message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative. | instruction | 0 | 2,474 | 12 | 4,948 |
Tags: implementation
Correct Solution:
```
import sys
import math
def period(x):
n = len(x)
res = []
for i in range(1, len(x)):
if all(i==j for i,j in zip(x, x[i:])):
res.append(i)
return res + [n]
def main():
[n] = map(int, next(sys.stdin).split())
a = list(map(int, next(sys.stdin).split()))
a = [0] + a
x = [j - i for i,j in zip(a, a[1:])]
periods = period(x)
print(len(periods))
print(' '.join(map(str,periods)))
if __name__ == '__main__':
main()
``` | output | 1 | 2,474 | 12 | 4,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineerin College
Date:31/05/2020
'''
import sys
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()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def read():
tc=1
if tc:
input=sys.stdin.readline
else:
sys.stdin=open('input1.txt', 'r')
sys.stdout=open('output1.txt','w')
def powermod(a,b,m):
a%=m
res=1
if(a==0):
return 0
while(b>0):
if(b&1):
res=(res*a)%m;
b=b>>1
a=(a*a)%m;
return res
def isprime(n):
if(n<=1):
return 0
for i in range(2,int(sqrt(n))+1):
if(n%i==0):
return 0
return 1
def solve():
n=ii()
a=li()
x=[a[0]]
for i in range(1,n):
x.append(a[i]-a[i-1])
f=n
ans=[]
for i in range(1,n):
if(x[i]==x[0]):
f=i
f2=1
for i in range(n):
if x[i]!=x[i%f]:
f2=0
break
if(f2):
x1=f
while(n>=x1):
ans.append(x1)
x1+=f
ans=list(set(ans))
if n not in ans:
ans.append(n)
ans.sort()
print(len(ans))
print(*ans)
if __name__ =="__main__":
read()
solve()
``` | instruction | 0 | 2,475 | 12 | 4,950 |
Yes | output | 1 | 2,475 | 12 | 4,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
def scand():
return int(input())
def scana():
return [int(x) for x in input().split()]
n=scand()
a=[0]+scana()
x=[str(a[i]-a[i-1]) for i in range(1,len(a))]
ans=[]
for i in range(1,n+1):
ok=True
for j in range(0,n):
if x[j]!=x[j%i]:
ok=False
break
if ok:
ans.append(str(i))
print(len(ans))
print(' '.join(ans))
``` | instruction | 0 | 2,476 | 12 | 4,952 |
Yes | output | 1 | 2,476 | 12 | 4,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a=[0]+a
b=[]
for i in range(n):
b.append(a[i+1]-a[i])
ans=[]
for k in range(1,n+1):
c=b[:k]*(n//k+1)
if c[:n]==b:
ans.append(k)
print(len(ans))
print(*ans)
``` | instruction | 0 | 2,477 | 12 | 4,954 |
Yes | output | 1 | 2,477 | 12 | 4,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
def ii():
return int(input())
def mi():
return map(int, input().split())
def li():
return list(mi())
n = ii()
a = [0] + li()
d = [a[i] - a[i - 1] for i in range(1, n + 1)]
ans = []
for i in range(1, n + 1):
ok = all(d[j] == d[j % i] for j in range(n))
if ok: ans.append(i)
print(len(ans))
print(*ans)
``` | instruction | 0 | 2,478 | 12 | 4,956 |
Yes | output | 1 | 2,478 | 12 | 4,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
n = int(input())
a = [0]
a1 = list(map(int, input().split()))
a = a + a1
ans = []
for k in range(1, n + 1):
x = [-1] * k
p = True
for j in range(n):
if x[j % k] == -1:
x[j % k] = a[j + 1] - a[j]
elif x[j % k] != a[j + 1] - a[j]:
p = False
break
if p:
ans.append(k)
print(len(ans))
print(*ans)
``` | instruction | 0 | 2,479 | 12 | 4,958 |
No | output | 1 | 2,479 | 12 | 4,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
diff=[a[0]]
for i in range(n-1):
diff.append(a[i+1]-a[i])
out=""
for i in range(1,n+1):
good=True
for j in range(n-i):
if diff[j]!=diff[j+i]:
good=False
if good:
out+=str(i)+" "
print(out[:-1])
``` | instruction | 0 | 2,480 | 12 | 4,960 |
No | output | 1 | 2,480 | 12 | 4,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
ar = [0]
n = int(input())
ls = list(map(int, input().split()))
for i in range(n):
ar.append(ls[i])
possible = [True] * n
ar2 = []
for i in range(1, n + 1):
val = ar[i] - ar[i - 1]
ar2.append(val)
for i in range(1, n + 1):
temp = [None] * i
for j in range(n):
if temp[j % i] is None or temp[j % i] == ar2[j]:
temp[j % i] = ar2[j]
else:
possible[i - 1] = False
break
s = ''
for i in range(n):
if possible[i]:
s += f'{i + 1} '
print(s)
``` | instruction | 0 | 2,481 | 12 | 4,962 |
No | output | 1 | 2,481 | 12 | 4,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, β¦, x_{k-1}.
Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 β€ i β€ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.
For example, if the x = [1, 2, 3] and n = 5, then:
* a_0 = 0,
* a_1 = x_{0mod 3}+a_0=x_0+0=1,
* a_2 = x_{1mod 3}+a_1=x_1+1=3,
* a_3 = x_{2mod 3}+a_2=x_2+3=6,
* a_4 = x_{3mod 3}+a_3=x_0+6=7,
* a_5 = x_{4mod 3}+a_4=x_1+7=9.
So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].
Now the boy hopes that he will be able to restore x from a! Knowing that 1 β€ k β€ n, help him and find all possible values of k β possible lengths of the lost array.
Input
The first line contains exactly one integer n (1 β€ n β€ 1000) β the length of the array a, excluding the element a_0.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^6).
Note that a_0 is always 0 and is not given in the input.
Output
The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.
The second line of the output should contain l integers β possible lengths of the lost array in increasing order.
Examples
Input
5
1 2 3 4 5
Output
5
1 2 3 4 5
Input
5
1 3 5 6 8
Output
2
3 5
Input
3
1 5 3
Output
1
3
Note
In the first example, any k is suitable, since a is an arithmetic progression.
Possible arrays x:
* [1]
* [1, 1]
* [1, 1, 1]
* [1, 1, 1, 1]
* [1, 1, 1, 1, 1]
In the second example, Bajtek's array can have three or five elements.
Possible arrays x:
* [1, 2, 2]
* [1, 2, 2, 1, 2]
For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.
In the third example, only k = n is good.
Array [1, 4, -2] satisfies the requirements.
Note that x_i may be negative.
Submitted Solution:
```
def recover(xs):
arr = [None] * len(xs)
prev = 0
for i, x in enumerate(xs):
arr[i] = x - prev
prev = x
return arr
def has_cycle(xs, k):
for i in range(k, len(xs)):
if xs[i] != xs[i % k]:
return False
return True
def min_cycle(xs):
for i in range(1, len(xs)):
if has_cycle(xs, i):
return i
return len(xs)
def solve(xs):
n = len(xs)
arr = recover(xs)
k = min_cycle(arr)
solution = list(range(k, n, k)) + [n]
# print(xs, arr, solution)
return solution
n = int(input())
xs = list(map(int, input().split()))[:n]
solution = solve(xs)
print(len(solution))
for x in solution:
print(x, end=" ")
print()
# assert recover([1,2,3,4,5]) == [1,1,1,1,1]
# assert recover([1,5,3]) == [1,4,-2]
# assert has_cycle([1,1,1], 1) == True
# assert has_cycle([1,1,1], 2) == True
# assert has_cycle([1,1,1], 3) == True
# assert has_cycle([1,2,1], 2) == True
# assert has_cycle([1,2,1], 1) == False
# assert has_cycle([1,2,1], 3) == True
# assert has_cycle([1,2,3], 1) == False
# assert has_cycle([1,2,3], 2) == False
# assert has_cycle([1,2,3,1,2], 3) == True
# assert min_cycle([1,2,3,1,2]) == 3
# assert min_cycle([1,2,3,4,1]) == 4
# assert min_cycle([1,2,3,4]) == 4
# assert min_cycle([1,1,1,1]) == 1
# assert solve([1,2,3,4,5]) == [1,2,3,4,5]
# assert solve([1,3,5,6,8]) == [3,5]
# assert solve([1,3,5,7,9]) == [5]
# assert solve([1,3,6,7,9]) == [3,5]
# assert solve([1,5,3]) == [3]
# assert solve([1,5,6]) == [2,3]
# assert solve([1,5,6]) == [2,3]
# assert solve([1]) == [1]
# assert solve([100]) == [1]
# assert solve([1,2,3]) == [1,2,3]
# assert solve([1,2,3]) == [1,2,3]
# assert solve([10,5,5,15,10]) == [3,5]
# assert solve([10,5,5,15,10,10]) == [3,6]
# assert solve([10,5,5,15,10,10,20,15,15]) == [3,6,9]
# assert solve([1000] * 1000) == [1000]
# assert solve(list(range(1, 1000 + 1))) == list(range(1, 1000 + 1))
# assert solve(list(range(1, 1000 + 1))) == list(range(1, 1000 + 1))
``` | instruction | 0 | 2,482 | 12 | 4,964 |
No | output | 1 | 2,482 | 12 | 4,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this task, Nastya asked us to write a formal statement.
