_s[half_k] for bunkatu_s in bunkatu_s_list]
tmp = bunkatu_n - max(Counter(aidas).values())
out += tmp
print(out)
mm = int(input())
for _ in range(mm):
answer()
```
| 5,912 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
from sys import stdin
inp = stdin.readline
t = int(inp().strip())
for c in range(t):
n, k = [int(x) for x in inp().strip().split()]
array = list(inp().strip())
d1 = {}
for i in range(n):
if i%k < k/2:
dIndex = k - 1 - i%k
else:
dIndex = i%k
if not d1.get(dIndex, 0):
d1[dIndex] = {array[i]:1}
elif not d1[dIndex].get(array[i],0):
d1[dIndex][array[i]] = 1
else:
d1[dIndex][array[i]] += 1
number = n//k
numberOfMoves = 0
for i in d1.values():
maximum = 0
for j in i.values():
if j > maximum:
maximum = j
numberOfMoves += (2 * number - maximum)
if k%2:
numberOfMoves -= number
print(numberOfMoves)
```
| 5,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
from collections import Counter
import sys
sys.setrecursionlimit(10 ** 5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
for _ in range(II()):
n,k=MI()
aa=[ord(c) for c in SI()]
fin=[False]*k
ans=0
for m1 in range(k):
if fin[m1]:continue
m2=(n-m1-1)%k
if m1==m2:
cnt = Counter(aa[m1::k])
fin[m1] = True
ans += n//k - max(cnt.values())
else:
cnt=Counter(aa[m1::k]+aa[m2::k])
fin[m1]=fin[m2]=True
ans+=2*n//k-max(cnt.values())
print(ans)
main()
```
| 5,914 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
from functools import lru_cache #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------
cnt = [[0]*26 for i in range(200005)]
par = [0]*200005
size = [0]*200005
def parent(v):
if v==par[v]:
return v
par[v]= parent(par[v])
return par[v]
def merge(u,v):
a = parent(u)
b = parent(v)
if a!=b:
if size[a]<size[b]:
a,b = b,a
par[b] = a
size[a] += size[b]
for i in range(26):
cnt[a][i]+=cnt[b][i]
for _ in range(int(input())):
n,k = mapi()
a = list(input().strip())
cnt = [[0]*26 for i in range(n+1)]
par = [0]*(n+1)
size = [0]*(n+1)
for i in range(n):
cnt[i][ord(a[i])-ord("a")]+=1
par[i] = i
size[i] = 1
for i in range(n):
if i < n-i-1:
merge(i,n-i-1)
if i+k < n:
merge(i,i+k)
res = 0
for i in range(n):
if par[i]==i:
mx = -1
for j in range(26):
mx = max(mx, cnt[i][j])
res+=size[i]-mx
print(res)
```
| 5,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
#vsInput()
for _ in range(Int()):
n,k=value()
s=input()
votes=defaultdict(lambda : defaultdict(int))
ans=defaultdict(lambda: ['',-1])
for i in range(0,n,k):
l=i
r=i+k-1
index=1
while(l<=r):
if(r!=l):
votes[index][s[l]]+=1
votes[index][s[r]]+=1
if(r==l):
votes[index][s[l]]+=1
if(votes[index][s[l]] > ans[index][1]):
ans[index][0]=s[l]
ans[index][1]=votes[index][s[l]]
if(votes[index][s[r]] > ans[index][1]):
ans[index][0]=s[r]
ans[index][1]=votes[index][s[r]]
l+=1
r-=1
index+=1
#print(votes)
#print(ans)
count=0
for i in range(0,n,k):
l=i
r=i+k-1
index=1
while(l<=r):
if(l!=r):
if(s[l]!=ans[index][0]):
count+=1
if(s[r]!=ans[index][0]):
count+=1
else:
if(s[l]!=ans[index][0]):
count+=1
l+=1
r-=1
index+=1
print(count)
```
| 5,916 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Tags: dfs and similar, dsu, greedy, implementation, strings
Correct Solution:
```
def solve(s, k):
n = len(s)
r = n//k
res = 0
for j in range(k//2):
d = [0] * 26
for i in range(r):
d[ord(s[i*k+j]) - 97] += 1
d[ord(s[i*k+k-1-j]) - 97] += 1
res += 2*r - max(d)
if k%2 == 1:
d = [0] * 26
for i in range(r):
d[ord(s[i*k+k//2]) - 97] += 1
res += r - max(d)
return res
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input()
print(solve(s, k))
```
| 5,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
from sys import stdin
inp = stdin.readline
t = int(inp().strip())
for c in range(t):
n, k = [int(x) for x in inp().strip().split()]
array = list(inp().strip())
d1 = {}
for i in range(n):
if i%k < k/2:
dIndex = k - 1 - i%k
else:
dIndex = i%k
if not d1.get(dIndex, 0):
d1[dIndex] = {array[i]:1}
elif not d1[dIndex].get(array[i],0):
d1[dIndex][array[i]] = 1
else:
d1[dIndex][array[i]] += 1
number = n//k
numberOfMoves = 0
for i in d1.keys():
maximum = 0
for j in d1[i].keys():
if d1[i][j] > maximum:
maximum = d1[i][j]
if k%2 == 1 and i == k//2:
numberOfMoves += (number - maximum)
else:
numberOfMoves += (2 * number - maximum)
print(numberOfMoves)
```
Yes
| 5,918 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
import sys
from collections import Counter
def input():
return sys.stdin.readline()[:-1]
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = input()
c = [Counter() for _ in range((k + 1)//2)]
for i in range(n):
c[min(i%k, k-1 - (i%k))][s[i]] += 1
ans = 0
for d in c:
ans += sum(d.values()) - max(d.values())
print(ans)
```
Yes
| 5,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split())
s = list(str(input()))
l1 = [0]*n
i = 0
c = 0
if n == 2:
if s[0] == s[1]:
print(0)
else:
print(1)
continue
while i<=n//2:
l2 = []
for j in range(i,n,k):
if l1[j] == 0:
l2.append(s[j])
l1[j] = 1
if l1[n-j-1] == 0:
l2.append(s[n-j-1])
l1[n-j-1] = 1
dict = {}
count, itm = 0, ''
for item in reversed(l2):
dict[item] = dict.get(item, 0) + 1
if dict[item] >= count :
count, itm = dict[item], item
aa = count
c+=(len(l2)-count)
for j in range(i,n):
if l1[j] == 1:
i+=1
if i>n//2:
break
else:
break
print(c)
```
Yes
| 5,920 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter
P=[2,3,5,7,11,13,17,19,23,29,31]
t=int(input())
for tests in range(t):
n,k=map(int,input().split())
S=input().strip()
A=[Counter() for i in range(k)]
for i in range(n):
A[min(i%k,(n-1-i)%k)][S[i]]+=1
#print(A)
ANS=0
for a in A:
if len(a)>1:
ANS+=sum(a.values())-max(a.values())
print(ANS)
```
Yes
| 5,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
from collections import defaultdict
for _ in range(int(input())):
N,K=map(int,input().split())
S=list(input())
Hash=defaultdict(lambda:0)
for i in range(N):
Hash[S[i]]+=1
count=0
for i in range(K):
if S[i]!=S[K-1-i]:
if Hash[S[i]]>Hash[S[K-1-i]]:
S[K-1-i]=S[i]
count+=1
else:
S[i]=S[K-1-i]
count+=1
for i in range(N-K-1):
if S[i+K]!=S[i]:
S[i+K]=S[i]
count+=1
print(count)
```
No
| 5,922 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
#GaandFatiPadiHaiIsliyeEasyQuestionKrnePadRheHai
#Contest668NeMaaChodDi
import math
def dfs(node):
vis[node]=1
temp[0]+=1
lis[ord(s[node])-97]+=1
for j in edge[node]:
if vis[j]==0:
dfs(j)
t=int(input())
for _ in range(t):
n,k=list(map(int,input().split()))
s=str(input())
if n==1:
print(0)
else:
vis=[0]*n
edge=[[] for i in range(n)]
for i in range(k,n):
edge[i%k].append(i)
p=math.ceil(n//2)
if n%2==0:
q=p-1
else:
q=p-2
for i in range(p,n):
if q%k!=i%k:
edge[q].append(i)
edge[i].append(q)
q-=1
ans=0
for i in range(n):
if vis[i]==0:
temp=[0]
lis=[0]*26
dfs(i)
a=max(lis)
ans+=temp[0]-a
print(ans)
```
No
| 5,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
import sys
def read(): return list(map(int,sys.stdin.readline().split()))
for _ in range(int(input())):
n, k = read()
s = sys.stdin.readline()
count = 0
for i in range(1,n-k+1):
a = {s[i-1],s[k+i-1],s[n-i]}
count += len(a)-1
for i in range(n-k+1,n):
if s[i-1] != s[n-i]:
count += 1
print(count)
```
No
| 5,924 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Word s of length n is called k-complete if
* s is a palindrome, i.e. s_i=s_{n+1-i} for all 1 ≤ i ≤ n;
* s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ n-k.
For example, "abaaba" is a 3-complete word, while "abccba" is not.
Bob is given a word s of length n consisting of only lowercase Latin letters and an integer k, such that n is divisible by k. He wants to convert s to any k-complete word.
To do this Bob can choose some i (1 ≤ i ≤ n) and replace the letter at position i with some other lowercase Latin letter.
So now Bob wants to know the minimum number of letters he has to replace to convert s to any k-complete word.
