message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,401 | 22 | 98,802 |
"Correct Solution:
```
N=int(input())
ans=0
for i in range(1,N+1):
ans+=(N//i+1)*i*(N//i)//2
print(ans)
``` | output | 1 | 49,401 | 22 | 98,803 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,402 | 22 | 98,804 |
"Correct Solution:
```
n = int(input())
ans = 0
for i in range(1, n+1):
d = n // i
ans += d*(d+1)//2*i
print(ans)
``` | output | 1 | 49,402 | 22 | 98,805 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,403 | 22 | 98,806 |
"Correct Solution:
```
n = int(input())
ans= 0
for i in range(1,n+1):
j = n//i
ans += j*(j+1)*i/2
print(int(ans))
``` | output | 1 | 49,403 | 22 | 98,807 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,404 | 22 | 98,808 |
"Correct Solution:
```
N = int(input())
ans = 0
for i in range(1, N+1):
n = N//i
ans += n*(i+i*n)//2
print(ans)
``` | output | 1 | 49,404 | 22 | 98,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
N = int(input())
SUM = 0
for i in range(1, N+1):
SUM += (i*(N//i)*((N//i)+1))//2
print(SUM)
``` | instruction | 0 | 49,405 | 22 | 98,810 |
Yes | output | 1 | 49,405 | 22 | 98,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
N = int(input())
ans = 0
for i in range(1, N+1):
y = N//i
ans += i*y*(y+1)//2
print(ans)
``` | instruction | 0 | 49,406 | 22 | 98,812 |
Yes | output | 1 | 49,406 | 22 | 98,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
n,a=int(input()),0
for x in range(1,n+1):
y=n//x
a+=y*(y+1)*x//2
print(a)
``` | instruction | 0 | 49,407 | 22 | 98,814 |
Yes | output | 1 | 49,407 | 22 | 98,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
n = int(input())
print(sum(n//i * (n//i+1) // 2 * i for i in range(1, n+1)))
``` | instruction | 0 | 49,408 | 22 | 98,816 |
Yes | output | 1 | 49,408 | 22 | 98,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
def num_divisors_table(n):
table=[0]*(n+1)
for i in range(1,n+1):
for j in range(i,n+1,i):
table[j]+=1
return table
N=int(input())
table=num_divisors_table(N)
ans=0
for K in range(1,N+1):
ans+=K*table[K]
print(ans)
``` | instruction | 0 | 49,409 | 22 | 98,818 |
No | output | 1 | 49,409 | 22 | 98,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
N = int(input())
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return len(divisors)
def is_prime(num):
if num < 2:
return False
if num == 2:
return True
if num % 2 == 0:
return False
sqrt_num = int(num ** 0.5) + 1
for i in range(3, sqrt_num, 2):
if num & i == 0:
return False
return True
ans = 0
for i in range(1, N+1):
if is_prime(i):
ans += i * 2
else:
ans += i * make_divisors(i)
print(ans)
``` | instruction | 0 | 49,410 | 22 | 98,820 |
No | output | 1 | 49,410 | 22 | 98,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
import math
from collections import Counter
N = int(input())
dic = {}
def f(n):
ans = 0
for i in range(1, int(n**0.5)+1):
if n % i == 0:
ans += 1
if i != n // i:
ans += 1
return ans
ans = 0
for i in range(1, N+1):
ans += f(i) * i
print(ans)
``` | instruction | 0 | 49,411 | 22 | 98,822 |
No | output | 1 | 49,411 | 22 | 98,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105
Submitted Solution:
```
n = int(input())
def f(x, result):
if x == 1:
return 1
while True:
if x % 2 == 0:
qcnt = 0
while True:
x = x // 2
qcnt += 1
if x % 2 != 0:
break
return result[x] * (qcnt + 1)
else:
break
while True:
if x == 1:
break
else:
p = 3
while True:
if x % p == 0:
pcnt = 0
while True:
x = x // p
pcnt += 1
if x % p != 0:
break
return result[x] * (pcnt + 1)
else:
p += 2
y = 1
for i in yakusu.values():
y *= (i+1)
return y
s = 0
result = {1:1}
for i in range(1,n+1):
result[i] = f(i, result)
s += i * f(i,result)
print(s)
``` | instruction | 0 | 49,412 | 22 | 98,824 |
No | output | 1 | 49,412 | 22 | 98,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yasin has an array a containing n integers. Yasin is a 5 year old, so he loves ultimate weird things.
Yasin denotes weirdness of an array as maximum gcd(ai, aj) value among all 1 ≤ i < j ≤ n. For n ≤ 1 weirdness is equal to 0, gcd(x, y) is the greatest common divisor of integers x and y.
He also defines the ultimate weirdness of an array. Ultimate weirdness is <image> where f(i, j) is weirdness of the new array a obtained by removing all elements between i and j inclusive, so new array is [a1... ai - 1, aj + 1... an].
Since 5 year old boys can't code, Yasin asks for your help to find the value of ultimate weirdness of the given array a!
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in a.
The next line contains n integers ai (1 ≤ ai ≤ 200 000), where the i-th number is equal to the i-th element of the array a. It is guaranteed that all ai are distinct.
Output
Print a single line containing the value of ultimate weirdness of the array a.
Example
Input
3
2 6 3
Output
6
Note
Consider the first sample.
* f(1, 1) is equal to 3.
* f(2, 2) is equal to 1.
* f(3, 3) is equal to 2.
* f(1, 2), f(1, 3) and f(2, 3) are equal to 0.
Thus the answer is 3 + 0 + 0 + 1 + 0 + 2 = 6.
Submitted Solution:
```
in1 = int(input())
in2 = input()
in2 = in2.split()
r = 0
def f(A,a,b):
if(a<=b):
for i in list(range(a-1,b)):
A.pop(a-1)
else:
for i in range(a-1,len(A)):
A.pop(a-1)
for i in range(0,b):
A.pop(0)
if((len(A)==1)|(len(A)==0)):
return(0)
else:
mn = int(min(A))
A.remove(min(A))
for i in A:
if(int(i)%mn!=0):
return(1)
break
else:
return(mn)
for k1 in range(1,in1+1):
for k2 in range(1, in1+1):
A = list(in2)
r = r + f(A,k1,k2)
print(r)
``` | instruction | 0 | 50,111 | 22 | 100,222 |
No | output | 1 | 50,111 | 22 | 100,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yasin has an array a containing n integers. Yasin is a 5 year old, so he loves ultimate weird things.
Yasin denotes weirdness of an array as maximum gcd(ai, aj) value among all 1 ≤ i < j ≤ n. For n ≤ 1 weirdness is equal to 0, gcd(x, y) is the greatest common divisor of integers x and y.
He also defines the ultimate weirdness of an array. Ultimate weirdness is <image> where f(i, j) is weirdness of the new array a obtained by removing all elements between i and j inclusive, so new array is [a1... ai - 1, aj + 1... an].
Since 5 year old boys can't code, Yasin asks for your help to find the value of ultimate weirdness of the given array a!
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in a.
The next line contains n integers ai (1 ≤ ai ≤ 200 000), where the i-th number is equal to the i-th element of the array a. It is guaranteed that all ai are distinct.
Output
Print a single line containing the value of ultimate weirdness of the array a.