An array a of length n and an array k of length n-1 are given. Two types of queries should be processed:
* increase a_i by x. Then if a_{i+1} < a_i + k_i, a_{i+1} becomes exactly a_i + k_i; again, if a_{i+2} < a_{i+1} + k_{i+1}, a_{i+2} becomes exactly a_{i+1} + k_{i+1}, and so far for a_{i+3}, ..., a_n;
* print the sum of the contiguous subarray from the l-th element to the r-th element of the array a.
It's guaranteed that initially a_i + k_i β€ a_{i+1} for all 1 β€ i β€ n-1.
Input
The first line contains a single integer n (2 β€ n β€ 10^{5}) β the number of elements in the array a.
The second line contains n integers a_1, a_2, β¦, a_n (-10^{9} β€ a_i β€ 10^{9}) β the elements of the array a.
The third line contains n-1 integers k_1, k_2, β¦, k_{n-1} (-10^{6} β€ k_i β€ 10^{6}) β the elements of the array k.
The fourth line contains a single integer q (1 β€ q β€ 10^{5}) β the number of queries.
Each of the following q lines contains a query of one of two types:
* if the query has the first type, the corresponding line contains the character '+' (without quotes), and then there are two integers i and x (1 β€ i β€ n, 0 β€ x β€ 10^{6}), it means that integer x is added to the i-th element of the array a as described in the statement.
* if the query has the second type, the corresponding line contains the character 's' (without quotes) and then there are two integers l and r (1 β€ l β€ r β€ n).
Output
For each query of the second type print a single integer in a new line β the sum of the corresponding subarray.
Examples
Input
3
1 2 3
1 -1
5
s 2 3
+ 1 2
s 1 2
+ 3 1
s 2 3
Output
5
7
8
Input
3
3 6 7
3 1
3
+ 1 3
+ 2 4
s 1 3
Output
33
Note
In the first example:
* after the first change a = [3, 4, 3];
* after the second change a = [3, 4, 4].
In the second example:
* after the first change a = [6, 9, 10];
* after the second change a = [6, 13, 14].
Submitted Solution:
```
n = int(input())
a = []
k = []
a = list(map(int, input().split()))
k = list(map(int, input().split()))
q = int(input())
for i in range(q):
qry = list(input().split())
if qry[0] == 's':
l = int(qry[1])
r = int(qry[2])
#if r < n and l > 1:
print(sum(a[l - 1:r]))
elif qry[0] == '+':
i = int(qry[1]) - 1
x = int(qry[2])
a[i] += x
if (i + 1) < n and a[i + 1] < a[i] + k[i]:
a[i + 1] = a[i] + k[i]
``` | instruction | 0 | 2,503 | 12 | 5,006 |
No | output | 1 | 2,503 | 12 | 5,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this task, Nastya asked us to write a formal statement.
An array a of length n and an array k of length n-1 are given. Two types of queries should be processed:
* increase a_i by x. Then if a_{i+1} < a_i + k_i, a_{i+1} becomes exactly a_i + k_i; again, if a_{i+2} < a_{i+1} + k_{i+1}, a_{i+2} becomes exactly a_{i+1} + k_{i+1}, and so far for a_{i+3}, ..., a_n;
* print the sum of the contiguous subarray from the l-th element to the r-th element of the array a.
It's guaranteed that initially a_i + k_i β€ a_{i+1} for all 1 β€ i β€ n-1.
Input
The first line contains a single integer n (2 β€ n β€ 10^{5}) β the number of elements in the array a.
The second line contains n integers a_1, a_2, β¦, a_n (-10^{9} β€ a_i β€ 10^{9}) β the elements of the array a.
The third line contains n-1 integers k_1, k_2, β¦, k_{n-1} (-10^{6} β€ k_i β€ 10^{6}) β the elements of the array k.
The fourth line contains a single integer q (1 β€ q β€ 10^{5}) β the number of queries.
Each of the following q lines contains a query of one of two types:
* if the query has the first type, the corresponding line contains the character '+' (without quotes), and then there are two integers i and x (1 β€ i β€ n, 0 β€ x β€ 10^{6}), it means that integer x is added to the i-th element of the array a as described in the statement.
* if the query has the second type, the corresponding line contains the character 's' (without quotes) and then there are two integers l and r (1 β€ l β€ r β€ n).
Output
For each query of the second type print a single integer in a new line β the sum of the corresponding subarray.
Examples
Input
3
1 2 3
1 -1
5
s 2 3
+ 1 2
s 1 2
+ 3 1
s 2 3
Output
5
7
8
Input
3
3 6 7
3 1
3
+ 1 3
+ 2 4
s 1 3
Output
33
Note
In the first example:
* after the first change a = [3, 4, 3];
* after the second change a = [3, 4, 4].
In the second example:
* after the first change a = [6, 9, 10];
* after the second change a = [6, 13, 14].
Submitted Solution:
```
import sys
n = int(sys.stdin.readline())
a = sys.stdin.readline().strip().split()
a = [int(x) for x in a]
k = sys.stdin.readline().strip().split()
k = [int(x) for x in k]
cache = {}
q = int(sys.stdin.readline())
st = 1
for i in range(q):
inp = sys.stdin.readline().split()
if inp[0]=='s':
s = 0
l = int(inp[1])
r = int(inp[2])
key = str(l)+'-'+str(r)
if r<st:
try:
print(cache[key])
except:
for pos in range(l-1,r):
s += a[pos]
cache[key] = s
print(cache[key])
else:
for pos in range(l-1,r):
s += a[pos]
cache[key] = s
print(cache[key])
else:
st = int(inp[1])
x = int(inp[2])
a[st-1]+=x
for pos in range(st,n):
tmp = a[pos-1]+k[pos-1]
#break
#print(tmp)
if a[pos]<tmp:
a[pos]=tmp#tmp#(a[pos-1]+k[pos-1])
#print(a)
'''
n = int(sys.stdin.readline())
inp = sys.stdin.readline().strip().split()
a = [int(x) for x in inp]
ans = -(10**9)
s = 0
for l in range(n-1):
l+=1
s = 0
for i in range(l-1,n-1):
print(s)
s = s+(abs(a[i]-a[i+1])*(-1)**(i-l))
if s>ans:
ans = s
print(ans)
'''
``` | instruction | 0 | 2,504 | 12 | 5,008 |
No | output | 1 | 2,504 | 12 | 5,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this task, Nastya asked us to write a formal statement.
An array a of length n and an array k of length n-1 are given. Two types of queries should be processed:
* increase a_i by x. Then if a_{i+1} < a_i + k_i, a_{i+1} becomes exactly a_i + k_i; again, if a_{i+2} < a_{i+1} + k_{i+1}, a_{i+2} becomes exactly a_{i+1} + k_{i+1}, and so far for a_{i+3}, ..., a_n;
* print the sum of the contiguous subarray from the l-th element to the r-th element of the array a.
It's guaranteed that initially a_i + k_i β€ a_{i+1} for all 1 β€ i β€ n-1.
Input
The first line contains a single integer n (2 β€ n β€ 10^{5}) β the number of elements in the array a.
The second line contains n integers a_1, a_2, β¦, a_n (-10^{9} β€ a_i β€ 10^{9}) β the elements of the array a.
The third line contains n-1 integers k_1, k_2, β¦, k_{n-1} (-10^{6} β€ k_i β€ 10^{6}) β the elements of the array k.
The fourth line contains a single integer q (1 β€ q β€ 10^{5}) β the number of queries.
Each of the following q lines contains a query of one of two types:
* if the query has the first type, the corresponding line contains the character '+' (without quotes), and then there are two integers i and x (1 β€ i β€ n, 0 β€ x β€ 10^{6}), it means that integer x is added to the i-th element of the array a as described in the statement.
* if the query has the second type, the corresponding line contains the character 's' (without quotes) and then there are two integers l and r (1 β€ l β€ r β€ n).
Output
For each query of the second type print a single integer in a new line β the sum of the corresponding subarray.
Examples
Input
3
1 2 3
1 -1
5
s 2 3
+ 1 2
s 1 2
+ 3 1
s 2 3
Output
5
7
8
Input
3
3 6 7
3 1
3
+ 1 3
+ 2 4
s 1 3
Output
33
Note
In the first example:
* after the first change a = [3, 4, 3];
* after the second change a = [3, 4, 4].