Note that Bob can do zero changes if the word s is already k-complete.
You are required to answer t test cases independently.
Input
The first line contains a single integer t (1 ≤ t≤ 10^5) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 10^5, n is divisible by k).
The second line of each test case contains a word s of length n.
It is guaranteed that word s only contains lowercase Latin letters. And it is guaranteed that the sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer, representing the minimum number of characters he has to replace to convert s to any k-complete word.
Example
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp
Output
2
0
23
16
Note
In the first test case, one optimal solution is aaaaaa.
In the second test case, the given word itself is k-complete.
Submitted Solution:
```
t = int(input())
for _ in range(t):
count=0
n,k = map(int, input().split())
l=input()
s=list(l)
for i in range(k):
d={}
j=0
while(i+(j*k)<n):
if(s[i+(j*k)] in d):
d[s[i+(j*k)]]+=1
else:
d[s[i+(j*k)]]=1
z=k-i-1
if(s[z+(j*k)] in d):
d[s[z+(j*k)]]+=1
else:
d[s[z+(j*k)]]=1
j+=1
p=max(d, key=d.get)
j=0
while(i+(j*k)<n):
if(s[i+(j*k)]!=p):
s[i+(j*k)]=p
count+=1
j+=1
#print(max(d, key=d.get),d,i)
m=''.join(s)
#print(count)
co=0
for i in range(n):
if(s[i]!=l[i]):
co+=1
print(co)
```
No
| 5,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=list(map(int,input().split()))
ans=""
if n[1]==0:
if n[0]==0:
print("1"*(n[2]+1))
else:
print("0"*(n[0]+1))
continue
cnt=0
for i in range(n[1]+1):
if cnt%2==0:
ans+="1"
else:
ans+="0"
cnt+=1
ans="1"*n[2]+ans[0]+"0"*n[0]+ans[1:]
print(ans)
```
| 5,926 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
for _ in range(int(input())):
n0, n1, n2 = map(int, input().split(' '))
if n0 > 0 and n2 > 0:
n1 -= 1
if n0 != 0:
print(0, end='')
for i in range(n0):
print(0, end='')
if n2 != 0:
print(1, end='')
for i in range(n2):
print(1, end='')
start = 1
if n2 != 0:
start = 0
if n0 == 0 and n2 == 0:
print(0, end='')
for i in range(n1):
print(start, end='')
start = (start + 1) % 2
print()
```
| 5,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def gift():
for _ in range(t):
n0,n1,n2=list(map(int,input().split()))
ans=''
if n0>0:
ans='0'+'0'*(n0)
if n1>0:
if ans=='':
ans+='0'
if n1%2==0:
ans='1'+ans
ans+='1'
ans+='01'*(n1//2-1)
else:
ans+='1'
ans+='01'*((n1+1)//2-1)
if ans=='':
ans+='1'
ans+='1'*(n2)
yield ans
if __name__ == '__main__':
t= int(input())
ans = gift()
print(*ans,sep='\n')
##100 55
##1 3 5
##33 22
##x+3*(k-x)=n
##n=3*k-2*x
##x=(3*k-n)//2
##1 3 1 3 1 3 49
```
| 5,928 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
for i in range(int(input())):
a, b, c = map(int, input().split())
s = ''
s1 = '1' * c
s0 = '0' * a
if b > 0:
s += '1'
for j in range(b):
if j % 2 == 0:
s += '0'
else:
s += '1'
s = s[0] + s0 + s[1::]
s = s1 + s
else:
s = s1 + s0
if a != 0:
s += '0'
else:
s += '1'
print(s)
```
| 5,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
tim = {'00': 0, '01': 1, '10': 1, '11': 2}
def binary_find():
def dfs(s,i,j,k):
if (i > n0) or (j > n1) or (k > n2):
return False
if (i == n0) and (j == n1) and (k == n2):
print(s)
return True
for new in range(2):
temp = s[-1] + str(new)
if temp == '00':
found = dfs(s + str(new),i+1,j,k)
elif temp in ['01','10']:
found = dfs(s + str(new),i,j+1,k)
else:
found = dfs(s + str(new),i,j,k+1)
if found :
return True
n0, n1, n2 = [int(x) for x in input().split()]
if not dfs('0',0,0,0):
dfs('1',0,0,0)
t = int(input())
for i in range(t):
binary_find()
```
| 5,930 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
for _ in range(II()):
n0,n1,n2=MI()
if n1==1:
ans="0"*(n0+1)+"1"*(n2+1)
elif n1 == n2 == 0:
ans = "0" * (n0 + 1)
elif n0==n2==0:
ans="0"
for i in range(n1):
ans+="0" if i%2 else "1"
elif n0 == n1 == 0:
ans = "1" * (n2 + 1)
elif n0 == 0:
ans = "1" * (n2 + 1)
for i in range(n1):
ans += "1" if i % 2 else "0"
elif n2==0:
ans="0"*(n0+1)
for i in range(n1):
ans+="0" if i%2 else "1"
else:
ans="0"*(n0+1)+"1"*(n2+1)
for i in range(n1-1):
ans += "1" if i % 2 else "0"
print(ans)
main()
```
| 5,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
t=int(input())
for you in range(t):
l=input().split()
n0=int(l[0])
n1=int(l[1])
n2=int(l[2])
if(n0==0 and n1==0):
for i in range(n2+1):
print(1,end="")
print()
elif(n1==0 and n2==0):
for i in range(n0+1):
print(0,end="")
print()
elif(n1%2==0):
for i in range(n1//2):
print("01",end="")
for i in range(n2):
print("1",end="")
for i in range(n0+1):
print("0",end="")
print()
else:
for i in range(n2):
print("1",end="")
for i in range((n1+1)//2):
print("10",end="")
for i in range(n0):
print("0",end="")
print()
```
| 5,932 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Tags: constructive algorithms, dfs and similar, math
Correct Solution:
```
# https://codeforces.com/problemset/problem/1352/F
def checkEven(n):
if n % 2 == 0:
return True
else:
return False
def main():
n = int(input())
arr = []
for _ in range(n):
temp = list(map(int, input().split()))
arr.append(temp)
for i in range(n):
reconstruct(arr[i][0], arr[i][1], arr[i][2])
def reconstruct(n0, n1, n2):
res = ""
if n1 == 0:
if n0 != 0:
res += "0" * (n0 + 1)
elif n2 != 0:
res += "1" * (n2 + 1)
else:
if checkEven(n1):
res += "10" * (n1 // 2)
res += "1"
else:
res += "10" * ((n1 + 1) // 2)
if n0 != 0:
res = "1" + "0" * n0 + res[1:]
if n2 != 0:
res = "1" * n2 + res
print(res)
if __name__ == "__main__":
main()
```
| 5,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
for _ in range(int(input())):
n0,n1,n2=map(int,input().split())
if n1%2==1:
print('0'*n0+'01'*(n1//2+1)+'1'*n2)
else:
if n1==0 and n2==0:
print('0'*(n0+1))
else:
print('10'*(n1//2)+'0'*n0+'1'*(n2+1))
```
Yes
| 5,934 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
def solve():
n0, n1, n2 = [int(x) for x in input().split()]
res = []
if not n1:
if n0:
res = ['0' for _ in range(n0 + 1)]
else:
res = ['1' for _ in range(n2 + 1)]
else:
res = ['0' if i & 1 else '1' for i in range(n1 + 1)]
res[1:1] = ['0' for _ in range(n0)]
res[0:0] = ['1' for _ in range(n2)]
print(''.join(res))
for _ in range(int(input())):
solve()
```
Yes
| 5,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
t = int(input())
for _ in range(t):
a, b, c = [int(x) for x in input().split()]
ans = ""
for i in range(b):
if i % 2 == 0:
ans += "1"
else:
ans += "0"
if b % 2 != 0:
ans = "1" * c + ans
ans += "0" * (a + 1)
if b % 2 == 0:
if b == 0:
if a > 0:
ans += "0" * (a + 1)
if c > 0:
ans += "1" * (c + 1)
else:
ans= "1" * c + ans
ans += "0" * a
ans += "1"
print(ans)
```
Yes
| 5,936 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
# t = int(input())
# n, k = map(int, input().split())
# a = list(map(int, input().split()))
# from collections import deque
#import copy
#import collections
#import sys
t = int(input())
for T in range(t):
n0, n1, n2 = map(int, input().split())
ans = ''
if n1==0:
if n0>0:
ans = '0'*(n0+1)
else:
ans = '1'*(n2+1)
else:
ans = '0'*(n0+1) + '1'*(n2+1)
flag = 0
for i in range(n1-1):
ans += str(flag)
flag ^= 1
print(ans)
```
Yes
| 5,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
n0 = '0'
n2 = '1'
t = int(input())
for _ in range(t):
output = ''
a, b, c = map(int, input().split())
if c: output += n2*(c+1)
if output:
for i in range(b):
output += str(i%2)
else:
if b:
for i in range(b+1):
output += str(i%2)
if b and output[-1] == '0':
output += n0*a
else:
if a: output += n0*(a+1)
print(output)
```
No
| 5,938 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
for _ in range((int)(input())):
zero,one,two=list(map(int,input().split()))
res=[]
if one%2==0:
if zero!=0:
res.append('0'*(zero+1))
one-=1
if two!=0:
res.append('1'*(two+1))
res.append('01'*(one))
else:
res.append('1'*(two))
if one!=0:
res.append('10'*(one//2+1))
res.append('0'*(zero))
print(''.join([i for i in res]))
```
No
| 5,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
for _ in range(int(input())):
n0, n1, n2 = map(int, input().split(' '))
if n0 != 0:
print(0, end='')
for i in range(n0):
print(0, end='')
if n2 != 0:
print(1, end='')
for i in range(n2):
print(1, end='')
start = 1
if n2 != 0:
start = 0
if n0 == 0 and n2 == 0:
print(0, end='')
for i in range(n1):
print(start, end='')
start = (start + 1) % 2
print()
```
No
| 5,940 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 ≤ n_0, n_1, n_2 ≤ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
n = int(input())
def check(x,y,z):
s=''
for _ in range(z):
s += ''.join('1')
for i in range(y):
if i%2!=0:
s+=''.join('1')
else:
s+=''.join('0')
for _ in range(x):
s+=''.join('0')
print(s)
for _ in range(n):
n0,n1,n2 = map(int,input().split())
check(n0,n1,n2)
```
No
| 5,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
from bisect import bisect_left, bisect_right
from sys import stdin, stdout
R = lambda : stdin.readline().strip()
RL = lambda f=None: list(map(f, R().split(' '))) if f else list(R().split(' '))
output = lambda x: stdout.write(str(x) + '\n')
output_list = lambda x: output(' '.join(map(str, x)))
n = int(R())
a = RL(int)
ass = []
for i in range(0, n, 2):
ass.append(a[i])
for i in range(1, n, 2):
ass.append(a[i])
ass += ass
k = (n+1)//2
sm = sum(ass[:k])
ans = sm
for i in range(k, len(ass)):
sm += ass[i]
sm -= ass[i-k]
ans = max(sm, ans)
output(ans)
```
| 5,942 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
n=int(input())
a=[int(x) for x in input().split()]
sumo=0
for i in range(0,n,2):
sumo+=a[i]
temo=sumo
for i in range(1,n,2):
temo+=a[i]-a[i-1]
if temo>sumo: sumo=temo
for i in range(0,n,2):
temo+=a[i]-a[i-1]
if temo>sumo: sumo=temo
print(sumo)
```
| 5,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
count = 0
for i in range(0, n, 2):
count += l[i]
m = count
for i in range(0, 2*n-2, 2):
count = count - l[i%n] + l[(i+1)%n]
if count > m: m = count
print(m)
```
| 5,944 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
n=int(input())
a=list(map(lambda x:int(x),input().split()))
new_a=[]
for i in range(0,len(a),2):
new_a.append(a[i])
for j in range(1,len(a),2):
new_a.append(a[j])
length=(n+1)//2
new_a=new_a+new_a[0:length-1]
window=new_a[0:length]
prefix_sum=[new_a[0]]
for i in range(1,len(new_a)):
prefix_sum.append(prefix_sum[i-1]+new_a[i])
s=prefix_sum[length-1]
answer=s
for i in range(length,len(new_a)):
s=s-new_a[i-length]+new_a[i]
answer=max(answer,s)
print(answer)
```
| 5,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
if n != 1:
maxCircle = 0
maxCircleCand = a[0]
for i in range(1,n,2):
maxCircleCand += a[i]
maxCircle = maxCircleCand
for i in range(1,n,2):
maxCircleCand -= a[i]
maxCircleCand += a[i+1]
if maxCircle < maxCircleCand:
maxCircle = maxCircleCand
maxCircleCand = 0
maxCircleCand += a[1]
for i in range(2,n,2):
maxCircleCand += a[i]
if maxCircle < maxCircleCand:
maxCircle = maxCircleCand
for i in range(2,n-1,2):
maxCircleCand -= a[i]
maxCircleCand += a[i+1]
if maxCircle < maxCircleCand:
maxCircle = maxCircleCand
print(maxCircle)
else:
print(a[0])
```
| 5,946 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
#!/usr/bin/pypy3
n = int(input())
a = list(map(int, input().split()))
c = 0
for i in range(0, n, 2):
c += a[i]
ans = c
for i in range(0, 2 * (n - 1), 2):
c = c + a[(i + 1) % n] - a[i % n]
ans = max(ans, c)
print(ans)
```
| 5,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
from sys import stdin, stdout
int_in = lambda: int(stdin.readline())
arr_in = lambda: [int(x) for x in stdin.readline().split()]
mat_in = lambda rows: [arr_in() for _ in range(rows)]
str_in = lambda: stdin.readline().strip()
out = lambda o: stdout.write("{}\n".format(o))
arr_out = lambda o: out(" ".join(map(str, o)))
bool_out = lambda o: out("YES" if o else "NO")
tests = lambda: range(1, int_in() + 1)
case_out = lambda i, o: out("Case #{}: {}".format(i, o))
# https://codeforces.com/contest/1372/problem/D
def solve(n, a):
if n == 1:
return a[0]
arr = []
for i in range(0, len(a), 2):
arr.append(a[i])
for i in range(1, len(a), 2):
arr.append(a[i])
arr += arr
need = (n + 1) // 2
curr = 0
for i in range(need):
curr += arr[i]
best = curr
for i in range(1, len(arr) - need):
curr -= arr[i - 1]
curr += arr[need + i - 1]
best = max(best, curr)
return best
if __name__ == "__main__":
n = int_in()
a = arr_in()
out(solve(n, a))
```
| 5,948 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Tags: brute force, dp, games, greedy
Correct Solution:
```
import sys
import math as mt
input=sys.stdin.readline
I=lambda:list(map(int,input().split()))
n,=I()
l=I()
ar=[]
for i in range(0,n,2):
ar.append(l[i])
for i in range(1,n,2):
ar.append(l[i])
k=(n+1)//2
s=sum(ar[:k])
ans=s
ar+=ar
for i in range(k,2*n):
s-=ar[(i-k)]
s+=ar[i]
ans=max(ans,s)
print(ans)
```
| 5,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
eacc = [0]
oacc = [0]
for i in range(n):
eacc.append(eacc[-1]+(a[i] if i%2==0 else 0))
oacc.append(oacc[-1]+(a[i] if i%2==1 else 0))
ans = 0
for i in range(n):
if i%2==0:
ans = max(ans, oacc[i]+eacc[n]-eacc[i])
else:
ans = max(ans, eacc[i]+oacc[n]-oacc[i])
print(ans)
```
Yes
| 5,950 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
import heapq
def solve(arr):
even_indicies = arr[0:len(arr):2]
odd_indicies = arr[1:len(arr):2]
modified_arr = even_indicies + odd_indicies + even_indicies
sum_len = (len(arr) + 1)//2
summation = sum(modified_arr[:sum_len])
max_sum = summation
for i in range(len(arr)-1):
summation = summation - modified_arr[i] + modified_arr[i+sum_len]
max_sum = max(summation, max_sum)
return max_sum
n = int(input())
arr = [int(i) for i in input().split(' ')]
print(solve(arr))
```
Yes
| 5,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a0=[a[i] for i in range(n) if i%2==0]
a1=[a[i] for i in range(n) if i%2==1]
a=a0+a1
l=n//2+1
ans=sum(a[:l])
ans_sub=sum(a[:l])
for i in range(n):
ans_sub-=a[i]
ans_sub+=a[(i+l)%n]
ans=max(ans,ans_sub)
print(ans)
```
Yes
| 5,952 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
from sys import stdin
input = stdin.buffer.readline
n = int(input())
*a, = map(int, input().split())
s1 = sum(a[i] for i in range(0, n, 2))
s2 = sum(a[i] for i in range(1, n, 2)) + a[0]
ans = max(s1, s2)
for i in range(0, n, 2):
s1 -= a[i] - a[(i + 1) % n]
ans = max(ans, s1)
for i in range(1, n, 2):
s2 -= a[i] - a[(i + 1) % n]
ans = max(ans, s2)
print(ans)
```
Yes
| 5,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
#from bisect import bisect_left as bl, bisect_right as br, insort
#from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
#INF = float('inf')
mod = int(1e9)+7
n=int(data())
a=mdata()
odd=[0]
even=[0]
for i in range(0,n-1,2):
odd.append(odd[-1]+a[i])
even.append(even[-1]+a[i+1])
odd.append(a[-1])
ans=sum(odd)
for i in range(1,n//2+1):
k=odd[i]+even[-1]-even[i-1]
ans=max(ans,k)
k=odd[-1]-odd[n//2-i-1]+even[n//2-i]
ans=max(ans,k)
out(ans)
```
No
| 5,954 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
from sys import stdin
input = stdin.buffer.readline
n = int(input())
*a, = map(int, input().split())
s1 = sum(a[i] for i in range(0, n, 2))
s2 = sum(a[i] for i in range(1, n, 2)) + a[0]
ans = max(s1, s2)
for i in range(1, n, 2):
s1 += a[i] - a[(i + 1) % n]
ans = max(ans, s1)
for i in range(0, n, 2):
s1 += a[i] - a[(i + 1) % n]
ans = max(ans, s1)
print(ans)
```
No
| 5,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
n = int(input())
l1 = [int(x) for x in input().split()]
temp=0
for i in range(len(l1)):
temp = max(temp,sum(l1[i::2])+max(l1[i-1::-2]))
print(temp)
```
No
| 5,956 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
def getn(arr , i):
if i == len(arr) -1:
i = -1
return arr[i-1] , arr[i+1]
n = int(input())
data = list(map(int , input().split()))
index = dict()
for i in range(len(data)):
index[data[i]] = i
sdata = sorted(data)
lisp = set()
total = 0
c = (n-1)//2
for i in range(len(sdata)):
if c>0:
if sdata[i] not in lisp:
ind = index[sdata[i]]
n1 , n2 = getn(data , ind)
lisp.add(n1)
lisp.add(n2)
total += sdata[i]
c-=1
print(sum(data) - total)
```
No
| 5,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
n, m = [int(i) for i in input().split(" ")]
a = [int(i) for i in input().split(" ")]
b = [int(i) for i in input().split(" ")]
ab = [b.copy() for i in a]
result = 0b1111111111
def isTrue(a, s):
return (a >> s) & 1
for s in range(9, -1, -1):
new_ab = [None for i in ab]
possible = True
for i, ca in enumerate(a):
if isTrue(ca, s):