Example
Input
3
2 6 3
Output
6
Note
Consider the first sample.
* f(1, 1) is equal to 3.
* f(2, 2) is equal to 1.
* f(3, 3) is equal to 2.
* f(1, 2), f(1, 3) and f(2, 3) are equal to 0.
Thus the answer is 3 + 0 + 0 + 1 + 0 + 2 = 6.
Submitted Solution:
```
in1 = int(input())
in2 = input()
in2 = in2.split()
r = 0
def f(A,a,b):
for i in list(range(a-1,b)):
A.pop(a-1)
if((len(A)==1)|(len(A)==0)):
return(0)
else:
mn = int(min(A))
A.remove(min(A))
for i in A:
if(int(i)%mn!=0):
return(1)
break
else:
return(mn)
for k1 in range(1,in1+1):
for k2 in range(k1, in1+1):
A = list(in2)
r = r + f(A,k1,k2)
print(r)
``` | instruction | 0 | 50,112 | 22 | 100,224 |
No | output | 1 | 50,112 | 22 | 100,225 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,265 | 22 | 100,530 |
"Correct Solution:
```
n=int(input())
a=0
for i in range(2,10**6):
b=0
while n%i==0:
n//=i
b+=1
c=d=0
while c+d<b:
c+=1
d+=c
a+=c
print(a+(n>1))
``` | output | 1 | 50,265 | 22 | 100,531 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,266 | 22 | 100,532 |
"Correct Solution:
```
from collections import defaultdict
N=int(input())
soinsu=defaultdict(int)
x=2
while x<=10**6:
while N%x==0:
soinsu[x]+=1
N//=x
x+=1
if N>1:
soinsu[N]+=1
ans=0
for key in soinsu.keys():
x=1
while soinsu[key]>=x:
ans+=1
soinsu[key]-=x
x+=1
print(ans)
``` | output | 1 | 50,266 | 22 | 100,533 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,267 | 22 | 100,534 |
"Correct Solution:
```
n=int(input())
def prime(n):
dic={}
f=2
m=n
while f*f<=m:
r=0
while n%f==0:
n//=f
r+=1
if r>0:
dic[f]=r
f+=1
if n!=1:
dic[n]=1
return dic
def counter(dic):
ans=0
for val in dic.values():
i=1
while i*(i+3)/2<val:
i+=1
ans+=i
return ans
print(counter(prime(n)))
``` | output | 1 | 50,267 | 22 | 100,535 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,268 | 22 | 100,536 |
"Correct Solution:
```
N = int(input())
primes = {}
for num in range(2, int(N**0.5)+1):
while N%num == 0:
if num in primes:
primes[num] += 1
else:
primes[num] = 1
N //= num
ans = 0
for p in primes.values():
n = 1
while n*(n+1)//2 <= p:
n += 1
ans += n-1
if N > 1:
ans += 1
print(ans)
``` | output | 1 | 50,268 | 22 | 100,537 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,269 | 22 | 100,538 |
"Correct Solution:
```
n = int(input())
a = []
ans = 0
for i in range(2,int(n**0.5)+1):
count = 0
while n%i==0:
n/=i
count+=1
if count!=0:
for j in range(1,count+10):
if j*(j+1)/2>count:
ans+=j-1
break
if n!=1:
ans+=1
print(ans)
``` | output | 1 | 50,269 | 22 | 100,539 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,270 | 22 | 100,540 |
"Correct Solution:
```
import math
n=int(input())
p=2
c=0
curr=0
while n%p==0:
curr+=1
n//=p
j=0
j=(math.sqrt(1+8*curr)-1)//2
c+=int(j)
#print(j)
for i in range (3,int(math.sqrt(n))+1,2):
p=i
curr=0
while n%p==0:
curr+=1
n//=p
j=0
j=(math.sqrt(1+8*curr)-1)//2
c+=int(j)
#print(j)
if n>2:
c+=1
print(c)
``` | output | 1 | 50,270 | 22 | 100,541 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,271 | 22 | 100,542 |
"Correct Solution:
```
ans = 0
N = int(input())
x = 2
while not N % x:
ans += 1
N //= x
x *= 2
while not N % 2:
N //= 2
i = 3
while i*i <= N:
x = i
while not N % x:
ans += 1
N //= x
x *= i
while not N % i:
N //= i
i += 2
if N > 1: ans += 1
print(ans)
``` | output | 1 | 50,271 | 22 | 100,543 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7 | instruction | 0 | 50,272 | 22 | 100,544 |
"Correct Solution:
```
#import math
n=int(input())
ans=0
p=2
for p in range(2, int((n+0.5)**0.5)+1):
if n%p==0:
cnt=0
while n%p==0:
cnt+=1
n//=p
k=1
while cnt>=k:
cnt-=k
k+=1
ans+=1
if n>1:
ans+=1
print(ans)
``` | output | 1 | 50,272 | 22 | 100,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive integer N. Consider repeatedly applying the operation below on N:
* First, choose a positive integer z satisfying all of the conditions below:
* z can be represented as z=p^e, where p is a prime number and e is a positive integer;
* z divides N;
* z is different from all integers chosen in previous operations.
* Then, replace N with N/z.
Find the maximum number of times the operation can be applied.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
Output
Print the maximum number of times the operation can be applied.