In the second example:
* after the first change a = [6, 9, 10];
* after the second change a = [6, 13, 14].
Submitted Solution:
```
import sys
n = int(sys.stdin.readline())
a = sys.stdin.readline().strip().split()
a = [int(x) for x in a]
k = sys.stdin.readline().strip().split()
k = [int(x) for x in k]
cache = {}
q = int(sys.stdin.readline())
st = 1
for i in range(q):
inp = sys.stdin.readline().split()
if inp[0]=='s':
s = 0
l = int(inp[1])
r = int(inp[2])
key = str(l)+'-'+str(r)
if r>st:
try:
print(cache[key])
except:
for pos in range(l-1,r):
s += a[pos]
cache[key] = s
print(cache[key])
else:
for pos in range(l-1,r):
s += a[pos]
cache[key] = s
print(cache[key])
else:
st = int(inp[1])
x = int(inp[2])
a[st-1]+=x
for pos in range(st,n):
tmp = a[pos-1]+k[pos-1]
#break
#print(tmp)
if a[pos]<tmp:
a[pos]=tmp#tmp#(a[pos-1]+k[pos-1])
#print(a)
'''
n = int(sys.stdin.readline())
inp = sys.stdin.readline().strip().split()
a = [int(x) for x in inp]
ans = -(10**9)
s = 0
for l in range(n-1):
l+=1
s = 0
for i in range(l-1,n-1):
print(s)
s = s+(abs(a[i]-a[i+1])*(-1)**(i-l))
if s>ans:
ans = s
print(ans)
'''
``` | instruction | 0 | 2,505 | 12 | 5,010 |
No | output | 1 | 2,505 | 12 | 5,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this task, Nastya asked us to write a formal statement.
An array a of length n and an array k of length n-1 are given. Two types of queries should be processed:
* increase a_i by x. Then if a_{i+1} < a_i + k_i, a_{i+1} becomes exactly a_i + k_i; again, if a_{i+2} < a_{i+1} + k_{i+1}, a_{i+2} becomes exactly a_{i+1} + k_{i+1}, and so far for a_{i+3}, ..., a_n;
* print the sum of the contiguous subarray from the l-th element to the r-th element of the array a.
It's guaranteed that initially a_i + k_i β€ a_{i+1} for all 1 β€ i β€ n-1.
Input
The first line contains a single integer n (2 β€ n β€ 10^{5}) β the number of elements in the array a.
The second line contains n integers a_1, a_2, β¦, a_n (-10^{9} β€ a_i β€ 10^{9}) β the elements of the array a.
The third line contains n-1 integers k_1, k_2, β¦, k_{n-1} (-10^{6} β€ k_i β€ 10^{6}) β the elements of the array k.
The fourth line contains a single integer q (1 β€ q β€ 10^{5}) β the number of queries.
Each of the following q lines contains a query of one of two types:
* if the query has the first type, the corresponding line contains the character '+' (without quotes), and then there are two integers i and x (1 β€ i β€ n, 0 β€ x β€ 10^{6}), it means that integer x is added to the i-th element of the array a as described in the statement.
* if the query has the second type, the corresponding line contains the character 's' (without quotes) and then there are two integers l and r (1 β€ l β€ r β€ n).
Output
For each query of the second type print a single integer in a new line β the sum of the corresponding subarray.
Examples
Input
3
1 2 3
1 -1
5
s 2 3
+ 1 2
s 1 2
+ 3 1
s 2 3
Output
5
7
8
Input
3
3 6 7
3 1
3
+ 1 3
+ 2 4
s 1 3
Output
33
Note
In the first example:
* after the first change a = [3, 4, 3];
* after the second change a = [3, 4, 4].
In the second example:
* after the first change a = [6, 9, 10];
* after the second change a = [6, 13, 14].
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
k=list(map(int,input().split()))
su=[0]*n
for i in range(n):
if(i>=1):
su[i]=su[i-1]+l[i-1]
try:
for _ in range(int(input())):
a,b,c=input().split()
b=int(b)
c=int(c)
if(a=='+'):
flag2=0
l[b-1]+=c
temp=b
while(temp<n):
su[temp]+=c
temp+=1
for i in range(b,n):
if(l[i]<l[i-1]+k[i-1]):
if(i<n-1):
su[i+1]+=l[i-1]+k[i-1]-l[i]
l[i]=l[i-1]+k[i-1]
else:
print(su[c-1]+l[c-1]-su[b-1])
except:
temp2=0
``` | instruction | 0 | 2,506 | 12 | 5,012 |
No | output | 1 | 2,506 | 12 | 5,013 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1] | instruction | 0 | 2,655 | 12 | 5,310 |
Tags: data structures, dp, greedy, math, number theory, two pointers
Correct Solution:
```
#!/usr/bin/env python3
import sys, getpass
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
M9 = 10**9 + 7 # 998244353
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
# ---------------------------- template ends here ----------------------------
primes = []
LARGE = int(10**3.5) + 10
# LARGE = 100
factors = [[] for _ in range(LARGE + 5)]
for i in range(2, LARGE):
if not factors[i]:
primes.append(i)
for j in range(i, LARGE, i):
factors[j].append(i)
factors = [tuple(x) for x in factors]
# log(primes)
# log(len(primes))
def factorize(x):
res = []
for p in primes:
cur = 0
while x%p == 0:
x = x//p
cur += 1
if cur%2:
res.append(p)
if x == 1:
return tuple(res)
return tuple(res + [x])
def solve_(lst, k):
# your solution here
assert k == 0
res = 1
cur = set()
for x in lst:
xfact = factorize(x)
# log(x, xfact)
if xfact in cur:
res += 1
cur = set()
cur.add(xfact)
else:
cur.add(xfact)
return res
# for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
for case_num in range(int(input())):
# read line as an integer
# k = int(input())
# read line as a string
# srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
_,k = list(map(int,input().split()))
lst = list(map(int,input().split()))
# read multiple rows
# mrr = read_matrix(k) # and return as a list of list of int
# arr = read_strings(k) # and return as a list of str
res = solve(lst, k) # include input here
# print result
# Google and Facebook - case number required
# print("Case #{}: {}".format(case_num+1, res))
# Other platforms - no case number required
print(res)
# print(len(res))
# print(*res) # print a list with elements
# for r in res: # print each list in a different line
# print(res)
# print(*res)
``` | output | 1 | 2,655 | 12 | 5,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1] | instruction | 0 | 2,656 | 12 | 5,312 |
Tags: data structures, dp, greedy, math, number theory, two pointers
Correct Solution:
```
import logging
import sys
from collections import Counter, defaultdict
from inspect import currentframe
# sys.setrecursionlimit(10 ** 6) #Pypyγ γ¨256MBδΈιγθΆ
γγ
input = sys.stdin.readline
logging.basicConfig(level=logging.DEBUG)
def dbg(*args):
id2names = {id(v): k for k, v in currentframe().f_back.f_locals.items()}
logging.debug(", ".join(id2names.get(id(arg), "???") +
" = " + repr(arg) for arg in args))
def prime_factorization(n):
"""
η΄ ε ζ°ε解 O(βN)
"""
i = 2
ret = []
while i * i <= n:
while n % i == 0:
ret.append(i)
n //= i
i += 1
if n > 1:
ret.append(n)
return ret
def solve():
ans = 1
# d = defaultdict(int)
banned_set = set()
n, k = map(int, input().split())
a = list(map(int, input().split()))
for ai in a:
pf = prime_factorization(ai)
c = Counter(pf)
num = 1
for k, v in c.items():
if v % 2 == 1:
num *= k
if num in banned_set:
ans += 1
banned_set = set()
banned_set.add(num)
else:
banned_set.add(num)
# d[num] += 1
return ans
# return max(d.values())
def main():
t = int(input())
for _ in range(t):
print(solve())
if __name__ == "__main__":
main()
``` | output | 1 | 2,656 | 12 | 5,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1] | instruction | 0 | 2,657 | 12 | 5,314 |
Tags: data structures, dp, greedy, math, number theory, two pointers
Correct Solution:
```
from math import ceil,floor,log
import sys,threading
from heapq import heappush,heappop
from collections import Counter,defaultdict,deque
import bisect
input=lambda : sys.stdin.readline().strip()
c=lambda x: 10**9 if(x=="?") else int(x)
def rec(a,l,r,m,i,l1):
# print(l,r)
if(l==r):
l1[l]=i
return
if(l>r):
return
l1[max(m[l:r+1],key=lambda x:a[x])]=i
rec(a,l,max(m[l:r+1],key=lambda x:a[x])-1,m,i+1,l1)