res = [cb for cb in ab[i] if not isTrue(cb, s)]
if len(res) != 0:
new_ab[i] = res
else:
possible = False
break
else:
new_ab[i] = ab[i]
if possible:
result ^= 1 << s
ab = new_ab
print(result)
```
| 5,958 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in range(513):
found_ans = True
for j in range(n):
ok = False
for k in range(m):
if (a[j]&b[k])|i == i:
ok = True
if not ok:
found_ans = False
break
if found_ans:
print(i)
break
```
| 5,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
def found(i,j):
for k in j:
if k|i<=i:
return True
return False
n,m = map(int,input().split())
a = list(set(list(map(int,input().split()))))
b = list(set(list(map(int,input().split()))))
c = []
for i in a:
c.append([])
for j in b:
c[-1].append(i&j)
for i in range(1024):
flag = 0
for j in c:
if not found(i,j):
flag = 1
break
if not flag:
ans = i
break
print (ans)
```
| 5,960 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
import os
from io import BytesIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
def lii(): return list(map(int, input().split()))
def ii(): return int(input())
n, m = lii()
A = lii()
B = lii()
for res in range(0, 2**9):
flag = True
for a in A:
found = False
for b in B:
c = a & b
if c | res == res:
found = True
break
if not found:
flag = False
break
if flag:
print(res)
break
else:
continue
```
| 5,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
from sys import stdin,stdout
from math import gcd,sqrt,factorial
from collections import deque,defaultdict
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
lcm=lambda x,y:(x*y)//gcd(x,y)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:0 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
sm=lambda x:(x**2+x)//2
N=10**9+7
def check(i):
for p in a:
flg=False
for k in b:
if (p&k)|i==i:flg=True
if not flg:return flg
return True
m,n=R()
a=L()
b=L()
for i in range(2**9):
if check(i):exit(print(i))
```
| 5,962 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
'''
___ ___ ___ ___ ___ ___
/\__\ /\ \ _____ /\ \ /\ \ /\ \ /\__\
/:/ _/_ \:\ \ /::\ \ \:\ \ ___ /::\ \ |::\ \ ___ /:/ _/_
/:/ /\ \ \:\ \ /:/\:\ \ \:\ \ /\__\ /:/\:\__\ |:|:\ \ /\__\ /:/ /\ \
/:/ /::\ \ ___ \:\ \ /:/ \:\__\ ___ /::\ \ /:/__/ /:/ /:/ / __|:|\:\ \ /:/ / /:/ /::\ \
/:/_/:/\:\__\ /\ \ \:\__\ /:/__/ \:|__| /\ /:/\:\__\ /::\ \ /:/_/:/__/___ /::::|_\:\__\ /:/__/ /:/_/:/\:\__\
\:\/:/ /:/ / \:\ \ /:/ / \:\ \ /:/ / \:\/:/ \/__/ \/\:\ \__ \:\/:::::/ / \:\~~\ \/__/ /::\ \ \:\/:/ /:/ /
\::/ /:/ / \:\ /:/ / \:\ /:/ / \::/__/ ~~\:\/\__\ \::/~~/~~~~ \:\ \ /:/\:\ \ \::/ /:/ /
\/_/:/ / \:\/:/ / \:\/:/ / \:\ \ \::/ / \:\~~\ \:\ \ \/__\:\ \ \/_/:/ /
/:/ / \::/ / \::/ / \:\__\ /:/ / \:\__\ \:\__\ \:\__\ /:/ /
\/__/ \/__/ \/__/ \/__/ \/__/ \/__/ \/__/ \/__/ \/__/
'''
"""
░░██▄░░░░░░░░░░░▄██
░▄▀░█▄░░░░░░░░▄█░░█░
░█░▄░█▄░░░░░░▄█░▄░█░
░█░██████████████▄█░
░█████▀▀████▀▀█████░
▄█▀█▀░░░████░░░▀▀███
██░░▀████▀▀████▀░░██
██░░░░█▀░░░░▀█░░░░██
███▄░░░░░░░░░░░░▄███
░▀███▄░░████░░▄███▀░
░░░▀██▄░▀██▀░▄██▀░░░
░░░░░░▀██████▀░░░░░░
░░░░░░░░░░░░░░░░░░░░
"""
import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt, pi, cos, sin, ceil, log2
from bisect import bisect_right, bisect_left
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)
def factorial(n):
if n == 0:
return 1
return (n * factorial(n - 1))
def ncr(n, r):
r = min(r, n - r)
numer = reduce(op.mul, range(n, n - r, -1), 1)
denom = reduce(op.mul, range(1, r + 1), 1)
return numer // denom # or / in Python 2
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return (list(factors))
def sumDigits(no):
return 0 if no == 0 else int(no % 10) + sumDigits(int(no / 10))
MOD = 1000000007
PMOD = 998244353
N = 10**5
T = 1
# T = int(stdin.readline())
for _ in range(T):
n, m = list(map(int, stdin.readline().rstrip().split()))
# n = int(stdin.readline())
a = list(map(int, stdin.readline().rstrip().split()))
b = list(map(int, stdin.readline().rstrip().split()))
# a = str(stdin.readline().strip('\n'))
# k = int(stdin.readline())
# c = list(map(int, stdin.readline().rstrip().split()))
ans = inf
for A in range(512):
for i in range(n):
isPossible = True
for j in range(m):
t = False
if (a[i] & b[j]) | A == A:
t = True
break
if not t:
isPossible = False
break
if isPossible:
ans = A
break
print(ans)
```
| 5,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
input()
a = [*map(int, input().split())]
b = [*map(int, input().split())]
for x in range(512):
r=1
for v in a:r&=any(v&j|x==x for j in b)
if r:
print(x);break
```
| 5,964 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
import sys
import math as mt
input=sys.stdin.buffer.readline
I=lambda:list(map(int,input().split()))
for tc in range(1):
n,m=I()
a=I()
b=I()
arr=[[a[i]&b[j] for j in range(m)] for i in range(n)]
arr.sort(key=lambda ar:min(ar),reverse=True)
#print(arr)
ans=0
for j in arr:
ans=min([i|ans for i in j])
print(ans)
```
| 5,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
n, m = map(int, input().split())
*a, = map(int, input().split())
*b, = map(int, input().split())
ans = 0
for v in range(2**9):
ok = True
for i in range(n):
found = False
for j in range(m):
if (v | (a[i] & b[j])) == v:
found = True
ok = ok & found
if not ok:
break
if ok:
ans = v
break
print(ans)
'''
mm = 0
for i in range(n):
tt = 2**10
for j in range(m):
t = a[i] & b[j]
if t < tt:
tt = t
if tt > mm:
mm = tt
ans = mm
for i in range(n):
c = 2**10
for j in range(m):
t = ans | (a[i] & b[j])
if t < c:
c = t
ans = c
print(ans)
'''
```
Yes
| 5,966 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, m = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
end = False
for i in range(512):
for j in a:
restart = True
for k in b:
if (j & k) | i == i:
restart = False
break
if restart: break
if restart: continue
else:
print(i)
break
```
Yes
| 5,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
from collections import Counter
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
import math
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
def check(A):
for i in range(n):
fnd = False
for j in range(m):
if (a[i]&b[j])|A == A:
fnd = True; break
if not fnd : return False
return True
n, m = gil()
a = gil()
b = gil()
for i in range(2**9 + 1):
if check(i):
print(i)
break
```
Yes
| 5,968 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
from functools import lru_cache
from sys import stdin, stdout
import sys
from math import *
# from collections import deque
# sys.setrecursionlimit(int(2e5))
input = stdin.readline
# print = stdout.write
# dp=[-1]*100000
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
ans=inf
for i in range(2**9):
c=0
for j in range(n):
for k in range(m):
if (a[j]&b[k])|i==i:
c+=1
break
if(c==n):
ans=min(ans,i)
print(ans)
```
Yes
| 5,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
info = list(map(int, input('').split(' ')))
array1 = list(map(int, input('').split(' ')))
array2 = list(map(int, input('').split(' ')))
proc = dict((elem1, [elem1 & elem2 for elem2 in array2]) for elem1 in array1)
final = 1000000000000000000
for i in range(len(array2)):
output = proc[array1[0]][i]
for j in range(1, len(array1)):
temp = list()
for k in range(len(array2)):
temp.append(output | proc[array1[j]][k])
temp.sort()
output = temp[0]
final = min(final, output)
print(output)
```
No
| 5,970 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
n,m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
A = {}
minn = 1024
prd = 0
for i in range(n):
minn = 1024
if A.get(a[i],0):
prd = prd|A.get([a[i]])
else:
for j in range(m):
# print(a[i]&b[j])
if minn>(a[i]&b[j]):
minn = a[i]&b[j]
I = i
J = j
prd = prd|minn
A[a[I]] = minn
print(prd)
```
No
| 5,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
import sys
import math as mt
input=sys.stdin.buffer.readline
I=lambda:list(map(int,input().split()))
for tc in range(1):
n,m=I()
a=I()
b=I()
arr=[[a[i]&b[j] for j in range(m)] for i in range(n)]
arr.sort(key=lambda ar:min(ar),reverse=True)
print(arr)
ans=0
for j in arr:
ans=min([i|ans for i in j])
print(ans)
```
No
| 5,972 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu likes bit operations. He wants to play a game with you.