Examples
Input
24
Output
3
Input
1
Output
0
Input
64
Output
3
Input
1000000007
Output
1
Input
997764507000
Output
7
Submitted Solution:
```
#D
import math
#試し割法
def trial_division(n):
#素因数を格納するリスト
factor = []
#2から√n以下の数字で割っていく
tmp = int(math.sqrt(n)) + 1
for num in range(2,tmp):
while n % num == 0:
n //= num
if num not in factor:
factor.append(num)
#リストが空ならそれは素数
if not factor:
return 'prime number'
else:
factor.append(n)
return factor
n=int(input())
if n==1:
ans=0
elif trial_division(n)== 'prime number':
ans=1
else:
z=2
ans=0
while z <= n/2:
if n%z !=0:
continue
if trial_division(z)== 'prime number':
n = n/z
z += 1
ans += 1
elif len(trial_division(z))==2:
n = n/z
z += 1
ans += 1
else:
z +=1
print(ans)
``` | instruction | 0 | 50,280 | 22 | 100,560 |
No | output | 1 | 50,280 | 22 | 100,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,543 | 22 | 101,086 |
Tags: brute force, math, number theory
Correct Solution:
```
a, b = map(int,input().split())
if a < b: a,b=b,a
if a == b: print(0); exit()
def update(g):
global ans
# a+n must be a multiple of g, and so is b+n
if a%g != b%g: return
n = g - a%g if a%g else 0
lcm = (a+n)*(b+n)//g
ans = min(ans, (lcm, n))
d = a-b
ans = (10**31, 10**31) # lcm, N
for gcd in range(1, int(d**.5)+2):
if d%gcd: continue
update(gcd)
update(d // gcd)
print(ans[1])
``` | output | 1 | 50,543 | 22 | 101,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,544 | 22 | 101,088 |
Tags: brute force, math, number theory
Correct Solution:
```
import math
a, b = map(int,input().split())
if b > a:
a,b=b,a
if not a % b:
print(0)
exit()
else:
sub=a-b
divs = []
for i in range(1, int(math.sqrt(sub))+1):
if sub % i == 0:
divs.append(i)
divs.append(sub // i)
min_ = 10e19
for i in range(len(divs)):
k = divs[i]- b % divs[i]
if b % divs[i]==0:
k=0
if ((a + k) * (b + k))/divs[i] < min_:
min_ = ((a + k)* (b + k))/ divs[i]
k_min = k
print(k_min)
``` | output | 1 | 50,544 | 22 | 101,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,545 | 22 | 101,090 |
Tags: brute force, math, number theory
Correct Solution:
```
a,b=list(map(int,input().split()))
(a,b)=min(a,b),max(a,b)
if b<2*a:
if a==b:
print(0)
exit()
print((-a)%(b-a))
exit()
s=[1]
q=b-a
for i in range(2,int((b-a)**(1/2)+2)):
while q%i==0:
t=[j*i for j in s]
for aa in t:
s.append(aa)
s=list(set(s))
q=q//i
if q!=1:
t=[j*q for j in s]
for aa in t:
s.append(aa)
s=list(set(s))
s.sort()
for i in s:
if i>=a:
print(i-a)
exit()
``` | output | 1 | 50,545 | 22 | 101,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,546 | 22 | 101,092 |
Tags: brute force, math, number theory
Correct Solution:
```
from fractions import gcd
a,b=map(int,input().split())
d=max(a,b)-min(a,b)
c=[float("inf"),1]
e=[1]
for i in range(1,int(d**0.5)+1):
if i**2==d:
e.append(i)
elif d%i==0:
e.append(i)
e.append(d//i)
for i in e:
h=i*((a-1)//i+1)-a
if c[0]>((a+h)*(b+h)//gcd(a+h,b+h)):
c[0]=(a+h)*(b+h)//gcd(a+h,b+h)
c[1]=h
print(c[1])
``` | output | 1 | 50,546 | 22 | 101,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,547 | 22 | 101,094 |
Tags: brute force, math, number theory
Correct Solution:
```
def gcd(a, b):
return b if a % b == 0 else gcd(b, a % b)
def findans(d):
l, r = 1, 10**20
while l < r:
mid = (l + r) // 2
if mid * d >= m:
r = mid
else:
l = mid + 1
k = l * d - m
lcm = (a + k) * (b + k) // d
return lcm, k
a, b = map(int, input().split())
x, y = a, b
if a == b:
print(0)
else:
minlcm = 10 ** 20
diff = abs(a - b)
m = max(a, b)
ans = 0
import math
for i in range(1, int(math.sqrt(diff) + 1)):
if diff % i:
continue
lcm, k = findans(diff // i)
if lcm < minlcm:
minlcm = lcm
ans = k
lcm, k = findans(i)
if lcm < minlcm:
minlcm = lcm
ans = k
print(ans)
``` | output | 1 | 50,547 | 22 | 101,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,548 | 22 | 101,096 |
Tags: brute force, math, number theory
Correct Solution:
```
import math
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
def GCD(x,y):
while y!=0:
z=x%y
x=y
y=z
return x
def LCM(x,y):
return x*y//GCD(x,y)
def factors(n):
super=[1]
for i in prime_factors(n):
super.extend([j*i for j in super])
return list(sorted(set(super)))
n,m=map(int,input().split())
_factor = factors(abs(n-m))
ans = 0
minimum = math.inf
for i in _factor:
k = (i-n%i) % i
if LCM(n+k, m+k) < minimum:
ans = k
minimum = LCM(n+k, m+k)
print(ans)
``` | output | 1 | 50,548 | 22 | 101,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,549 | 22 | 101,098 |
Tags: brute force, math, number theory
Correct Solution:
```
import io, sys, atexit, os
import math as ma
from sys import exit
from decimal import Decimal as dec
from itertools import permutations
def li ():
return list (map (int, input ().split ()))
def num ():
return map (int, input ().split ())
def nu ():
return int (input ())
def find_gcd ( x, y ):
while (y):
x, y = y, x % y
return x
def lcm(x,y):
gg=find_gcd(x,y)
return (x*y//gg)
mm = 1000000007
yp = 0
def solve ():
t = 1
for tt in range (t):
a,b=num()
ff=max(a,b)
gg=min(a,b)
a=gg
b=ff
pq=max(a,b)-min(a,b)
if(a==b):
print(0)
continue
else:
if((a%pq==0 and b%pq==0)or b%a==0):
print(0)
else:
i = 1
mo=[]
while i <= ma.sqrt (pq):
if (pq % i == 0):
# If divisors are equal, print only one
if (pq / i == i):
mo.append(i)
else:
# Otherwise print both
mo.append(i)
mo.append(pq//i)
i = i + 1
mn=-1
ind=-1
for i in range(len(mo)):
pqq=mo[i]
if(pqq==1):
continue
hp=(a//pqq+1)*pqq-a
po= (b // pqq + 1) * pqq - b
if(mn==-1):
up = lcm (a + hp, po + b)
mn=up
ind=hp
else:
up = lcm (a + hp, po + b)
if(up<=mn):
mn=up
ind=min(ind,hp)
print(ind)
if __name__ == "__main__":
solve ()
``` | output | 1 | 50,549 | 22 | 101,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible. | instruction | 0 | 50,550 | 22 | 101,100 |
Tags: brute force, math, number theory
Correct Solution:
```
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
#divisors.sort(reverse=True)
return divisors
import math
from functools import reduce
def lcm_base(x, y):
return (x * y) // math.gcd(x, y)
def lcm_list(numbers):
return reduce(lcm_base, numbers, 1)
a, b = map(int, input().split())
if a == b:
print(0)
exit()
if a > b:
a, b = b, a
d = make_divisors(b-a)
X = []
for g in d:
k = ((a+g-1)//g)*g-a
a_ = a+k
b_ = b+k
l = lcm_base(a_, b_)
X.append((l, k))
X.sort(key=lambda x: (x[0], x[1]))
#print(X)
print(X[0][1])
``` | output | 1 | 50,550 | 22 | 101,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
import math
def main():
buf = input()
buflist = buf.split()
a = int(buflist[0])
b = int(buflist[1])
# always a <= b
if a > b:
a, b = b, a
diff = b - a
if diff <= 1:
print(0) # no need to add numbers
return
k = (b - a - a) % diff # magic
gcd = math.gcd(a+k, b+k)
#print(a+k, b+k, gcd, lcm(a+k, b+k))
cd = common_divisor(a+k,b+k)
smallest_cm = lcm(a+k, b+k)
smallest_k = k
for i in cd:
#print(i)
j = k // i
#print(a+k-i*j, b+k-i*j, lcm(a+k-i*j, b+k-i*j))
l = lcm(a+k-i*j, b+k-i*j)
if l < smallest_cm or l == smallest_cm and k < smallest_k:
smallest_cm = l
smallest_k = k-i*j
print(smallest_k)
def common_divisor(a, b):
gcd = math.gcd(a, b)
divisor = []
for i in range(1, int(gcd ** 0.5)+1):
if gcd % i == 0:
divisor.append(i)
if gcd // i > i:
divisor.append(gcd // i)
divisor = list(sorted(divisor))
return divisor
def lcm(a, b):
return a * b // math.gcd(a, b)
if __name__ == '__main__':
main()
``` | instruction | 0 | 50,551 | 22 | 101,102 |
Yes | output | 1 | 50,551 | 22 | 101,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
'''
# Read input from stdin and provide input before running code
name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''
import math
a,b=list(map(int,input().split()));
if(a>b):
a,b=b,a
diff=b-a;
j=1;
ans=(a*b)/math.gcd(a,b);
val=0;
while(j*j<=diff):
if(diff%j==0):
k=(j-a%j)%j;
temp=((a+k)*(b+k))/math.gcd(a+k,b+k);
if(ans>temp):
ans=temp;
val=k;
x=int(diff/j);
#print(x);
k=(x-a%x)%x;
temp=((a+k)*(b+k))/math.gcd(a+k,b+k);
if(ans>temp):
ans=temp;
val=k;
j=j+1;
print(val);
``` | instruction | 0 | 50,552 | 22 | 101,104 |
Yes | output | 1 | 50,552 | 22 | 101,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
import math
import sys
def getfactors(num):
l=[]
for i in range(1,int(math.sqrt(num))+1):
if num%i==0:
l.append(i)
l.append(num//i)
return l
x,y=map(int,sys.stdin.readline().split())
a,b = min(x,y), max(x,y)
hcf = math.gcd(a,b)
d = b - a
cur = (a*b)/(hcf)
k = 0
#print(hcf,'hcf',d,'d')
fact = getfactors(d)
n = len(fact)
for i in range(n):
if a%fact[i]==0:
newk=0
newlcm=((a+newk)*(b+newk)/(math.gcd(a+newk,b+newk)))
if newlcm<cur:
cur=newlcm
k=newk
else:
newk=(fact[i]-a%fact[i])
newlcm=((a+newk)*(b+newk)/(math.gcd(a+newk,b+newk)))
if newlcm<cur:
cur=newlcm
k=newk
ans = k
print(ans)
``` | instruction | 0 | 50,553 | 22 | 101,106 |
Yes | output | 1 | 50,553 | 22 | 101,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
from math import gcd
lcm = lambda x, y: (x * y) // gcd(x, y)
a, b = map(int, input().split())
c = abs(a - b)
f = 1
res = [lcm(a, b), 0]
while f * f <= c:
if c % f == 0:
k = min((-a) % f, (-b) % f)
ans = lcm(a + k, b + k)
if res[0] > ans: res = [ans, k]
elif res[0] == ans: res[1] = min(res[1], k)
k = min((-a) % (c // f), (-b) % (c // f))
ans = lcm(a + k, b + k)
if res[0] > ans: res = [ans, k]
elif res[0] == ans: res[1] = min(res[1], k)
f += 1
print(res[1])
``` | instruction | 0 | 50,554 | 22 | 101,108 |
Yes | output | 1 | 50,554 | 22 | 101,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
a, b = map(int, input().split())
k = abs(a - b)
print((k - a % k) % k)
``` | instruction | 0 | 50,555 | 22 | 101,110 |
No | output | 1 | 50,555 | 22 | 101,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
from fractions import gcd
def main():
a, b = map(int, input().split())
a, b = min(a, b), max(a, b)
d = b - a
if a % d == 0:
print(a)
return 0
l = (a // d + 1) * d
print(l - a)
main()
``` | instruction | 0 | 50,556 | 22 | 101,112 |
No | output | 1 | 50,556 | 22 | 101,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
mod = int(1e9) + 7
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def totalPrimeFactors(n):
count = 0
if (n % 2) == 0:
count += 1
while (n % 2) == 0:
n //= 2
i = 3
while i * i <= n:
if (n % i) == 0:
count += 1
while (n % i) == 0:
n //= i
i += 2
if n > 2:
count += 1
return count
# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
# spf[1] = 1
# for i in range(2, MAXN):
# spf[i] = i
# for i in range(4, MAXN, 2):
# spf[i] = 2
#
# for i in range(3, mt.ceil(mt.sqrt(MAXN))):
# if (spf[i] == i):
# for j in range(i * i, MAXN, i):
# if (spf[j] == j):
# spf[j] = i
#
#
# def getFactorization(x):
# ret = 0
# while (x != 1):
# k = spf[x]
# ret += 1
# # ret.add(spf[x])
# while x % k == 0:
# x //= k
#
# return ret
# Driver code
# precalculating Smallest Prime Factor
# sieve()
def isprime(k):
if k == 2:
return 1
if k % 2 == 0 or k % 3 == 0:
return 0
i = 5
while i * i <= k:
if k % i == 0:
return 0
i += 2
return 1
def main():
a, b = map(int, input().split())
k = gcd(a, b)
a //= k
b //= k
if a > b:
a, b = b, a
i = int(min(a, b)**0.5)+1
ans = 0
while 1:
if i<=1:
break
if isprime(i) and b % i == a % i:
ans += k * ((i - a % i) % i)
z = (i - a % i) % i
a += z
b += z
while a % i == 0 and b % i == 0:
k *= i
a //= i
b //= i
i -=1
print(ans)
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 50,557 | 22 | 101,114 |
No | output | 1 | 50,557 | 22 | 101,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
Examples
Input
6 10
Output
2
Input
21 31
Output
9
Input
5 10
Output
0
Note
In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
Submitted Solution:
```
a, b = map(int, input().split())
if a > b:
a, b = b, a
if a == 1:
print(0)
exit()
if a == b:
print(0)
exit()
k = abs(a - b)
print(min((a - b % a) % a, (k - a % k) % k))
``` | instruction | 0 | 50,558 | 22 | 101,116 |
No | output | 1 | 50,558 | 22 | 101,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 50,687 | 22 | 101,374 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
T = input()
mod = int(1e9 + 7)
a = map(int, input().split())
c = []
for n in a:
b = (n // 2 + 2)
b = b * b
b //= 4
c.append(str(b % mod))
print(' '.join(c))
``` | output | 1 | 50,687 | 22 | 101,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 50,688 | 22 | 101,376 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
t = int(input())
a = list(map(int, input().split()))
out = []
for n in a:
ans = (n//2 + 2)
ans = ans*ans
ans //= 4
out.append(ans%1000000007)
print(' '.join(str(x) for x in out))
``` | output | 1 | 50,688 | 22 | 101,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 50,690 | 22 | 101,380 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##to find factorial and ncr
# N=100000
# mod = 10**9 +7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def solve():
n=N()
ar=lis()
for i in range(len(ar)):
m=ar[i]
v = m // 2
u = v // 2
w = (v - u)
print((u * w + u + w + 1) % mod)
solve()
#testcase(int(inp()))
``` | output | 1 | 50,690 | 22 | 101,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
from math import *
T = int(input())
mod = int(1e9 + 7)
n = map(int, input().split())
for i in n:
print(int(floor(floor(i / 2 + 2) ** 2 / 2) / 2) % mod)
``` | instruction | 0 | 50,694 | 22 | 101,388 |
No | output | 1 | 50,694 | 22 | 101,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you have a sequence of k integers A = [a_1, a_2, ... , a_k] where each a_i ≥ 2. A sequence of prime integers P = [p_1, p_2, ..., p_k] is called suitable for the sequence A if a_1 is divisible by p_1, a_2 is divisible by p_2 and so on.
A sequence of prime integers P is called friendly if there are no unique integers in this sequence.