rec(a,max(m[l:r+1],key=lambda x:a[x])+1,r,m,i+1,l1)
class node:
def __init__(self,x,y):
self.a=[x,y]
def __lt__(self,b):
return b.a[0]<self.a[0]
def __repr__(self):
return str(self.a[0])+" "+str(self.a[1])
def main():
prime=[1]*(10**5)
p=[]
for i in range(2,10**5):
if(prime[i]):
for j in range(i*i,10**5,i):
prime[j]=0
p.append(i)
for _ in range(int(input())):
n,k=list(map(int,input().split()))
a=list(map(int,input().split()))
s=set()
ans=1
for m in a:
i=m
j=0
t=1
for j in range(10**5):
k=0
while(i%p[j]==0):
i//=p[j]
k+=1
if(k%2==1):
t*=p[j]
if(p[j]*p[j]>i):
if(i>2):
t*=i
break
if(t in s):
ans+=1
s=set()
s.add(t)
# print(s)
print(ans)
main()
``` | output | 1 | 2,657 | 12 | 5,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1] | instruction | 0 | 2,658 | 12 | 5,316 |
Tags: data structures, dp, greedy, math, number theory, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
max_ = 10 ** 4
tf = [True] * (max_ + 1)
tf[0] = False
tf[1] = False
for i in range(2, int((max_ + 1) ** 0.5 + 1)):
if not tf[i]: continue
for j in range(i * i, max_ + 1, i):
tf[j] = False
primes = []
for i in range(2, max_ + 1):
if tf[i]:
primes.append(i)
def main():
n, k = map(int, input().split())
alst = list(map(int, input().split()))
ans = 1
se = set()
for a in alst:
for p in primes:
pp = p * p
if pp > a:
break
while a % pp == 0:
a //= pp
if a in se:
ans += 1
se = set()
se.add(a)
print(ans)
for _ in range(int(input())):
main()
``` | output | 1 | 2,658 | 12 | 5,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1] | instruction | 0 | 2,660 | 12 | 5,320 |
Tags: data structures, dp, greedy, math, number theory, two pointers
Correct Solution:
```
import functools
import sys
input = iter(sys.stdin.read().splitlines()).__next__
MAX_A = 10**7
MAX_SQRT = 3163 # ceil sqrt MAX_A
# factors = [[] for _ in range(MAX_A+1)] # list of unique prime factors for each i
# for i in range(2, len(factors)):
# if not factors[i]:
# for j in range(i, MAX_A+1, i):
# factors[j].append(i)
# def get_unmatched_primes(x):
# """ primes for which x has odd exponent in prime factorization """
# # sieve too slow in the first place
# exp_parities = [0] * len(factors)
# for i, factor in enumerate(factors[x]):
# while not x % factor:
# x //= factor
# exp_parities[i] ^= 1
# return [factor for i, factor in enumerate(factors[x]) if exp_parities[i]]
# unmatched_prime_product = [1]*(MAX_A+1) # in prime factorization of index, product of primes with odd exponent
# squares = {i**2 for i in range(3163)}
is_prime = [True] * (MAX_SQRT + 1) # primes above 3163 aren't going to be squared in any a_i
for i in range(2, len(is_prime)):
if is_prime[i]:
for j in range(2*i, len(is_prime), i):
is_prime[j] = False
primes_under_sqrt = [i for i in range(2, len(is_prime)) if is_prime[i]]
def eliminate_squares(x):
for prime in primes_under_sqrt:
square = prime**2
while x % square == 0:
x //= square
return x
def solve():
n, k = map(int, input().split())
assert k == 0
a = list(map(int, input().split()))
num_segments = 1
current_remainders = set() # product of unsquared factors
for a_i in a:
a_i_remainder = eliminate_squares(a_i)
if a_i_remainder in current_remainders:
num_segments += 1
current_remainders = set()
current_remainders.add(a_i_remainder)
return num_segments
t = int(input())
output = []
for _ in range(t):
output.append(solve())
print(*output, sep="\n")
``` | output | 1 | 2,660 | 12 | 5,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1] | instruction | 0 | 2,661 | 12 | 5,322 |
Tags: data structures, dp, greedy, math, number theory, two pointers
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
from collections import Counter
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
for _ in range(int(input()) if True else 100000):
#n = int(input())
#n, k = _+4, 3
n, k = map(int, input().split())
#c, d = map(int, input().split())
b = list(map(int, input().split()))
ans = 1
#s = input()
st = set()
for i in b:
fp = prime_factors(i)
pro = 1
for x in fp:
if fp[x] % 2:
pro *= x
if pro in st:
ans += 1
st = set()
st.add(pro)
else:
st.add(pro)
print(ans)
``` | output | 1 | 2,661 | 12 | 5,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1]
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
ans = 1
ct = 0
while n % 2 == 0:
n = n // 2
ct+=1
if(ct&1):
ans*=2
for i in range(3, int(math.sqrt(n))+1, 2):
ct = 0
while n % i == 0:
n = n // i
ct+=1
if(ct&1):
ans*=i
if n > 2:
ans*=n
return ans
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
t = int(input())
for _ in range(t):
n, k = mapin()
a = [int(x) for x in input().split()]
for i in range(n):
a[i] = primeFactors(a[i])
dd = set()
ans = 0
# print(a)
for i in range(n):
if(a[i] not in dd):
dd.add(a[i])
else:
ans+=1
dd = set()
dd.add(a[i])
if(len(dd) >0):
ans+=1
dd = set()
ans2 = 0
for i in range(n):
if(a[n-i-1] not in dd):
dd.add(a[n-i-1])
else:
ans2+=1
dd = set()
dd.add(a[n-i-1])
if(len(dd) >0):
ans2+=1
print(min(ans, ans2))
``` | instruction | 0 | 2,663 | 12 | 5,326 |
Yes | output | 1 | 2,663 | 12 | 5,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1]
Submitted Solution:
```
q=[]
for j in range(2, 4000):
q.append(j**2)
t=int(input())
for i in range(t):
n, k=[int(j) for j in input().split()]
w=[int(j) for j in input().split()]
m=[]
for j in w:
c=0
while c<len(q) and j>=q[c]:
while j%q[c]==0:
j//=q[c]
c+=1
m.append(j)
print(m)
res=1
r=set()
for j in m:
if j in r:
res+=1
r=set([j])
else:
r|={j}
print(res)
``` | instruction | 0 | 2,666 | 12 | 5,332 |
No | output | 1 | 2,666 | 12 | 5,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1]
Submitted Solution:
```
import math
def is_square(n):
sn = math.sqrt(n)
return (sn * sn == n)
def square_free(lst):
for i in range(len(lst)):
for j in range(len(lst)):
if i == j:
continue
else:
if is_square(lst[i] * lst[j]):
return False
else:
continue
return True
t = int(input())
for i in range(t):
n = list(map(int,input().split()))
lst = list(map(int,input().split()))
if len(lst) == 1:
print(1)
continue
cur = 0
ans = 1
for i in range(1,len(lst)):
if square_free(lst[cur:i+1]):
continue
else:
ans += 1
cur = i
print(ans)
``` | instruction | 0 | 2,667 | 12 | 5,334 |
No | output | 1 | 2,667 | 12 | 5,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1]
Submitted Solution:
```
import sys
LI=lambda:list(map(int,sys.stdin.readline().split()))
MI=lambda:map(int,sys.stdin.readline().split())
SI=lambda:sys.stdin.readline().strip('\n')
II=lambda:int(sys.stdin.readline())
ok=[1]*10001
primes=[]
for i in range(2,10001):
if ok[i]:
primes.append(i)
for j in range(i+i,10001,i):
ok[j]=0
for _ in range(II()):
n,k=MI()
s=set()
ans=1
for v in MI():
d=v
x=1
for p in primes:
if v==1:
break
cnt=0
while v%p==0:
cnt+=1
v//=p
if cnt:
cnt=1 if cnt%2 else 2
x*=p**cnt
if d==v:
x=d
if x in s:
# print(s, x)
s.clear()
ans+=1
s.add(x)
print(ans)
``` | instruction | 0 | 2,668 | 12 | 5,336 |
No | output | 1 | 2,668 | 12 | 5,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The only difference is that in this version k = 0.
There is an array a_1, a_2, β¦, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.
Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.
What is the minimum number of continuous segments you should use if you will make changes optimally?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n, k (1 β€ n β€ 2 β
10^5, k = 0).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^7).
It's guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case print a single integer β the answer to the problem.