Boboniu gives you two sequences of non-negative integers a_1,a_2,…,a_n and b_1,b_2,…,b_m.
For each i (1≤ i≤ n), you're asked to choose a j (1≤ j≤ m) and let c_i=a_i\& b_j, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). Note that you can pick the same j for different i's.
Find the minimum possible c_1 | c_2 | … | c_n, where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The first line contains two integers n and m (1≤ n,m≤ 200).
The next line contains n integers a_1,a_2,…,a_n (0≤ a_i < 2^9).
The next line contains m integers b_1,b_2,…,b_m (0≤ b_i < 2^9).
Output
Print one integer: the minimum possible c_1 | c_2 | … | c_n.
Examples
Input
4 2
2 6 4 0
2 4
Output
2
Input
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
Output
0
Input
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
Output
147
Note
For the first example, we have c_1=a_1\& b_2=0, c_2=a_2\& b_1=2, c_3=a_3\& b_1=0, c_4 = a_4\& b_1=0.Thus c_1 | c_2 | c_3 |c_4 =2, and this is the minimal answer we can get.
Submitted Solution:
```
n,m=map(int, input().split(' '))
a_list=list(map(int, input().split(' ')))
b_list=list(map(int, input().split(' ')))
c_list=[[] for i in range(len(a_list))]
for a in range(len(a_list)):
for b in b_list:
c_list[a].append(a_list[a]&b)
minimum=[]
for x in range(len(a_list)):
minimum.append(min(c_list[x]))
maximum=max(minimum)
for x in range(len(a_list)):
temp=c_list[x][0]|maximum
for y in range(len(b_list)):
if(temp>c_list[x][y]|maximum):
temp=c_list[x][y]|maximum
maximum=temp
print(maximum)
```
No
| 5,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
import math
import random
def cnt(x):
fac = {}
y = x
while (x > 1):
i = 2
flag = False
while (i*i <= x):
if (x % i == 0):
fac[i] = 0
while (x % i == 0):
x //= i
fac[i] += 1
flag = True
break
i += 1
if (not flag):
fac[x] = 1
break
f = set()
i = 2
while (i*i <= y):
if (y % i == 0):
f.add(i)
if (y // i != i):
f.add(y//i)
i += 1
f.add(y)
return fac,f
t = int(input())
while (t):
n = int(input())
mp,f = cnt(n)
primes = list(mp.keys())
if (len(primes) == 1):
for x in f:
print(x,end=' ')
print('\n0')
elif (len(primes) == 2):
if (mp[primes[0]] == 1 and mp[primes[1]] == 1):
print(primes[0], primes[1], n)
print(1)
else:
a,b = primes
print(a,a*b,b,end=' ')
f.discard(a)
f.discard(a*b)
f.discard(b)
for i in range(2,mp[b]+1):
print(b**i,end=' ')
for i in range(1,mp[b]+1):
for j in range(1,mp[a]+1):
if (i != 1 or j != 1):
print(b**i * a**j, end=' ')
for i in range(2,mp[a]+1):
print(a**i,end=' ')
print('\n0')
else:
for i in range(len(primes)):
a,b = primes[i],primes[(i+1)%len(primes)]
f.discard(a)
f.discard(a*b)
ff = {}
for p in primes:
ff[p] = set()
for x in list(f):
if (x % p == 0):
ff[p].add(x)
f.discard(x)
for i in range(len(primes)):
a,b = primes[i],primes[(i+1)%len(primes)]
print(a,end=' ')
for v in ff[a]:
print(v,end=' ')
print(a*b,end=' ')
print('\n0')
t -= 1
```
| 5,974 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for you in range(t):
n=int(input())
z=n
primes=[]
i=2
while(i*i<=z):
if(z%i==0):
primes.append(i)
while(z%i==0):
z=z//i
i+=1
if(z!=1):
primes.append(z)
hashi=dict()
for i in primes:
hashi[i]=[]
hashinew=dict()
new=[]
k=len(primes)
hasho=dict()
if(k>2):
for i in range(k):
new.append(primes[i]*primes[(i+1)%k])
hasho[primes[i]*primes[(i+1)%k]]=1
if(k==2):
hasho[primes[0]*primes[1]]=1
i=2
while(i*i<=n):
if(n%i==0):
num1=i
num2=n//i
if(num1 not in hasho):
for j in primes:
if(num1%j==0):
break
hashi[j].append(num1)
if(num2!=num1 and num2 not in hasho):
for j in primes:
if(num2%j==0):
break
hashi[j].append(num2)
i+=1
for j in primes:
if(n%j==0):
break
hashi[j].append(n)
done=dict()
if(len(primes)==1):
for i in hashi[primes[0]]:
print(i,end=" ")
print()
print(0)
continue
if(len(primes)==2):
if(primes[0]*primes[1]==n):
print(primes[0],primes[1],n)
print(1)
else:
for i in hashi[primes[0]]:
print(i,end=" ")
for i in hashi[primes[1]]:
print(i,end=" ")
print(primes[0]*primes[1],end=" ")
print()
print(0)
continue
for i in range(k):
for j in hashi[primes[i]]:
print(j,end=" ")
ko=primes[i]*primes[(i+1)%k]
print(ko,end=" ")
print()
print(0)
```
| 5,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
from itertools import accumulate
from functools import lru_cache
M = mod = 998244353
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')]
def isprime(n):
for j in range(2, int(n ** 0.5) + 1):
if n % j == 0:return 0
return 1
for _ in range(val()):
n = val()
l1 = factors(n)[1:]
l = []
for j in l1:
if isprime(j):l.append(j)
l1 = set(l1)
l1 -= set(l)
# print(l, l1)
d = defaultdict(set)
for j in range(len(l)):
for i in sorted(list(l1)):
if i % l[j] == 0 and i % l[j - 1] == 0:
d[tuple(sorted([l[j], l[j - 1]]))].add(i)
l1.remove(i)
break
# print(l, l1)
for j in range(len(l)):
for i in sorted(list(l1)):
if i % l[j] == 0 and i % l[j - 1] == 0:
d[tuple(sorted([l[j], l[j - 1]]))].add(i)
l1.remove(i)
# print(l, l1, d)
only = defaultdict(list)
for j in range(len(l)):
for i in sorted(list(l1)):
if i % l[j] == 0:
only[l[j]].append(i)
l1.remove(i)
fin = []
if len(l) == 2:
fin.append(l[0])
for j in only[l[0]]:fin.append(j)
for i in range(len(l)):
for j in list(d[tuple(sorted([l[i], l[(i + 1) % len(l)]]))]):
fin.append(j)
d[tuple(sorted([l[i], l[(i + 1) % len(l)]]))].remove(j)
if i != len(l) - 1:break
if i != len(l) - 1:
fin.append(l[i + 1])
for j in only[l[i + 1]]:
fin.append(j)
else:
fin.append(l[0])
for j in only[l[0]]:fin.append(j)
for i in range(len(l)):
for j in d[tuple(sorted([l[i], l[(i + 1) % len(l)]]))]:
fin.append(j)
if i != len(l) - 1:
fin.append(l[i + 1])
for j in only[l[i + 1]]:
fin.append(j)
ans = 0
for i in range(len(fin)):
if math.gcd(fin[i], fin[i - 1]) == 1:ans += 1
print(*fin)
print(ans)
```
| 5,976 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
import sys
import math
ii = lambda: sys.stdin.readline().strip()
idata = lambda: [int(x) for x in ii().split()]
def solve():
n = int(ii())
simple = []
dividers = []
slov = {}
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
dividers += [i]
if n // i != i:
dividers += [n // i]
dividers += [1]
dividers.sort()
for i in range(len(dividers) - 1, -1, -1):
flag = 1
for j in range(len(simple)):
if (n // dividers[i]) % simple[j] == 0 and flag:
flag = 0
slov[simple[j]] += [n // dividers[i]]
if flag:
simple += [n // dividers[i]]
slov[n // dividers[i]] = []
dividers += [n]
simple.sort()
if len(simple) == 2:
if n % pow(simple[0], 2) != 0 and n % pow(simple[1], 2) != 0:
print(*dividers[1:])
print(1)
else:
slov[simple[0]].remove(n)
ans = [simple[0]] + slov[simple[0]] + [simple[1]] + slov[simple[1]] + [n]
print(*ans)
print(0)
return
if len(simple) >= 2:
s = simple[0] * simple[-1]
slov[simple[0]].remove(s)
ans = []
for j in range(len(simple)):
ans += [simple[j]]
d = simple[(j + 1) % len(simple)] * simple[j]
for i in slov[simple[j]]:
if i != d and i != simple[j]:
ans += [i]
ans += [d]
print(*ans)
print(0)
for t in range(int(ii())):
solve()
```
| 5,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
from collections import defaultdict, deque
T = int(input())
for _ in range(T):
n = int(input())
d, p = defaultdict(int), 2
while p*p <= n:
while n % p == 0:
d[p], n = d[p]+1, n//p
p += 1
if n > 1: d[n] = 1
ls = [1]
for u, cc in d.items():
m, p = len(ls), 1
for _ in range(cc):
p *= u
for v in reversed(ls[0:m]):
ls.append(v*p)
if len(ls) > 2: ls[-1], ls[-2] = ls[-2], ls[-1]
move = (len(d) == 2 and max(d.values()) == 1)
print(*ls[1:])
print(1 if move else 0)
```
| 5,978 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
import random
from math import gcd, sqrt, floor, ceil