A sequence A is called ideal, if each sequence P that is suitable for A is friendly as well (i. e. there is no sequence P that is suitable for A, but not friendly). For example, the sequence [2, 4, 16] is ideal, while the sequence [2, 4, 6] is not ideal (there exists a sequence P = [2, 2, 3] which is suitable for A, but not friendly).
You are given n different integers x_1, x_2, ..., x_n. You have to choose exactly k of them in such a way that they form an ideal sequence, or report that it is impossible. Note that no integer can be chosen more than once.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (2 ≤ x_i ≤ 10^{18}).
Output
If it is impossible to choose exactly k integers from x_1, x_2, ..., x_n in such a way that the chosen integers form an ideal sequence, print 0.
Otherwise, print k pairwise distinct integers — the elements of the chosen ideal sequence. If there are multiple answers, print any of them.
Examples
Input
3 3
2 4 6
Output
0
Input
3 3
2 4 16
Output
2 4 16
Input
4 3
2 4 6 16
Output
2 4 16 | instruction | 0 | 50,711 | 22 | 101,422 |
Tags: binary search, greedy, math, number theory
Correct Solution:
```
from math import gcd
def t_prime(n):
if n == 1:
return False
for p in [2,3,5,7,11,13,17,19,23]:
if p < n and pow(p,n-1,n) != 1:
return False
return True
primes = [i for i in range(2,10**5) if t_prime(i)]
pset = set(primes)
n, k = map(int, input().split())
l = list(map(int, input().split()))
if k == 1:
print(0)
exit()
for i in range(n):
for j in range(i):
u, v = l[i], l[j]
poss = gcd(u,v)
poss2 = max(u,v)//poss
smol = min(poss,poss2)
if t_prime(smol) and smol not in pset:
primes.append(smol)
pset.add(smol)
powers = set()
count = 0
outLs = []
pgood = []
for p in primes:
curr = []
fp = [v for v in l if v % p == 0]
for v in fp:
v2 = v
while v2 % p == 0:
v2 //= p
if v2 == 1:
curr.append(v)
powers.add(v)
if len(curr) > 1:
count += len(curr)
outLs.append(curr)
pgood.append(p)
order = [(len(lis), lis) for lis in outLs]
order.sort(key = lambda x: x[0])
if len(order) == 0:
print(0)
exit()
if order[-1][0] == 2 and k % 2 and count > k:
extra = -1
need = -1
last = []
for v in l:
if v in powers:
continue
v2 = v
primesn = []
for p in pgood:
add = 1
while v2 % p == 0:
v2 //= p
if add:
primesn.append(p)
add = 0
if v2 == 1 and (need == -1 or need > len(primesn)):
extra = v
last = primesn
need = len(last)
assert need >= 2
if need == -1 or 2 * need + 1 > k:
print(0)
exit()
other = []
out = [extra]
for a,b in outLs:
works = False
for p in last:
if a % p == 0:
works = True
break
if works:
out.append(a)
out.append(b)
else:
other.append(a)
other.append(b)
assert len(out) == 2 * need + 1
assert (k - 2 * need - 1) % 2 == 0
ret = out + other[:(k - 2*need - 1)]
assert len(ret) == k
print(' '.join(map(str,ret)))
exit()
out = []
need = k
for i in range(len(order)):
assert need != 1
lis = order[i][1]
if len(lis) < need - 1 or len(lis) == need or (len(lis) == need - 1 and i == len(order) - 1):
out += lis
need -= len(lis)
elif len(lis) == need - 1:
if len(lis) > 2:
out += lis[:-1]
need -= (len(lis) - 1)
assert need == 2
else:
out += lis[:need]
need = 0
assert need + len(out) == k
assert need >= 0
assert need == 0 or len(out) == count
for v in l:
if need == 0:
break
if v in powers:
continue
v2 = v
for p in pgood:
while v2 % p == 0:
v2 //= p
if v2 == 1:
out.append(v)
need -= 1
if need == 0:
print(' '.join(map(str,out)))
exit()
else:
print(0)
``` | output | 1 | 50,711 | 22 | 101,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you have a sequence of k integers A = [a_1, a_2, ... , a_k] where each a_i ≥ 2. A sequence of prime integers P = [p_1, p_2, ..., p_k] is called suitable for the sequence A if a_1 is divisible by p_1, a_2 is divisible by p_2 and so on.
A sequence of prime integers P is called friendly if there are no unique integers in this sequence.
A sequence A is called ideal, if each sequence P that is suitable for A is friendly as well (i. e. there is no sequence P that is suitable for A, but not friendly). For example, the sequence [2, 4, 16] is ideal, while the sequence [2, 4, 6] is not ideal (there exists a sequence P = [2, 2, 3] which is suitable for A, but not friendly).
You are given n different integers x_1, x_2, ..., x_n. You have to choose exactly k of them in such a way that they form an ideal sequence, or report that it is impossible. Note that no integer can be chosen more than once.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (2 ≤ x_i ≤ 10^{18}).
Output
If it is impossible to choose exactly k integers from x_1, x_2, ..., x_n in such a way that the chosen integers form an ideal sequence, print 0.
Otherwise, print k pairwise distinct integers — the elements of the chosen ideal sequence. If there are multiple answers, print any of them.
Examples
Input
3 3
2 4 6
Output
0
Input
3 3
2 4 16
Output
2 4 16
Input
4 3
2 4 6 16
Output
2 4 16 | instruction | 0 | 50,712 | 22 | 101,424 |
Tags: binary search, greedy, math, number theory
Correct Solution:
```
from math import gcd
def t_prime(n):
if n == 1: return False
for p in [2,3,5,7,11,13,17,19,23]:
if p < n and pow(p,n-1,n) != 1: return False
return True
primes = [i for i in range(2,10**5) if t_prime(i)];pset = set(primes);n, k = map(int, input().split());l = list(map(int, input().split()))
if k == 1: print(0);exit()
for i in range(n):
for j in range(i):
u, v = l[i], l[j];poss = gcd(u,v);poss2 = max(u,v)//poss;smol = min(poss,poss2)
if t_prime(smol) and smol not in pset:primes.append(smol);pset.add(smol)
powers = set();count = 0;outLs = [];pgood = []
for p in primes:
curr = []; fp = [v for v in l if v % p == 0]
for v in fp:
v2 = v
while v2 % p == 0: v2 //= p
if v2 == 1: curr.append(v); powers.add(v)
if len(curr) > 1: count += len(curr); outLs.append(curr); pgood.append(p)
order = [(len(lis), lis) for lis in outLs];order.sort(key = lambda x: x[0])
if len(order) == 0: print(0); exit()
if order[-1][0] == 2 and k % 2 and count > k:
extra = -1; need = -1; last = []
for v in l:
if v in powers: continue
v2 = v; primesn = []
for p in pgood:
add = 1
while v2 % p == 0:
v2 //= p
if add: primesn.append(p)
add = 0
if v2 == 1 and (need == -1 or need > len(primesn)): extra = v; last = primesn; need = len(last); assert need >= 2
if need == -1 or 2 * need + 1 > k: print(0); exit()
other = []; out = [extra]
for a,b in outLs:
works = False
for p in last:
if a % p == 0: works = True; break
if works: out.append(a); out.append(b)
else: other.append(a); other.append(b)
assert len(out) == 2 * need + 1; assert (k - 2 * need - 1) % 2 == 0; ret = out + other[:(k - 2*need - 1)]; assert len(ret) == k; print(' '.join(map(str,ret))); exit()
out = [];need = k
for i in range(len(order)):
assert need != 1;lis = order[i][1]
if len(lis) < need - 1 or len(lis) == need or (len(lis) == need - 1 and i == len(order) - 1): out += lis; need -= len(lis)
elif len(lis) == need - 1:
if len(lis) > 2: out += lis[:-1]; need -= (len(lis) - 1); assert need == 2
else: out += lis[:need]; need = 0
assert need + len(out) == k;assert need >= 0;assert need == 0 or len(out) == count
for v in l:
if need == 0: break
if v in powers: continue
v2 = v
for p in pgood:
while v2 % p == 0: v2 //= p
if v2 == 1: out.append(v); need -= 1
if need == 0: print(' '.join(map(str,out))); exit()
else: print(0)
``` | output | 1 | 50,712 | 22 | 101,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you have a sequence of k integers A = [a_1, a_2, ... , a_k] where each a_i ≥ 2. A sequence of prime integers P = [p_1, p_2, ..., p_k] is called suitable for the sequence A if a_1 is divisible by p_1, a_2 is divisible by p_2 and so on.