Example
Input
3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1
Output
3
2
1
Note
In the first test case the division may be as follows:
* [18, 6]
* [2, 4]
* [1]
Submitted Solution:
```
# SHRi GANESHA author: Kunal Verma #
import os
import sys
from bisect import bisect_right
from collections import Counter, defaultdict
from heapq import *
from io import BytesIO, IOBase
from math import gcd, inf, sqrt, ceil
def lcm(a,b):
return (a*b)//gcd(a,b)
'''
mod = 10 ** 9 + 7
fac = [1]
for i in range(1, 2 * 10 ** 5 + 1):
fac.append((fac[-1] * i) % mod)
fac_in = [pow(fac[-1], mod - 2, mod)]
for i in range(2 * 10 ** 5, 0, -1):
fac_in.append((fac_in[-1] * i) % mod)
fac_in.reverse()
def comb(a, b):
if a < b:
return 0
return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''
#MAXN = 100001
#spf = [0 for i in range(MAXN)]
def getFactorization(x):
ret = list()
while (x != 1):
ret.append(spf[x])
x = x // spf[x]
return ret
def printDivisors(n):
i = 2
z=[1,n]
while i <= sqrt(n):
if (n % i == 0):
if (n / i == i):
z.append(i)
else:
z.append(i)
z.append(n//i)
i = i + 1
return z
def cheaker(a):
j=1
p=False
for i in a:
if i!=(j):
break
j+=1
q=len(a)
for i in a[::-1]:
if i!=q:
break
q-=1
for i in range(j-1,q):
if a[i]==i+1:
return False
return True
def create(n, x,f):
pq = len(bin(n)[2:])
if f==0:
tt=min
else:
tt=max
dp = [[inf] * n for _ in range(pq)]
dp[0] = x
for i in range(1, pq):
for j in range(n - (1 << i) + 1):
dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
return dp
def enquiry(l, r, dp,f):
if l > r:
return inf if not f else -inf
if f==1:
tt=max
else:
tt=min
pq1 = len(bin(r - l + 1)[2:])- 1
return tt(dp[pq1][l], dp[pq1][r - (1 << pq1) + 1])
def SieveOfEratosthenes(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
x=[]
for i in range(2,n+1):
if prime[i] :
x.append(i)
return x
def main():
zz = SieveOfEratosthenes(10 ** 4)
for _ in range(int(input())):
n,k=map(int,input().split())
a=[int(X) for X in input().split()]
p=[]
for i in a:
p.append(1)
for j in range(len(zz)):
c=0
while(i%zz[j]==0):
i=i//zz[j]
c+=1
if c%2:
p[-1]*=zz[j]
if i==1:
break
z=Counter()
an=1
j=0
# print(p)
for i in range(n):
if z[p[i]]!=0:
an+=1
while (j<i):
z[p[j]]-=1
j+=1
z[p[i]]+=1
else:
z[p[i]]+=1
print(an)
#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()
``` | instruction | 0 | 2,669 | 12 | 5,338 |
No | output | 1 | 2,669 | 12 | 5,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,686 | 12 | 5,372 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
from collections import defaultdict as df
import sys
input = lambda : sys.stdin.readline().rstrip()
def compare(a,b):
n1 = len(a) - 1
n2 = len(b) - 1
more = False
ans = []
while True:
if n2 <= 1 and n1 <= 1:
break
if n2 <= 1:
ans.append(a[n1])
n1 -= 1
elif n1 <= 1:
more = True
ans.append('0')
n2 -= 1
else:
if a[n1] == b[n2]:
ans.append('0')
elif a[n1] == '1' and b[n2] == '0':
ans.append('1')
else:
more = True
ans.append('0')
n1 -= 1
n2 -= 1
return ans,more
t = int(input())
for _ in range(t):
#input()
#n = len(s)
n = int(input())
#k,n,m = map(int, input().split())
arr = list(map(int, input().split()))
ans = [0]
ini = bin(arr[0])
for i in range(1, n):
cmp = bin(arr[i])
res,more = compare(ini, cmp)
#print(res, more)
res.reverse()
res = int(''.join(res),2)
ini = bin(res^arr[i])
ans.append(res)
print(*ans)
``` | output | 1 | 2,686 | 12 | 5,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,687 | 12 | 5,374 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
t = int(input())
for j in range(t):
n = int(input())
x = list(map(int, input().split()))
ans = [0]
for i in range(1, n):
k = 0
otv = 0
while 2 ** k <= max(x[i], x[i-1]):
if 2 ** k & x[i-1] == 2**k and 2 ** k & x[i] == 0:
otv += 2 ** k
x[i] = x[i] ^ 2**k
k += 1
ans.append(otv)
print(*ans)
``` | output | 1 | 2,687 | 12 | 5,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,688 | 12 | 5,376 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
#####################################
import atexit, io, sys, collections, math, heapq, fractions,copy, os, functools
import sys
import random
import collections
from io import BytesIO, IOBase
##################################### python 3 START
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")
##################################### python 3 END
def f(ais):
r = [0 for _ in range(len(ais))]
for i in range(1, len(ais)):
ai = ais[i]
cur = 0
prev = r[i - 1] ^ ais[i-1]
for j in range(33):
if (prev >> j) & 1:
if (ai >> j) & 1 == 0:
cur += 1 << j
r[i] = cur
return r
for _ in range(int(input())):
n = int(input())
ais = list(map(int, input().split()))
print (' '.join(map(str,f(ais))))
``` | output | 1 | 2,688 | 12 | 5,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,689 | 12 | 5,378 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
def foo(a,b):
c,d=0,0
while b:
if b&1:
c=c|((a&1)^1)<<d
else:
c+=0
d+=1
a=a>>1
b=b>>1
return c
for _ in range(int(input())):
n=int(input())
x=list(map(int,input().split()))
ans=[0]
rambo=-1
for i in range(1,n):
rambo+=2
ans.append(foo(x[i],(x[i-1]^ans[i-1])))
if rambo != 0:
for i in range(len(ans)):
print(ans[i],end=" ")
print()
``` | output | 1 | 2,689 | 12 | 5,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,690 | 12 | 5,380 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
def solution(s):
res = [0]
a = s[0]
i = 1
while i < len(s):
a = a | s[i]
curr = a ^ s[i]
res.append(curr)
a = curr ^ s[i]
i += 1
return res
if __name__ == '__main__':
n = int(input())
for i in range(n):
input()
# k, _ = tuple(map(int, input().split()))
a = list(map(int, input().split()))
print(*solution(a))
``` | output | 1 | 2,690 | 12 | 5,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,691 | 12 | 5,382 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
T=int(input());
for t in range(T):
n=int(input());
x=list(map(int,input().split()));
y=[];
y.append(0);
for i in range(n-1):
b=x[i]|x[i+1];
y.append(abs(x[i+1]-b));
x[i+1]=b;
print(*y);
``` | output | 1 | 2,691 | 12 | 5,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,692 | 12 | 5,384 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
# Problem: D. Co-growing Sequence
# Contest: Codeforces - Codeforces Round #731 (Div. 3)
# URL: https://codeforces.com/contest/1547/problem/D
# Memory Limit: 512 MB
# Time Limit: 2000 ms
#
# Powered by CP Editor (https://cpeditor.org)
# ____ ____ ___ # skeon19
#| | / | | | |\ | #
#|____ |/ |___ | | | \ | # EON_KID
# | |\ | | | | \ | #
# ____| | \ ___ |____ |___| | \| # Soul_Silver
from collections import defaultdict
import sys,io,os
INP=sys.stdin.readline
inp=lambda:[*map(int,INP().encode().split())]
sinp=lambda:[*map(str,INP().split())]
out=sys.stdout.write
#from functools import reduce
#from bisect import bisect_right,bisect_left
#from sortedcontainers import SortedList, SortedSet, SortedDict
#import sympy #Prime number library
#import heapq
def main():
for _ in range(inp()[0]):
n=inp()[0]
d=defaultdict(int)
l=inp()
y=[0]
ans=[l[0]]
for i in range(1,n):
y.append((ans[i-1]&l[i])^ans[i-1])
ans.append(y[i]^l[i])
print(*y)
###############################################################
def SumOfExpOfFactors(a):
fac = 0
lis=SortedList()
if not a&1:
lis.add(2)
while not a&1:
a >>= 1
fac += 1
for i in range(3,int(a**0.5)+1,2):
if not a%i:
lis.add(i)
while not a%i:
a //= i
fac += 1
if a != 1:
lis.add(a)
fac += 1
return fac,lis
###############################################################
div=[0]*(1000001)
def NumOfDivisors(n): #O(nlog(n))
for i in range(1,n+1):
for j in range(i,n+1,i):
div[j]+=1
###############################################################
def primes1(n): #return a list having prime numbers from 2 to n
""" Returns a list of primes < n """
sieve = [True] * (n//2)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]