#__________________________________________________#
# brute force solution_____________________________#
def solve_1(n):
d = divisors(n)
order = [0] * len(d)
moves = [999999999] * 1
mark = [False] * len(d)
permutation(d, [0] * len(d), 0, moves, mark, order)
if gcd(order[0], order[-1]) == 1:
moves[0] += 1
return order, moves[0]
def permutation(d, p, pos, moves, mark, order):
if pos == len(d):
m = 0
for i in range(len(p) - 1):
if gcd(p[i], p[i + 1]) == 1:
m += 1
if m < moves[0]:
moves[0] = m
for i in range(len(p)):
order[i] = p[i]
#print(p)
return
for i in range(len(d)):
if not mark[i]:
mark[i] = True
p[pos] = d[i]
permutation(d, p, pos + 1, moves, mark, order)
mark[i] = False
return
def divisors(n):
d = []
for i in range(2, int(sqrt(n)) + 1):
if(n % i == 0):
d.insert(floor(len(d)/2), i)
if(i * i != n):
d.insert(ceil(len(d)/2), n//i)
d.append(n)
return d
#__________________________________________________#
# solution_________________________________________#
def solve_2(n):
d = divisors(n)
moves = 0
if len(d) == 1:
return d, 0
order = sort(d)
if(gcd(order[0], order[len(order) - 1]) == 1):
moves += 1
return order, moves
def sort(d):
order = []
order.append(d[0])
count = len(d)
d.remove(d[0])
j = 0
while(len(order) < count):
if int(gcd(d[j], order[-1])) != 1:
order.append(d[j])
d.remove(d[j])
j = 0
else: j += 1
return order
#__________________________________________________#
# main_____________________________________________#
def main():
t = int(input())
for i in range(t):
n = int(input())
#order, moves = solve_1(n) #solucion_fuerza_bruta
order, moves = solve_2(n) #solucion_eficiente
print(' '.join(str(j) for j in order))
print(moves)
return
# _________________________________________________#
# random cases tester______________________________#
def random_cases():
n = random.randint(2, 50)
print('Case: \n' + "n = " + str(n))
order, moves = solve_1(n) #solucion_fuerza_bruta
#order, moves = solve_2(n) #solucion_eficiente
print(' '.join(str(i) for i in order)) #solucion
print(moves)
return
#__________________________________________________#
# tester___________________________________________#
def tester():
for i in range(1, 20):
n = random. randint(4, 50)
order_1, moves_1 = solve_1(n)
order_2, moves_2 = solve_2(n)
print('Case ' + str(i) + ':\n' + 'n = ' + str(n))
print('solucion fuerza bruta : \n' + ' '.join(str(i) for i in order_1) + '\n' + str(moves_1)) #solucion_fuerza_bruta
print('solucion eficiente : \n' + ' '.join(str(i) for i in order_2) + '\n' + str(moves_2)) #solucion_eficiente
for i in range(min(len(order_1), len(order_2))):
if(order_1[i] != order_2[i] or len(order_1) != len(order_2) or moves_1 != moves_2):
print(Color.RED + 'Wrong Answer\n' + Color.RESET)
return
print(Color.GREEN + 'Accepted\n' + Color.RESET)
return
main() #entrada_de_casos_por_consola
#random_cases() #casos_random
#tester() #tester_automatico
```
| 5,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineering College
'''
from os import path
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
mod=1000000007
# mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('a')
def ceil(a,b):
return (a+b-1)//b
file=1
def f():
sys.stdout.flush()
def solve():
for _ in range(ii()):
n = ii()
n1 = n
primedivisor = []
for i in range(2,int(sqrt(n))+1):
c = 0
while(n%i==0):
c +=1
n //=i
if c > 0:
primedivisor.append([i,c])
if n>1:
primedivisor.append([n,1])
cnt = len(primedivisor)
# only one prime divisor
if cnt == 1:
p = primedivisor[0][0]
q = primedivisor[0][1]
for i in range(1,q+1):
print(p**i,end=" ")
print()
print(0)
continue
p = primedivisor[0][0]
q = primedivisor[0][1]
ans = []
# divisor -> p^1,p^2,p^2,p^3.....p^q
for j in range(1,q+1):
ans.append(p**j)
f = 0
for i in range(1,cnt):
p = primedivisor[i][0]
q = primedivisor[i][1]
curlen = len(ans)
lcm = p * primedivisor[i-1][0]
ans.append(lcm)
# divisor -> p^2,p^2,p^3.....p^q
for j in range(2,q+1):
ans.append(p**j)
for j in range(curlen):
for k in range(1,q+1):
x = p**k * ans[j]
if x == lcm:
continue
if x==n1:
f = 1
continue
ans.append(x)
ans.append(p)
if len(ans) == 3:
print(*ans)
print(1)
else:
if(f==0):
ans[-2],ans[-1] = ans[-1],ans[-2]
else:
ans.append(n1)
print(*ans)
print(0)
if __name__ =="__main__":
if path.exists('input.txt'):
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output.txt','w')
else:
input=sys.stdin.readline
solve()
```
| 5,980 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from collections import Counter
def main():
for _ in range(int(input())):
n = int(input())
fac = Counter()
div = {n}
x = n
for i in range(2,int(x**0.5)+1):
if not x%i:
div.add(i)
div.add(x//i)
while not n%i:
fac[i] += 1
n //= i
if n != 1:
fac[n] += 1
x = list(fac.keys())
if len(fac) == 1:
print(' '.join(map(str,div)))
print(0)
elif sum(fac.values()) == 2 and len(fac) == 2:
print(x[0],x[0]*x[1],x[1])
print(1)
else:
fir = []
for i in range(len(x)):
su = x[i]*x[i-1]
for j in div:
if not j%su:
fir.append(j)
break
div.remove(fir[-1])
ans = []
for ind,val in enumerate(x):
ans.append(fir[ind])
nu = 0
for j in div:
if not j%val:
nu += 1
ans.append(j)
for j in range(-1,-1-nu,-1):
div.remove(ans[j])
print(' '.join(map(str,ans)))
print(0)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
```
| 5,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
pfacs = []
m, i = n, 2
while i * i <= m:
while m % i == 0:
pfacs.append(i)
m //= i
i += 1
if m > 1:
pfacs.append(m)
pr = list(set(pfacs))
facs = {1}
for p in pfacs:
facs |= set(p * x for x in facs)
facs.remove(1)
def get():
global facs
if len(pr) == 1:
return facs, 0
if len(pr) == 2:
p, q = pr
if len(pfacs) == 2:
return [p, q, n], 1
pmul = set(x for x in facs if x % p == 0)
facs -= pmul
pmul -= {p, p * q, n}
facs.remove(q)
return [p, *pmul, p * q, q, *facs, n], 0
ans = []
for p, q in zip(pr, pr[1:] + [pr[0]]):
pmul = set(x for x in facs if x % p == 0)
if p == pr[0]:
pmul.remove(p * pr[-1])
facs -= pmul
pmul -= {p, p * q}
ans += [p, *pmul, p * q]
return ans, 0
ans, c = get()
print(*ans)
print(c)
```
Yes
| 5,982 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
def gen(i, cur):
global dvs, used
if i == len(kk):
if (ohne != 1 or cur != 1) and (ok or not used[cur * ohne]):
dvs.append(cur * ohne)
return
gen(i + 1, cur)
for j in range(kk[i]):
cur *= pp[i]
gen(i + 1, cur)
gans = []
for _ in range(int(input())):
n = int(input())
pp = []
kk = []
i = 2
cnt = []
while i * i <= n:
if n % i == 0:
pp.append(i)
kk.append(0)
while n % i == 0:
kk[-1] += 1
n //= i
i += 1
if n != 1:
pp.append(n)
kk.append(1)
dvs = []
ohne = 1
ok = True
gen(0, 1)
if len(pp) == 1:
gans.append(' '.join(map(str, dvs)))
gans.append(str(0))
elif len(pp) == 2 and kk[0] == kk[1] == 1:
gans.append(' '.join(map(str, dvs)))
gans.append(str(1))
elif len(pp) == 2:
used = dict()
for i in range(len(dvs)):
used[dvs[i]] = False
ans = []
ok = False
used[pp[0] * pp[1]] = True
aaa = [pp[0] * pp[1]]
if kk[0] > 1:
used[pp[0] * pp[0] * pp[1]] = True
aaa.append(pp[0] * pp[0] * pp[1])
else:
used[pp[0] * pp[1] * pp[1]] = True
aaa.append(pp[0] * pp[1] * pp[1])
for i in range(len(pp)):
dvs = []
ans.append(aaa[i])
kk[i] -= 1
ohne = pp[i]
gen(0, 1)
for j in range(len(dvs)):
used[dvs[j]] = True
ans.append(dvs[j])
gans.append(' '.join(map(str, ans)))
gans.append(str(0))
else:
used = dict()
for i in range(len(dvs)):
used[dvs[i]] = False
ans = []
ok = False
for i in range(len(pp)):
used[pp[i - 1] * pp[i]] = True
for i in range(len(pp)):
dvs = []
ans.append(pp[i - 1] * pp[i])
kk[i] -= 1
ohne = pp[i]
gen(0, 1)
for j in range(len(dvs)):
used[dvs[j]] = True
ans.append(dvs[j])