A sequence of prime integers P is called friendly if there are no unique integers in this sequence.
A sequence A is called ideal, if each sequence P that is suitable for A is friendly as well (i. e. there is no sequence P that is suitable for A, but not friendly). For example, the sequence [2, 4, 16] is ideal, while the sequence [2, 4, 6] is not ideal (there exists a sequence P = [2, 2, 3] which is suitable for A, but not friendly).
You are given n different integers x_1, x_2, ..., x_n. You have to choose exactly k of them in such a way that they form an ideal sequence, or report that it is impossible. Note that no integer can be chosen more than once.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (2 ≤ x_i ≤ 10^{18}).
Output
If it is impossible to choose exactly k integers from x_1, x_2, ..., x_n in such a way that the chosen integers form an ideal sequence, print 0.
Otherwise, print k pairwise distinct integers — the elements of the chosen ideal sequence. If there are multiple answers, print any of them.
Examples
Input
3 3
2 4 6
Output
0
Input
3 3
2 4 16
Output
2 4 16
Input
4 3
2 4 6 16
Output
2 4 16 | instruction | 0 | 50,713 | 22 | 101,426 |
Tags: binary search, greedy, math, number theory
Correct Solution:
```
from math import gcd
def t_prime(n):
if n == 1: return False
for p in [2,3,5,7,11,13,17,19,23]:
if p < n and pow(p,n-1,n) != 1: return False
return True
primes = [i for i in range(2,10**5) if t_prime(i)];pset = set(primes);n, k = map(int, input().split());l = list(map(int, input().split()))
if k == 1: print(0);exit()
for i in range(n):
for j in range(i):
u, v = l[i], l[j];poss = gcd(u,v);poss2 = max(u,v)//poss;smol = min(poss,poss2)
if t_prime(smol) and smol not in pset:primes.append(smol);pset.add(smol)
powers = set();count = 0;outLs = [];pgood = []
for p in primes:
curr = []; fp = [v for v in l if v % p == 0]
for v in fp:
v2 = v
while v2 % p == 0: v2 //= p
if v2 == 1: curr.append(v); powers.add(v)
if len(curr) > 1: count += len(curr); outLs.append(curr); pgood.append(p)
order = [(len(lis), lis) for lis in outLs];order.sort(key = lambda x: x[0])
if len(order) == 0: print(0); exit()
if order[-1][0] == 2 and k % 2 and count > k:
extra = -1
need = -1
last = []
for v in l:
if v in powers:
continue
v2 = v
primesn = []
for p in pgood:
add = 1
while v2 % p == 0:
v2 //= p
if add:
primesn.append(p)
add = 0
if v2 == 1 and (need == -1 or need > len(primesn)):
extra = v
last = primesn
need = len(last)
assert need >= 2
if need == -1 or 2 * need + 1 > k:
print(0)
exit()
other = []
out = [extra]
for a,b in outLs:
works = False
for p in last:
if a % p == 0:
works = True
break
if works:
out.append(a)
out.append(b)
else:
other.append(a)
other.append(b)
assert len(out) == 2 * need + 1
assert (k - 2 * need - 1) % 2 == 0
ret = out + other[:(k - 2*need - 1)]
assert len(ret) == k
print(' '.join(map(str,ret)))
exit()
out = []
need = k
for i in range(len(order)):
assert need != 1
lis = order[i][1]
if len(lis) < need - 1 or len(lis) == need or (len(lis) == need - 1 and i == len(order) - 1):
out += lis
need -= len(lis)
elif len(lis) == need - 1:
if len(lis) > 2:
out += lis[:-1]
need -= (len(lis) - 1)
assert need == 2
else:
out += lis[:need]
need = 0
assert need + len(out) == k
assert need >= 0
assert need == 0 or len(out) == count
for v in l:
if need == 0: break
if v in powers: continue
v2 = v
for p in pgood:
while v2 % p == 0: v2 //= p
if v2 == 1: out.append(v); need -= 1
if need == 0:
print(' '.join(map(str,out)))
exit()
else:
print(0)
``` | output | 1 | 50,713 | 22 | 101,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you have a sequence of k integers A = [a_1, a_2, ... , a_k] where each a_i ≥ 2. A sequence of prime integers P = [p_1, p_2, ..., p_k] is called suitable for the sequence A if a_1 is divisible by p_1, a_2 is divisible by p_2 and so on.
A sequence of prime integers P is called friendly if there are no unique integers in this sequence.
A sequence A is called ideal, if each sequence P that is suitable for A is friendly as well (i. e. there is no sequence P that is suitable for A, but not friendly). For example, the sequence [2, 4, 16] is ideal, while the sequence [2, 4, 6] is not ideal (there exists a sequence P = [2, 2, 3] which is suitable for A, but not friendly).
You are given n different integers x_1, x_2, ..., x_n. You have to choose exactly k of them in such a way that they form an ideal sequence, or report that it is impossible. Note that no integer can be chosen more than once.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (2 ≤ x_i ≤ 10^{18}).
Output
If it is impossible to choose exactly k integers from x_1, x_2, ..., x_n in such a way that the chosen integers form an ideal sequence, print 0.
Otherwise, print k pairwise distinct integers — the elements of the chosen ideal sequence. If there are multiple answers, print any of them.