##############################################################
def GCD(a,b):
if(b==0):
return a
else:
return GCD(b,a%b)
##############################################################
#
# 1
# / \
# 2 3
# /\ / \ \ \
#4 5 6 7 8 9
# \
# 10
class Graph:
def __init__(self):
self.Graph = defaultdict(list)
def addEdge(self, u, v):
self.Graph[u].append(v)
self.Graph[v].append(u)
#DFS Graph / tree
def DFSUtil(self, v, visited):
visited.add(v)
#print(v, end=' ')
for neighbour in self.Graph[v]:
if neighbour not in visited: #not needed if its a tree
self.DFSUtil(neighbour, visited)
def DFS(self, v):
visited = set() #not needed if its a tree
self.DFSUtil(v, visited) #Visited not needed if its a tree
#BFS Graph / tree
def BFS(self, s):
# Mark all the vertices as not visited
visited = set()
# Create a queue for BFS
queue = []
queue.append(s)
visited.add(s)
while queue:
s = queue.pop(0)
#print (s, end = " ")
for i in self.Graph[s]:
if i not in visited:
queue.append(i)
visited.add(i)
'''
g = Graph()
g.addEdge(1, 2)
g.addEdge(1, 3)
g.addEdge(2, 4)
g.addEdge(2, 5)
g.addEdge(3, 6)
g.addEdge(3, 7)
g.addEdge(3, 8)
g.addEdge(3, 9)
g.addEdge(6,10)
g.DFS(1)
g.BFS(1) '''
##############################################################
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# https://github.com/cheran-senthil/PyRival/blob/master/pyrival/data_structures/SortedList.py
###################################################################################################
if __name__ == '__main__':
main()
``` | output | 1 | 2,692 | 12 | 5,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0 | instruction | 0 | 2,693 | 12 | 5,386 |
Tags: bitmasks, constructive algorithms, greedy
Correct Solution:
```
n = int(input())
for k in range(n):
nn = input()
a = list(map(int, input().split()))
ans = [0]
for i in range(1, len(a)):
st = 1
an = 0
while st <= a[i - 1]:
if ((a[i - 1] // st) % 2 - (a[i] // st) % 2) == 1:
an += st
a[i] += st
st *= 2
ans.append(an)
print(*ans, sep=' ')
``` | output | 1 | 2,693 | 12 | 5,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
from itertools import combinations_with_replacement
import sys
from sys import stdin
import math
import bisect
#Find Set LSB = (x&(-x)), isPowerOfTwo = (x & (x-1))
# 1<<x =2^x
#x^=1<<pos flip the bit at pos
def BinarySearch(a, x):
i = bisect.bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
else:
return -1
def iinput():
return int(input())
def minput():
return map(int,input().split())
def linput():
return list(map(int,input().split()))
def fiinput():
return int(stdin.readline())
def fminput():
return map(int,stdin.readline().strip().split())
def flinput():
return list(map(int,stdin.readline().strip().split()))
for _ in range(iinput()):
n=iinput()
list1=linput()
list2=[0]
for i in range(1,n):
prev=bin(list1[i-1])
nex=bin(list1[i])
l1=len(prev)
l2=len(nex)
if(l2<l1):
nex=nex[0:2]+"0"*(l1-l2)+nex[2:]
if(l1<l2):
prev=prev[0:2]+"0"*(l2-l1)+prev[2:]
# print(prev,nex)
y="0b"
for j in range(2,len(prev)):
if(prev[j]=="0"):
y=y+"0"
elif(prev[j]=="1" and nex[j]=="1"):
y=y+"0"
elif(prev[j]=="1" and nex[j]=="0"):
y=y+"1"
list1[i]=list1[i]^int(y,2)
# print(y)
list2.append(int(y,2))
print(*list2)
# print(x)
# y="0b"
# for j in range(2,len(x)):
# if(x[j]=="0"):
# y=y+"1"
# else:
# y=y+"0"
# y=int(y,2)
# list2.append(y)
# print(*list2)
``` | instruction | 0 | 2,694 | 12 | 5,388 |
Yes | output | 1 | 2,694 | 12 | 5,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
# Test
#####################################################################################################################
def minCo_growingSequenceWith(array):
sequence, prevValue = '0', array[0]
for value in array[1:]:
if prevValue & value != prevValue:
n = prevValue&(prevValue^value)
prevValue = value^n
sequence += f' {n}'
else:
prevValue = value
sequence += ' 0'
return sequence
def testCase_1547d():
input()
return tuple(map(int, input().split(' ')))
testCases = (testCase_1547d() for x in range(int(input())))
tuple(print(minCo_growingSequenceWith(testCase)) for testCase in testCases)
``` | instruction | 0 | 2,695 | 12 | 5,390 |
Yes | output | 1 | 2,695 | 12 | 5,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
import sys
from math import log, floor
#comment these out later
#sys.stdin = open("in.in", "r")
#sys.stdout = open("out.out", "w")
def main():
inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
t = inp[ii]; ii += 1
for _ in range(t):
n = inp[ii]; ii += 1
ar = inp[ii:ii+n]; ii += n
dream = [ar[0]]
top = []
for o in range(n-1):
dream.append((dream[-1] ^ ar[o+1]) | dream[-1])
for i in range(n):
top.append(dream[i] ^ ar[i])
print(" ".join(list(map(str, top))))
main()
``` | instruction | 0 | 2,696 | 12 | 5,392 |
Yes | output | 1 | 2,696 | 12 | 5,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
for i in range(int(input())):
n=int(input())
x=list(map(int,input().split()))
y=[0]
z=x[0]^0
for i in range(1,n):
a=0
for j in range(30):
if z//(2**j)%2:
a+=(1-x[i]//(2**j)%2)*(2**j)
y.append(a)
z=x[i]^a
print(*y)
``` | instruction | 0 | 2,697 | 12 | 5,394 |
Yes | output | 1 | 2,697 | 12 | 5,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
def ans(a,n):
arr = [0]
for i in range(1,n):
x_prev = a[i-1]+arr[i-1]
x_curr = a[i]
if(x_prev>=x_curr):
arr.append(x_prev-x_curr)
continue #correct
#gotta find LS zero
i = 0
temp = x_prev
x = 0
flag = False
while(temp>0):
z = temp%2
if z==0:
flag = True
break
temp = temp//2
i+=1
if flag==True:
if x_prev+2**i>=x_curr:
x = x_prev+2**i-x_curr
else:
x = 2**(i+1)-1-x_curr
else:
x = 2**(i+1)-1-x_curr
arr.append(x)
return(arr)
n = int(input())
for i in range(n):
m = int(input())
b = input().split()
a = []
for i in b:
a.append(int(i))
x = ans(a,m)
for i in x:
print(i, end = ' ')
print()
``` | instruction | 0 | 2,698 | 12 | 5,396 |
No | output | 1 | 2,698 | 12 | 5,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
arrX = list(map(int, input().split()))
print(' '.join(['0'] * n))
``` | instruction | 0 | 2,699 | 12 | 5,398 |
No | output | 1 | 2,699 | 12 | 5,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
dp=[0 for i in range(n)]
def f():
prev=dp[0]^l[0]
j=-1
for i in range(1,n):
t=j+1
s=1<<t
while prev&(s):
t+=1
s<<=1
temp=prev^l[i]
temp2=(prev+s)^l[i]
if temp2<temp:
dp[i]=temp2
prev+=s
j=t
else:
dp[i]=temp
f()
for i in range(n):
print(dp[i],end=" ")
print()
``` | instruction | 0 | 2,700 | 12 | 5,400 |
No | output | 1 | 2,700 | 12 | 5,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.
For example, the following four sequences are growing:
* [2, 3, 15, 175] β in binary it's [10_2, 11_2, 1111_2, 10101111_2];
* [5] β in binary it's [101_2];
* [1, 3, 7, 15] β in binary it's [1_2, 11_2, 111_2, 1111_2];
* [0, 0, 0] β in binary it's [0_2, 0_2, 0_2].
The following three sequences are non-growing:
* [3, 4, 5] β in binary it's [11_2, 100_2, 101_2];
* [5, 4, 3] β in binary it's [101_2, 100_2, 011_2];
* [1, 2, 4, 8] β in binary it's [0001_2, 0010_2, 0100_2, 1000_2].
Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 β y_1, x_2 β y_2, ..., x_n β y_n is growing where β denotes [bitwise XOR](http://tiny.cc/bry9uz).
You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.
The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 β€ k β€ n such that a_i = b_i for any 1 β€ i < k but a_k < b_k.
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains an integer n (1 β€ n β€ 2 β
10^5) β length of the sequence x_i.
The second line contains n integers x_1, x_2, ..., x_n (0 β€ x_i < 2^{30}) β elements of the sequence x_i.
It is guaranteed that the sum of n overall all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print n integers y_1, y_2, ..., y_n (0 β€ y_i < 2^{30}) β lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.