gans.append(' '.join(map(str, ans)))
gans.append(str(0))
print('\n'.join(gans))
```
Yes
| 5,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = 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()
# ------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# sys.setrecursionlimit(300000)
# from heapq import *
# from collections import deque as dq
from math import ceil,floor,sqrt,pow
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
for i in range(N()):
n = N()
nn,k = n,2
p,d = [],[i for i in range(2,int(sqrt(n))+1) if n%i==0]
d+=[n//i for i in d]
d = set(d)
d.add(n)
# print(d)
for i in d:
if nn%i==0:
p.append(i)
while nn%i==0:
nn//=i
# print(d)
# print(p)
if len(p)==1:
print_list(d)
print(0)
elif len(p)==2:
if len(d)==3:
print_list(d)
print(1)
else:
a,b = p[0],p[1]
res = [a*b,a]
d-={a*b,a,n}
for i in d:
if i%a==0:
res.append(i)
res.append(n)
for i in d:
if i%a>0:
res.append(i)
print_list(res)
print(0)
else:
res = []
s = 0
pp = [p[i]*p[i-1] for i in range(1,len(p))]+[p[0]*p[-1]]
d-=set(pp)
d-=set(p)
for i in range(len(p)):
k = p[i]
for t in d:
if t%k==0:
res.append(t)
d-=set(res[s:])
s = len(res)
res.append(p[i])
res.append(pp[i])
print_list(res)
print(0)
```
Yes
| 5,984 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
from itertools import product
def p_factorization_t(n):
if n == 1: return []
pf_cnt = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i == 0:
cnt = 0
while temp%i == 0:
cnt += 1
temp //= i
pf_cnt.append((i,cnt))
if temp != 1: pf_cnt.append((temp,1))
return pf_cnt
def main():
ansl = []
for _ in range(int(input())):
n = int(input())
facs = p_factorization_t(n)
# print(facs)
if len(facs) == 1:
p,cnt = facs[0]
al = []
for i in range(1,cnt+1):
al.append(pow(p,i))
print(*al)
print(0)
ff = []
pd = {}
ps = []
for p,cnt in facs:
row = []
for i in range(0,cnt+1):
row.append(pow(p,i))
ff.append(row)
pd[p] = []
ps.append(p)
vals = [1]
for row in ff:
new_vals = []
for v in vals:
for p in row:
new_vals.append(p*v)
if p != 1:
pd[row[1]].append(v*p)
vals = new_vals[:]
if len(facs) >= 3:
al = []
for i in range(len(ps)):
cval = -1
if i > 0:
cval = (ps[i]*ps[i-1])
al.append(cval)
else:
cval = (ps[i]*ps[-1])
for v in pd[ps[i]]:
if v != cval:
al.append(v)
print(*al)
print(0)
elif len(facs) == 2:
al = []
for i in range(len(ps)):
cval = -1
if i > 0:
cval = (ps[i]*ps[i-1])
al.append(cval)
else:
cval = (ps[i]*ps[-1])
for v in pd[ps[i]]:
if v != cval:
al.append(v)
print(*al)
if facs[0][1] == 1 and facs[1][1] == 1:
print(1)
else:
print(0)
# elif len(facs) == 2:
if __name__ == "__main__":
main()
```
Yes
| 5,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
depth=32000
primes=[True]*depth
primes[0]=primes[1]=False
prime_list=[0]*3432
k=0
for i in range(depth):
if primes[i]:
prime_list[k]=i
k+=1
for j in range(i*i,depth,i):
primes[j]=False
t=int(input())
for i in range(t):
flag=True
n=int(input())
divisors=[0]*200000
num_of_div=0
k=n
j=0
current_lenght=0
''' Разлагаем на множители'''
while k>1 and j<3432:
degree=0
while k%prime_list[j]==0:
degree+=1
k=k//prime_list[j]
if degree:
num_of_div+=1
current=1
if degree>1:
flag=False
if current_lenght==0 and degree:
for index in range(degree-1):
current*=prime_list[j]
divisors[index]=current
print(current, end=' ')
current*=prime_list[j]
if k==1:
print(current)
else:
print(current, end=' ')
divisors[degree-1]=current
current_lenght=degree
elif degree and k>1:
for m in range(1,degree+1):
current*=prime_list[j]
for index in range(current_lenght):
divisors[m*current_lenght+index]=\
divisors[current_lenght-index-1]*current
print(divisors[m*current_lenght+index], end=' ')
print(prime_list[j], end=' ')
current_lenght=degree*current_lenght+1
elif degree and k==1:
for m in range(1,degree):
current*=prime_list[j]
for index in range(current_lenght):
divisors[m*current_lenght+index]=\
divisors[current_lenght-index-1]*current
print(divisors[m*current_lenght+index], end=' ')
if degree>1:
print(prime_list[j], end=' ')
m=degree
current*=prime_list[j]
for index in range(current_lenght-1):
divisors[m*current_lenght+index]=\
divisors[current_lenght-index-1]*current
print(divisors[m*current_lenght+index], end=' ')
print(current, end=' ')
print(current*divisors[0])
j+=1
'''Дописываем последний делитель'''
if k!=1:
for index in range(current_lenght-1,0,-1):
print(k*divisors[index], end=' ')
num_of_div+=1
print(k,end=' ')
print(divisors[0]*k)
if num_of_div==2 and flag:
print(1)
else:
print(0)
```
No
| 5,986 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
import sys
input = sys.stdin.readline
import math
def yaku(x):
xr=math.ceil(math.sqrt(x))
LIST=[]
for i in range(1,xr+1):
if x%i==0:
LIST.append(i)
LIST.append(x//i)
L=int(math.sqrt(x))
FACT=dict()
for i in range(2,L+2):
while x%i==0:
FACT[i]=FACT.get(i,0)+1
x=x//i
if x!=1:
FACT[x]=FACT.get(x,0)+1
return sorted(set(LIST)),sorted(set(FACT))
t=int(input())
for tests in range(t):
n=int(input())
L,SS=yaku(n)
USE=[0]*len(L)
USE[0]=1
USE[-1]=1
ANS=[n]
for i in range(len(SS)-1):
NEXT=SS[i]*SS[i+1]
ANSX=[]
for j in range(len(L)):
if L[j]==NEXT:
USE[j]=1
if USE[j]==0 and L[j]%SS[i]==0:
USE[j]=1
ANSX.append(L[j])
if NEXT!=n:
ANSX.append(NEXT)
ANS+=ANSX
ANS.append(SS[-1])
SC=0
for i in range(len(ANS)):
if math.gcd(ANS[i],ANS[i-1])==1:
SC+=1
print(*ANS)
print(SC)
```
No
| 5,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
def gcd(a, b):
while b: a, b = b, a % b
return a
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 7, 61] if n < 1<<32 else [2, 3, 5, 7, 11, 13, 17] if n < 1<<48 else [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = y * y % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i * i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += i % 2 + (3 if i % 3 == 1 else 1)
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(pf):
ret = [1]
for p in pf:
ret_prev = ret
ret = []
for i in range(pf[p]+1):
for r in ret_prev:
ret.append(r * (p ** i))
return sorted(ret)
T = int(input())
for _ in range(T):
N = int(input())
pf = primeFactor(N)
dv = divisors(pf)
if len(pf) == 2 and len(dv) == 4:
print(*dv[1:])
print(1)
continue
if len(pf) == 1:
print(*dv[1:])
print(0)
continue
lpf = list(pf)
# print("lpf =", lpf)
X = [[] for _ in range(len(pf))]
S = {1}
for i, p in enumerate(lpf):
# print("i, p, pf[p] =", i, p, pf[p])
X[i].append(lpf[i-1] * p)
S.add(lpf[i-1] * p)
for j in range(1, pf[p] + 1):
X[i].append(p ** j)
S.add(p ** j)
for a in dv:
if a not in S:
for i, p in enumerate(lpf):
if a % p == 0:
X[i].append(a)
break
# print("X =", X)
ANS = []
for x in X:
for y in x:
ANS.append(y)
print(*ANS)
print(0)
```
No
| 5,988 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
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")
#-------------------game starts now-----------------------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#-------------------------------------------------------------------------
for _ in range (int(input())):
n=int(input())
s=set()
d=set()
a=list()
c=1
d.add(n)
for i in range (2,int(math.sqrt(n))+1):
if n%i==0:
d.add(i)
if i*i!=n:
d.add(n//i)
copy=n
while copy%2==0:
copy//=2
s.add(2)
for i in range (3,int(math.sqrt(copy))+1):
if copy%i==0:
while copy%i==0:
copy//=i
s.add(i)
if copy>2:
s.add(copy)
#print(s)
#print(d)
b=list(s)
if len(b)==2 and b[0]*b[1]==n:
print(*b,b[0]*b[1])
print(1)
else:
for i in s:
a.append([i])
d.remove(i)
#print(a)
for i in range (len(a)):
if i==len(a)-1 and len(s)!=1:
c=a[i][0]*a[0][1]
if c in d:
a[i]=[c]+a[i]
d.remove(c)
else:
c=a[i][0]*a[(i+1)%(len(s))][0]
if c in d:
a[i]=[c]+a[i]
d.remove(c)
#print(a)
for i in range (len(a)):