Examples
Input
3 3
2 4 6
Output
0
Input
3 3
2 4 16
Output
2 4 16
Input
4 3
2 4 6 16
Output
2 4 16 | instruction | 0 | 50,714 | 22 | 101,428 |
Tags: binary search, greedy, math, number theory
Correct Solution:
```
from math import gcd
def t_prime(n):
if n == 1:
return False
for p in [2,3,5,7,11,13,17,19,23]:
if p < n and pow(p,n-1,n) != 1:
return False
return True
primes = [i for i in range(2,10**5) if t_prime(i)]
pset = set(primes)
n, k = map(int, input().split())
l = list(map(int, input().split()))
if k == 1:
print(0)
exit()
for i in range(n):
for j in range(i):
u, v = l[i], l[j]
poss = gcd(u,v)
poss2 = max(u,v)//poss
smol = min(poss,poss2)
if t_prime(smol) and smol not in pset:
primes.append(smol)
pset.add(smol)
powers = set()
count = 0
outLs = []
pgood = []
for p in primes:
curr = []
fp = [v for v in l if v % p == 0]
for v in fp:
v2 = v
while v2 % p == 0:
v2 //= p
if v2 == 1:
curr.append(v)
powers.add(v)
if len(curr) > 1:
count += len(curr)
outLs.append(curr)
pgood.append(p)
order = [(len(lis), lis) for lis in outLs]
order.sort(key = lambda x: x[0])
if len(order) == 0:
print(0)
exit()
if order[-1][0] == 2 and k % 2 and count > k:
extra = -1
need = -1
last = []
for v in l:
if v in powers:
continue
v2 = v
primesn = []
for p in pgood:
add = 1
while v2 % p == 0:
v2 //= p
if add:
primesn.append(p)
add = 0
if v2 == 1 and (need == -1 or need > len(primesn)):
extra = v
last = primesn
need = len(last)
assert need >= 2
if need == -1 or 2 * need + 1 > k:
print(0)
exit()
other = []
out = [extra]
for a,b in outLs:
works = False
for p in last:
if a % p == 0:
works = True
break
if works:
out.append(a)
out.append(b)
else:
other.append(a)
other.append(b)
assert len(out) == 2 * need + 1
assert (k - 2 * need - 1) % 2 == 0
ret = out + other[:(k - 2*need - 1)]
assert len(ret) == k
print(' '.join(map(str,ret)))
exit()
out = []
need = k
for i in range(len(order)):
assert need != 1
lis = order[i][1]
if len(lis) < need - 1 or len(lis) == need or (len(lis) == need - 1 and i == len(order) - 1):
out += lis
need -= len(lis)
elif len(lis) == need - 1:
if len(lis) > 2:
out += lis[:-1]
need -= (len(lis) - 1)
assert need == 2
else:
out += lis[:need]
need = 0
assert need + len(out) == k
assert need >= 0
assert need == 0 or len(out) == count
for v in l:
if need == 0: break
if v in powers: continue
v2 = v
for p in pgood:
while v2 % p == 0: v2 //= p
if v2 == 1: out.append(v); need -= 1
if need == 0:
print(' '.join(map(str,out)))
exit()
else:
print(0)
``` | output | 1 | 50,714 | 22 | 101,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you have a sequence of k integers A = [a_1, a_2, ... , a_k] where each a_i ≥ 2. A sequence of prime integers P = [p_1, p_2, ..., p_k] is called suitable for the sequence A if a_1 is divisible by p_1, a_2 is divisible by p_2 and so on.
A sequence of prime integers P is called friendly if there are no unique integers in this sequence.
A sequence A is called ideal, if each sequence P that is suitable for A is friendly as well (i. e. there is no sequence P that is suitable for A, but not friendly). For example, the sequence [2, 4, 16] is ideal, while the sequence [2, 4, 6] is not ideal (there exists a sequence P = [2, 2, 3] which is suitable for A, but not friendly).
You are given n different integers x_1, x_2, ..., x_n. You have to choose exactly k of them in such a way that they form an ideal sequence, or report that it is impossible. Note that no integer can be chosen more than once.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (2 ≤ x_i ≤ 10^{18}).
Output
If it is impossible to choose exactly k integers from x_1, x_2, ..., x_n in such a way that the chosen integers form an ideal sequence, print 0.
Otherwise, print k pairwise distinct integers — the elements of the chosen ideal sequence. If there are multiple answers, print any of them.
Examples
Input
3 3
2 4 6
Output
0
Input
3 3
2 4 16
Output
2 4 16
Input
4 3
2 4 6 16
Output
2 4 16 | instruction | 0 | 50,715 | 22 | 101,430 |
Tags: binary search, greedy, math, number theory
Correct Solution:
```
from math import gcd
def t_prime(n):
if n == 1: return False
for p in [2,3,5,7,11,13,17,19,23]:
if p < n and pow(p,n-1,n) != 1: return False
return True
primes = [i for i in range(2,10**5) if t_prime(i)];pset = set(primes);n, k = map(int, input().split());l = list(map(int, input().split()))
if k == 1: print(0);exit()
for i in range(n):
for j in range(i):
u, v = l[i], l[j];poss = gcd(u,v);poss2 = max(u,v)//poss;smol = min(poss,poss2)
if t_prime(smol) and smol not in pset:primes.append(smol);pset.add(smol)
powers = set();count = 0;outLs = [];pgood = []
for p in primes:
curr = []; fp = [v for v in l if v % p == 0]
for v in fp:
v2 = v
while v2 % p == 0: v2 //= p
if v2 == 1: curr.append(v); powers.add(v)
if len(curr) > 1: count += len(curr); outLs.append(curr); pgood.append(p)
order = [(len(lis), lis) for lis in outLs];order.sort(key = lambda x: x[0])
if len(order) == 0: print(0); exit()
if order[-1][0] == 2 and k % 2 and count > k:
extra = -1; need = -1; last = []
for v in l:
if v in powers: continue
v2 = v; primesn = []
for p in pgood:
add = 1
while v2 % p == 0:
v2 //= p
if add: primesn.append(p)
add = 0
if v2 == 1 and (need == -1 or need > len(primesn)): extra = v; last = primesn; need = len(last); assert need >= 2
if need == -1 or 2 * need + 1 > k: print(0); exit()
other = []; out = [extra]
for a,b in outLs:
works = False
for p in last:
if a % p == 0: works = True; break
if works: out.append(a); out.append(b)
else: other.append(a); other.append(b)
assert len(out) == 2 * need + 1; assert (k - 2 * need - 1) % 2 == 0; ret = out + other[:(k - 2*need - 1)]; assert len(ret) == k; print(' '.join(map(str,ret))); exit()
out = [];need = k
for i in range(len(order)):
assert need != 1
lis = order[i][1]
if len(lis) < need - 1 or len(lis) == need or (len(lis) == need - 1 and i == len(order) - 1):
out += lis
need -= len(lis)
elif len(lis) == need - 1:
if len(lis) > 2:
out += lis[:-1]
need -= (len(lis) - 1)
assert need == 2
else:
out += lis[:need]
need = 0
assert need + len(out) == k
assert need >= 0
assert need == 0 or len(out) == count
for v in l:
if need == 0: break
if v in powers: continue
v2 = v
for p in pgood:
while v2 % p == 0: v2 //= p
if v2 == 1: out.append(v); need -= 1
if need == 0:
print(' '.join(map(str,out)))
exit()
else:
print(0)
``` | output | 1 | 50,715 | 22 | 101,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you have a sequence of k integers A = [a_1, a_2, ... , a_k] where each a_i ≥ 2. A sequence of prime integers P = [p_1, p_2, ..., p_k] is called suitable for the sequence A if a_1 is divisible by p_1, a_2 is divisible by p_2 and so on.
A sequence of prime integers P is called friendly if there are no unique integers in this sequence.