Example
Input
5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0
Output
0 0 0 0
0 1 3 7
0 1 0 3 2
0 2 0 14
0
Submitted Solution:
```
from math import gcd, floor, inf, log2, sqrt, log10
from bisect import bisect_right, bisect_left
from statistics import mode
power_of_two = lambda number: int(log2(number))
cases = int(input())
for _ in range(cases):
n = int(input())
arr = list(map(int, input().split()))
ans = [0]
prev = 0
first = arr[0]
if first == 0:
mx = 0
else:
mx = 2**(power_of_two(first)+1)-1
for num in arr[1:]:
if num == 0:
ans.append(mx)
continue
x = power_of_two(num)
x = max(x, prev)
prev = x
mx = 2**(x+1)-1
y = mx - num
ans.append(y)
print(*ans)
``` | instruction | 0 | 2,701 | 12 | 5,402 |
No | output | 1 | 2,701 | 12 | 5,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levko has an array that consists of integers: a1, a2, ... , an. But he doesnβt like this array at all.
Levko thinks that the beauty of the array a directly depends on value c(a), which can be calculated by the formula:
<image> The less value c(a) is, the more beautiful the array is.
Itβs time to change the world and Levko is going to change his array for the better. To be exact, Levko wants to change the values of at most k array elements (it is allowed to replace the values by any integers). Of course, the changes should make the array as beautiful as possible.
Help Levko and calculate what minimum number c(a) he can reach.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 2000). The second line contains space-separated integers a1, a2, ... , an ( - 109 β€ ai β€ 109).
Output
A single number β the minimum value of c(a) Levko can get.
Examples
Input
5 2
4 7 4 7 4
Output
0
Input
3 1
-100 0 100
Output
100
Input
6 3
1 2 3 7 8 9
Output
1
Note
In the first sample Levko can change the second and fourth elements and get array: 4, 4, 4, 4, 4.
In the third sample he can get array: 1, 2, 3, 4, 5, 6.
Submitted Solution:
```
def f(A, k1, i):
global Min
if i == n + 1:
Sum = 0
for i in range(1, n):
Sum = max(abs(A[i + 1] - A[i]), Sum)
if Sum == 0:
print(A)
if Sum < Min:
Min = Sum
return
if k1 < k:
a = A[i]
A[i] = A[i - 1]
x1 = f(A, k1 + 1, i + 1)
A[i] = a
x2 = f(A, k1, i + 1)
n, k = map(int, input().split())
A1 = [0] + list(map(int, input().split())) + [0]
Min = 10**10
f(A1, 0, 1)
print(Min)
``` | instruction | 0 | 2,791 | 12 | 5,582 |
No | output | 1 | 2,791 | 12 | 5,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levko has an array that consists of integers: a1, a2, ... , an. But he doesnβt like this array at all.
Levko thinks that the beauty of the array a directly depends on value c(a), which can be calculated by the formula:
<image> The less value c(a) is, the more beautiful the array is.
Itβs time to change the world and Levko is going to change his array for the better. To be exact, Levko wants to change the values of at most k array elements (it is allowed to replace the values by any integers). Of course, the changes should make the array as beautiful as possible.
Help Levko and calculate what minimum number c(a) he can reach.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 2000). The second line contains space-separated integers a1, a2, ... , an ( - 109 β€ ai β€ 109).
Output
A single number β the minimum value of c(a) Levko can get.
Examples
Input
5 2
4 7 4 7 4
Output
0
Input
3 1
-100 0 100
Output
100
Input
6 3
1 2 3 7 8 9
Output
1
Note
In the first sample Levko can change the second and fourth elements and get array: 4, 4, 4, 4, 4.
In the third sample he can get array: 1, 2, 3, 4, 5, 6.
Submitted Solution:
```
import sys
import math
MAX = int(1e9)
MIN = int(1e9)
N = 2000
def reset(dp):
for i in range(1, N):
# Can always change all preceding entries
dp[i] = i
def check(n, k, arr, c, dp):
# dp[i]: minimum # of changes to make cost under c and i is unchanged.
reset(dp)
for i in range(1, n):
for j in range(0, i):
# Check if I can update the dp if arr[i] and arr[j] are kept unchanged
# Number of changeable spaces I have
have = i - j - 1
if c:
# Number of changeable spaces I need
needed = math.ceil(abs(arr[i] - arr[j]) / c) - 1
else:
# Can only change if arr[i] and arr[j] are the same if c == 0
needed = have + int(arr[i] != arr[j])
# Can only update dp if I have enough changeable spaces
if needed <= have:
dp[i] = min(dp[i], dp[j] + have)
return dp[n-1] <= k
def main():
# Parse input
lines = sys.stdin.readlines()
l1, l2 = lines[0], lines[1]
n, k = [int(x) for x in l1.strip().split(' ')]
arr = [int(x) for x in l2.strip().split(' ')]
dp = [0 for _ in range(N)]
# Binary search on c
lo = 0
hi = 2 * MAX
while lo < hi:
c = (lo + hi) // 2
valid = check(n, k, arr, c, dp)
if valid:
hi = c
else:
lo = c + 1
print(lo)
if __name__ == '__main__':
main()
``` | instruction | 0 | 2,792 | 12 | 5,584 |
No | output | 1 | 2,792 | 12 | 5,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levko has an array that consists of integers: a1, a2, ... , an. But he doesnβt like this array at all.
Levko thinks that the beauty of the array a directly depends on value c(a), which can be calculated by the formula:
<image> The less value c(a) is, the more beautiful the array is.
Itβs time to change the world and Levko is going to change his array for the better. To be exact, Levko wants to change the values of at most k array elements (it is allowed to replace the values by any integers). Of course, the changes should make the array as beautiful as possible.
Help Levko and calculate what minimum number c(a) he can reach.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 2000). The second line contains space-separated integers a1, a2, ... , an ( - 109 β€ ai β€ 109).
Output
A single number β the minimum value of c(a) Levko can get.
Examples
Input
5 2
4 7 4 7 4
Output
0
Input
3 1
-100 0 100
Output
100
Input
6 3
1 2 3 7 8 9
Output
1
Note
In the first sample Levko can change the second and fourth elements and get array: 4, 4, 4, 4, 4.
In the third sample he can get array: 1, 2, 3, 4, 5, 6.
Submitted Solution:
```
def f(A, k1, i):
global Min
if i == n + 1:
Sum = 0
for i in range(1, n):
Sum = max(abs(A[i + 1] - A[i]), Sum)
if Sum < Min:
Min = Sum
return
if k1 < k:
a = A[i]
A[i] = A[i - 1]
if i == 1:
A[i] = A[i + 1]
x1 = f(A, k1 + 1, i + 1)
A[i] = a
x2 = f(A, k1, i + 1)
n, k = map(int, input().split())
A1 = [0] + list(map(int, input().split())) + [0]
Min = 10**10
f(A1, 0, 1)
print(Min)
``` | instruction | 0 | 2,793 | 12 | 5,586 |
No | output | 1 | 2,793 | 12 | 5,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levko has an array that consists of integers: a1, a2, ... , an. But he doesnβt like this array at all.
Levko thinks that the beauty of the array a directly depends on value c(a), which can be calculated by the formula:
<image> The less value c(a) is, the more beautiful the array is.
Itβs time to change the world and Levko is going to change his array for the better. To be exact, Levko wants to change the values of at most k array elements (it is allowed to replace the values by any integers). Of course, the changes should make the array as beautiful as possible.
Help Levko and calculate what minimum number c(a) he can reach.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 2000). The second line contains space-separated integers a1, a2, ... , an ( - 109 β€ ai β€ 109).
Output
A single number β the minimum value of c(a) Levko can get.
Examples
Input
5 2
4 7 4 7 4
Output
0
Input
3 1
-100 0 100
Output
100
Input
6 3
1 2 3 7 8 9
Output
1
Note
In the first sample Levko can change the second and fourth elements and get array: 4, 4, 4, 4, 4.
In the third sample he can get array: 1, 2, 3, 4, 5, 6.
Submitted Solution:
```
def f(A, k1, i):
global Min
if i == n + 1:
Sum = 0
for i in range(1, n):
Sum = max(abs(A[i + 1] - A[i]), Sum)
if Sum < Min:
Min = Sum
return
if k1 < k:
a = A[i]
A[i] = A[i - 1]
x1 = f(A, k1 + 1, i + 1)
A[i] = a
x2 = f(A, k1, i + 1)
n, k = map(int, input().split())
A1 = [0] + list(map(int, input().split())) + [0]
Min = 10**10
f(A1, 0, 1)
print(Min)
``` | instruction | 0 | 2,794 | 12 | 5,588 |
No | output | 1 | 2,794 | 12 | 5,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers of length n and integer number k. You should print any integer number x in the range of [1; 10^9] (i.e. 1 β€ x β€ 10^9) such that exactly k elements of given sequence are less than or equal to x.
Note that the sequence can contain equal elements.
If there is no such x, print "-1" (without quotes).