t=d.copy()
for j in t:
if j%a[i][0]==0:
a[i].append(j)
d.remove(j)
#print(a)
for i in range (len(a)):
print(*a[i][1:],a[i][0],end=' ')
print()
print(0)
```
No
| 5,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Tags: combinatorics, dp, math, two pointers
Correct Solution:
```
mod = 998244353
eps = 10**-9
def main():
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split())) + [0]
A.sort()
dp = [[0] * (i+1) for i in range(N+1)]
dp[0][0] = 1
l = 0
for i in range(1, N+1):
for ll in range(l+1, i):
if A[ll] * 2 <= A[i]:
l = ll
else:
break
for j in range(1, l+2):
dp[i][j] = (dp[l][j-1] + (dp[i][j-1] * (l-j+2))%mod)%mod
for j in range(i):
dp[i][j] = (dp[i-1][j] + dp[i][j])%mod
print(dp[-1][-1])
if __name__ == '__main__':
main()
```
| 5,990 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Tags: combinatorics, dp, math, two pointers
Correct Solution:
```
from collections import defaultdict
import sys
input = sys.stdin.readline
n = int(input())
a = map(int, input().split())
mod = 998244353
d = defaultdict(int)
for x in a:
d[x] += 1
d[0] = 0
b = list(d.items())
b.sort()
m = len(b)
ba = [0] * m
cn = [0] * (m + 1)
k = h = 0
for i, x in enumerate(b):
while h < m and x[0] >= b[h][0] * 2:
h += 1
ba[i] = h - 1
while k < m and x[0] * 2 > b[k][0]:
k += 1
cn[k] += x[1]
for i in range(m):
cn[i+1] += cn[i]
dp = [0] * m
dp[0] = 1
b = [x[1] for x in b]
for i in range(n):
ndp = [0] * m
for j in range(1, m):
if cn[j] >= i - 1:
ndp[j] = dp[j] * (cn[j] - i + 1) % mod
dp[j] += dp[j-1]
if dp[j] >= mod:
dp[j] -= mod
for j in range(1, m):
ndp[j] += dp[ba[j]] * b[j]
ndp[j] %= mod
dp = ndp
print(sum(dp) % mod)
```
| 5,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Tags: combinatorics, dp, math, two pointers
Correct Solution:
```
from collections import Counter
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
mod = 998244353
a.sort()
dp = [1] + [0] * n
for i in range(1, n + 1):
x, pt = 1, i - 2
while pt >= 0 and 2 * a[pt] > a[i - 1]:
x = x * (n - pt - 2) % mod
pt -= 1
dp[i] = (dp[i - 1] * (n - i) + dp[pt + 1] * x) % mod
print(dp[-1])
```
| 5,992 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Tags: combinatorics, dp, math, two pointers
Correct Solution:
```
from collections import Counter
import sys
input = sys.stdin.readline
n = int(input())
a = map(int, input().split())
mod = 998244353
d = Counter(a)
d[0] = 0
b = list(d.items())
b.sort()
m = len(b)
ba = [0] * m
cn = [0] * (m + 1)
k = h = 0
for i, x in enumerate(b):
while h < m and x[0] >= b[h][0] * 2:
h += 1
ba[i] = h - 1
while k < m and x[0] * 2 > b[k][0]:
k += 1
cn[k] += x[1]
for i in range(m):
cn[i+1] += cn[i]
dp = [0] * m
dp[0] = 1
b = [x[1] for x in b]
for i in range(n):
ndp = [0] * m
for j in range(1, m):
if cn[j] >= i - 1:
ndp[j] = dp[j] * (cn[j] - i + 1) % mod
dp[j] += dp[j-1]
if dp[j] >= mod:
dp[j] -= mod
for j in range(1, m):
ndp[j] += dp[ba[j]] * b[j]
ndp[j] %= mod
dp = ndp
print(sum(dp) % mod)
```
| 5,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Tags: combinatorics, dp, math, two pointers
Correct Solution:
```
M = 998244353
n = int(input())
l = sorted(map(int, input().split()))[::-1]
out = [0] * n
big = 0
if l[0] >= 2 * l[1]:
out[1] = 1
big = 1
for i in range(2, n):
new = [0] * n
bigN = 0
for j in range(i):
if l[j] >= 2 * l[i]:
big += out[j]
else:
new[j] += out[j] * (i - 1)
new[j] %= M
new[i] = big
bigN = (i * big) % M
out = new
big = bigN
print((big + sum(out))%M)
```
| 5,994 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Tags: combinatorics, dp, math, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
mod = 998244353
a.sort()
dp = [1] + [0] * n
for i in range(1, n + 1):
x, pt = 1, i - 2
while pt >= 0 and 2 * a[pt] > a[i - 1]:
x = x * (n - pt - 2) % mod
pt -= 1
dp[i] = (dp[i - 1] * (n - i) + dp[pt + 1] * x) % mod
print(dp[-1])
```
| 5,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Submitted Solution:
```
import sys, bisect
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
mod = 998244353
a.sort()
lim = [-1] + [0] * n
dp = [1] + [0] * n
f = [1] + [0] * n
f1 = [0] * (n + 1)
for i in range(1, n + 1):
f[i] = i * f[i - 1] % mod
f1[n] = pow(f[n], mod - 2, mod)
for i in range(n - 1, 0, -1):
f1[i] = f1[i + 1] * (i + 1) % mod
def perm(n, m):
return f[n] * f1[n - m] % mod
for i in range(n):
lim[i + 1] = bisect.bisect(a, a[i] / 2)
for i in range(1, n + 1):
for j in range(i):
if lim[i] > lim[j]:
dp[i] += dp[j] * perm(n - 2 - lim[j], lim[i] - lim[j] - 1)
dp[i] %= mod
if lim[-1] == n - 1:
print(int(dp[-1]))
else:
print(0)
```
No
| 5,996 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Submitted Solution:
```
import sys, bisect
input = sys.stdin.readline
import functools
n = int(input())
a = list(map(int, input().split()))
mod = 998244353
a.sort()
lim = [-1] + [0] * n
dp = [1] + [0] * n
f = [1] + [0] * n
f1 = [0] * (n + 1)
for i in range(1, n + 1):
f[i] = i * f[i - 1] % mod
f1[n] = pow(f[n], mod - 2, mod)
for i in range(n - 1, -1, -1):
f1[i] = f1[i + 1] * (i + 1) % mod
# @functools.lru_cache(None)
def perm(n, m):
return f[n] // f[n - m] % mod
for i in range(n):
l, r = -1, i + 1
while r - l > 1:
mid = (l + r) // 2
if a[mid] * 2 > a[i]:
r = mid
else:
l = mid
lim[i + 1] = r
for i in range(1, n + 1):
for j in range(i):
if lim[i] > lim[j]:
dp[i] += dp[j] * perm(n - 2 - lim[j], lim[i] - lim[j] - 1)
dp[i] %= mod
if lim[-1] == n - 1:
print(int(dp[-1]))
else:
print(0)
```
No
| 5,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Submitted Solution:
```
import sys, bisect
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
mod = 998244353
a.sort()
lim = [-1] + [0] * n
dp = [1] + [0] * n
f = [1] + [0] * n
f1 = [0] * (n + 1)
for i in range(1, n + 1):
f[i] = i * f[i - 1] % mod
f1[n] = pow(f[n], mod - 2, mod)
for i in range(n - 1, -1, -1):
f1[i] = f1[i + 1] * (i + 1) % mod
def perm(n, m):
return f[n] * f1[n - m] % mod
for i in range(n):
lim[i + 1] = bisect.bisect(a, a[i] / 2)
for i in range(1, n + 1):
for j in range(i):
if lim[i] > lim[j]:
dp[i] += dp[j] * perm(n - 2 - lim[j], lim[i] - lim[j] - 1)
dp[i] %= mod
if lim[-1] == n - 1:
print(int(dp[-1]))
else:
print(0)
```
No
| 5,998 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n fishermen have just returned from a fishing vacation. The i-th fisherman has caught a fish of weight a_i.
Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 1 to n). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.
Suppose a fisherman shows a fish of weight x, and the maximum weight of a previously shown fish is y (y = 0 if that fisherman is the first to show his fish). Then:
* if x ≥ 2y, the fisherman becomes happy;
* if 2x ≤ y, the fisherman becomes sad;
* if none of these two conditions is met, the fisherman stays content.
Let's call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 5000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of emotional orders, taken modulo 998244353.
Examples
Input
4
1 1 4 9
Output
20
Input
4
4 3 2 1
Output
0
Input
3
4 2 1
Output
6
Input
8
42 1337 13 37 420 666 616 97
Output
19200
Submitted Solution:
```
import sys, bisect
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
mod = 998244353
a.sort()
lim = [-1] + [0] * n
dp = [1] + [0] * n
def perm(n, m):
if not m:
return 1
ans = 1
for i in range(n, n - m, -1):
ans *= i
return ans % mod
for i in range(n):
lim[i + 1] = bisect.bisect(a, a[i] / 2)
for i in range(1, n + 1):
for j in range(i):
if lim[i] > lim[j]:
dp[i] += dp[j] * perm(n - 2 - lim[j], lim[i] - lim[j] - 1)
dp[i] %= mod
print(int(dp[-1]))
```
No
| 5,999 |
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