Input
The first line of the input contains integer numbers n and k (1 β€ n β€ 2 β
10^5, 0 β€ k β€ n). The second line of the input contains n integer numbers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9) β the sequence itself.
Output
Print any integer number x from range [1; 10^9] such that exactly k elements of given sequence is less or equal to x.
If there is no such x, print "-1" (without quotes).
Examples
Input
7 4
3 7 5 1 10 3 20
Output
6
Input
7 2
3 7 5 1 10 3 20
Output
-1
Note
In the first example 5 is also a valid answer because the elements with indices [1, 3, 4, 6] is less than or equal to 5 and obviously less than or equal to 6.
In the second example you cannot choose any number that only 2 elements of the given sequence will be less than or equal to this number because 3 elements of the given sequence will be also less than or equal to this number. | instruction | 0 | 3,060 | 12 | 6,120 |
Tags: sortings
Correct Solution:
```
n,k = map(int,input().split())
a = [int(i) for i in input().split()]
a = sorted(a)
count = 0
if n < k:
print(-1)
exit()
if k == 0:
ans = a[0]-1
else:
ans = a[k-1]
for i in range(n):
if a[i]<= ans:
count+=1
if count != k or ans<1:
print(-1)
else:
print(ans)
``` | output | 1 | 3,060 | 12 | 6,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers of length n and integer number k. You should print any integer number x in the range of [1; 10^9] (i.e. 1 β€ x β€ 10^9) such that exactly k elements of given sequence are less than or equal to x.
Note that the sequence can contain equal elements.
If there is no such x, print "-1" (without quotes).
Input
The first line of the input contains integer numbers n and k (1 β€ n β€ 2 β
10^5, 0 β€ k β€ n). The second line of the input contains n integer numbers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9) β the sequence itself.
Output
Print any integer number x from range [1; 10^9] such that exactly k elements of given sequence is less or equal to x.
If there is no such x, print "-1" (without quotes).
Examples
Input
7 4
3 7 5 1 10 3 20
Output
6
Input
7 2
3 7 5 1 10 3 20
Output
-1
Note
In the first example 5 is also a valid answer because the elements with indices [1, 3, 4, 6] is less than or equal to 5 and obviously less than or equal to 6.
In the second example you cannot choose any number that only 2 elements of the given sequence will be less than or equal to this number because 3 elements of the given sequence will be also less than or equal to this number. | instruction | 0 | 3,061 | 12 | 6,122 |
Tags: sortings
Correct Solution:
```
n, k = map(int, input().split())
a = sorted(list(map(int, input().split())))
if k == 0:
if min(a) == 1:
print(-1)
else:
print(1)
exit()
c = a[k - 1] + 1
counter = 0
for el in a:
if el <= c:
counter += 1
if counter == k and c <= 10 ** 9:
print(c)
exit()
c = a[k - 1]
counter = 0
for el in a:
if el <= c:
counter += 1
if counter == k and c <= 10 ** 9:
print(c)
exit()
print(-1)
``` | output | 1 | 3,061 | 12 | 6,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers of length n and integer number k. You should print any integer number x in the range of [1; 10^9] (i.e. 1 β€ x β€ 10^9) such that exactly k elements of given sequence are less than or equal to x.
Note that the sequence can contain equal elements.
If there is no such x, print "-1" (without quotes).
Input
The first line of the input contains integer numbers n and k (1 β€ n β€ 2 β
10^5, 0 β€ k β€ n). The second line of the input contains n integer numbers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9) β the sequence itself.
Output
Print any integer number x from range [1; 10^9] such that exactly k elements of given sequence is less or equal to x.
If there is no such x, print "-1" (without quotes).
Examples
Input
7 4
3 7 5 1 10 3 20
Output
6
Input
7 2
3 7 5 1 10 3 20
Output
-1
Note
In the first example 5 is also a valid answer because the elements with indices [1, 3, 4, 6] is less than or equal to 5 and obviously less than or equal to 6.
In the second example you cannot choose any number that only 2 elements of the given sequence will be less than or equal to this number because 3 elements of the given sequence will be also less than or equal to this number. | instruction | 0 | 3,062 | 12 | 6,124 |
Tags: sortings
Correct Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
zzz=lambda:[int(i) for i in stdin.readline().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=True
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
"""
###########################---START-CODING---##############################
num=1
for _ in range( num ):
n,k=zz()
lst=[1]+szzz()
p=lst[k]
if k==n:
print(p)
continue
print(p if lst[k+1]-lst[k] else -1)
``` | output | 1 | 3,062 | 12 | 6,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers of length n and integer number k. You should print any integer number x in the range of [1; 10^9] (i.e. 1 β€ x β€ 10^9) such that exactly k elements of given sequence are less than or equal to x.
Note that the sequence can contain equal elements.
If there is no such x, print "-1" (without quotes).
Input
The first line of the input contains integer numbers n and k (1 β€ n β€ 2 β
10^5, 0 β€ k β€ n). The second line of the input contains n integer numbers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9) β the sequence itself.
Output
Print any integer number x from range [1; 10^9] such that exactly k elements of given sequence is less or equal to x.
If there is no such x, print "-1" (without quotes).
Examples
Input
7 4
3 7 5 1 10 3 20
Output
6
Input
7 2
3 7 5 1 10 3 20
Output
-1
Note
In the first example 5 is also a valid answer because the elements with indices [1, 3, 4, 6] is less than or equal to 5 and obviously less than or equal to 6.
In the second example you cannot choose any number that only 2 elements of the given sequence will be less than or equal to this number because 3 elements of the given sequence will be also less than or equal to this number. | instruction | 0 | 3,063 | 12 | 6,126 |
Tags: sortings
Correct Solution:
```
n,k=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
a.append(1)
a.sort()
if k==n:
print(a[k])
else:
if a[k+1]-a[k]==0:
print(-1)
else:
print(a[k])
``` | output | 1 | 3,063 | 12 | 6,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers of length n and integer number k. You should print any integer number x in the range of [1; 10^9] (i.e. 1 β€ x β€ 10^9) such that exactly k elements of given sequence are less than or equal to x.
Note that the sequence can contain equal elements.
If there is no such x, print "-1" (without quotes).
Input
The first line of the input contains integer numbers n and k (1 β€ n β€ 2 β
10^5, 0 β€ k β€ n). The second line of the input contains n integer numbers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9) β the sequence itself.
Output
Print any integer number x from range [1; 10^9] such that exactly k elements of given sequence is less or equal to x.
If there is no such x, print "-1" (without quotes).
Examples
Input
7 4
3 7 5 1 10 3 20
Output
6
Input
7 2
3 7 5 1 10 3 20
Output
-1
Note
In the first example 5 is also a valid answer because the elements with indices [1, 3, 4, 6] is less than or equal to 5 and obviously less than or equal to 6.
In the second example you cannot choose any number that only 2 elements of the given sequence will be less than or equal to this number because 3 elements of the given sequence will be also less than or equal to this number. | instruction | 0 | 3,064 | 12 | 6,128 |
Tags: sortings
Correct Solution:
```
n,k = map(int,input().split())
l = map(int,input().split())
l = sorted(l)
if k == 0:
if l[0] == 1:
print(-1)
else:
print(1)
elif k == n:
print(1000000000)
else:
if l[k] == l[k-1]:
print(-1)
else:
print(l[k-1])
``` | output | 1 | 3,064 | 12 | 6,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers of length n and integer number k. You should print any integer number x in the range of [1; 10^9] (i.e. 1 β€ x β€ 10^9) such that exactly k elements of given sequence are less than or equal to x.
Note that the sequence can contain equal elements.
If there is no such x, print "-1" (without quotes).
Input
The first line of the input contains integer numbers n and k (1 β€ n β€ 2 β
10^5, 0 β€ k β€ n). The second line of the input contains n integer numbers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^9) β the sequence itself.
Output
Print any integer number x from range [1; 10^9] such that exactly k elements of given sequence is less or equal to x.
If there is no such x, print "-1" (without quotes).
Examples
Input
7 4
3 7 5 1 10 3 20
Output
6
Input
7 2
3 7 5 1 10 3 20
Output
-1
Note
In the first example 5 is also a valid answer because the elements with indices [1, 3, 4, 6] is less than or equal to 5 and obviously less than or equal to 6.
In the second example you cannot choose any number that only 2 elements of the given sequence will be less than or equal to this number because 3 elements of the given sequence will be also less than or equal to this number. | instruction | 0 | 3,065 | 12 | 6,130 |
Tags: sortings
Correct Solution:
```
n,k=(int(i) for i in input().split())
a=[int(i) for i in input().split()]
a.sort()
if k==0:
print(-1+a[0]*(a[0]>1))
quit()
if a[k-1] not in a[k:]:
print(a[k-1])
quit()
print(-1)
``` | output | 1 | 3,065 | 12 | 6,131 |
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