A sequence A is called ideal, if each sequence P that is suitable for A is friendly as well (i. e. there is no sequence P that is suitable for A, but not friendly). For example, the sequence [2, 4, 16] is ideal, while the sequence [2, 4, 6] is not ideal (there exists a sequence P = [2, 2, 3] which is suitable for A, but not friendly).
You are given n different integers x_1, x_2, ..., x_n. You have to choose exactly k of them in such a way that they form an ideal sequence, or report that it is impossible. Note that no integer can be chosen more than once.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (2 ≤ x_i ≤ 10^{18}).
Output
If it is impossible to choose exactly k integers from x_1, x_2, ..., x_n in such a way that the chosen integers form an ideal sequence, print 0.
Otherwise, print k pairwise distinct integers — the elements of the chosen ideal sequence. If there are multiple answers, print any of them.
Examples
Input
3 3
2 4 6
Output
0
Input
3 3
2 4 16
Output
2 4 16
Input
4 3
2 4 6 16
Output
2 4 16
Submitted Solution:
```
from collections import defaultdict
from math import sqrt, floor, ceil
inf = float("inf")
b = 10**6
sieve = [True] * (b + 1)
smallPrimes = set()
for i in range(2, b + 1):
#print (i)
#print (i)
if sieve[i]:
smallPrimes.add(i)
else:
continue
j = 2 * i
while j < b + 1:
sieve[j] = False
j += i
smallPrimesPowered = dict()
for smallPrime in smallPrimes:
cur = smallPrime*smallPrime*smallPrime
smallPrimesPowered[cur] = smallPrime
while cur <= 10**18:
cur *= smallPrime
smallPrimesPowered[cur] = smallPrime
splitted = [int(amit) for amit in input().split(" ")]
n = splitted[0]
k = splitted[1]
arr = [int(amit) for amit in input().split(" ")]
freq = defaultdict(lambda: set())
def isPrime(n):
if int(n) != n:
return False
divided = False
for i in range(2, ceil(sqrt(n)) + 1):
if n % i == 0:
divided = True
break
return (not divided)
for item in arr:
if item in smallPrimes:
freq[item].add(item)
elif sqrt(item) in smallPrimes or isPrime(sqrt(item)):
freq[int(sqrt(item))].add(item)
elif item in smallPrimesPowered:
freq[smallPrimesPowered[item]].add(item)
importantPrimes = defaultdict(lambda: set())
sumOfGroups = 0
non2 = False
for item in freq:
if len(freq[item]) >= 2:
importantPrimes[item] = freq[item]
sumOfGroups += len(importantPrimes[item])
if len(freq[item]) != 2:
non2 = True
#print (importantPrimes)
if k <= sumOfGroups:
if non2 == False:
if k % 2 == 1:
#print ("hi")
minNumImportantPrimeFactors = inf
minNumImportantPrimeFactorsVal = None
minNumImportantPrimeFactorsSet = set()
for item in arr:
numImportantPrimeFactors = 0
numImportantPrimeFactorsSet = set()
cur = item
for pf in importantPrimes:
divided = False
while cur % pf == 0:
divided = True
cur /= pf
if divided:
numImportantPrimeFactorsSet.add(pf)
numImportantPrimeFactors += 1
if cur != 1:
continue
if 2 <= numImportantPrimeFactors < minNumImportantPrimeFactors:
minNumImportantPrimeFactors = numImportantPrimeFactors
minNumImportantPrimeFactorsVal = item
minNumImportantPrimeFactorsSet = numImportantPrimeFactorsSet
if minNumImportantPrimeFactorsVal is None:
print (0)
else:
#print(freq)
ans = set()
#print (minNumImportantPrimeFactorsVal)
#print (minNumImportantPrimeFactorsSet)
if minNumImportantPrimeFactors <= (k-1)//2:
for pf in minNumImportantPrimeFactorsSet:
for val in importantPrimes[pf]:
ans.add(val)
ans.add(minNumImportantPrimeFactorsVal)
for pf in importantPrimes:
if pf in minNumImportantPrimeFactorsSet or len(ans) >= k:
continue
for val in importantPrimes[pf]:
ans.add(val)
if len(ans) == k:
res = ""
for item in ans:
res += str(item) + " "
print (res)
else:
#print ("nooo")
print ("0")
else:
ans = set()
for pf in importantPrimes:
if len(ans) >= k:
continue
for val in importantPrimes[pf]:
ans.add(val)
res = ""
for item in ans:
res += str(item) + " "
print (res)
else:
biggestGroupVal = None
biggestGroupSize = -inf
for pf in importantPrimes:
if len(importantPrimes[pf]) > biggestGroupSize:
biggestGroupSize = len(importantPrimes[pf])
biggestGroupVal = pf
ans = set()
canDiscard = set()
for pf in importantPrimes:
if pf == biggestGroupVal:
continue
if len(ans) >= k - biggestGroupSize:
break
added = 0
for val in importantPrimes[pf]:
ans.add(val)
added += 1
if added > 2:
canDiscard.add(val)
addedFromBiggest = 0
for val in importantPrimes[biggestGroupVal]:
ans.add(val)
addedFromBiggest += 1
if addedFromBiggest > 2:
canDiscard.add(val)
if addedFromBiggest >= 2 and len(ans) >= k:
break
while len(ans) > k and len(canDiscard) > 0:
discarded = canDiscard.pop()
ans.discard(discarded)
if len(ans) == 1 or len(ans) != k:
print (0)
else:
res = ""
for item in ans:
res += str(item) + " "
print (res)
else:
ans = set()
for pf in importantPrimes:
for val in importantPrimes[pf]:
ans.add(val)
rejected = set()
for item in arr:
if len(ans) >= k:
break
cur = item
for pf in importantPrimes:
divided = False
while cur % pf == 0:
divided = True
cur /= pf
if cur == 1:
# entirely divisible by important primes
ans.add(item)
else:
rejected.add(item)
if len(ans) == k:
res = ""
for item in ans:
res += str(item) + " "
print (res)
else:
if (n == k == 1000):
print ("actual size of ans: " + str(len(ans)))
res = ""
for pf in importantPrimes:
res += str(pf) + " "
if len(res) > 100:
print (res)
res = ""
print ("rejected: " + str(rejected))
print (0)
``` | instruction | 0 | 50,716 | 22 | 101,432 |
No | output | 1 | 50,716 | 22 | 101,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you have a sequence of k integers A = [a_1, a_2, ... , a_k] where each a_i ≥ 2. A sequence of prime integers P = [p_1, p_2, ..., p_k] is called suitable for the sequence A if a_1 is divisible by p_1, a_2 is divisible by p_2 and so on.
A sequence of prime integers P is called friendly if there are no unique integers in this sequence.
A sequence A is called ideal, if each sequence P that is suitable for A is friendly as well (i. e. there is no sequence P that is suitable for A, but not friendly). For example, the sequence [2, 4, 16] is ideal, while the sequence [2, 4, 6] is not ideal (there exists a sequence P = [2, 2, 3] which is suitable for A, but not friendly).
You are given n different integers x_1, x_2, ..., x_n. You have to choose exactly k of them in such a way that they form an ideal sequence, or report that it is impossible. Note that no integer can be chosen more than once.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (2 ≤ x_i ≤ 10^{18}).
Output
If it is impossible to choose exactly k integers from x_1, x_2, ..., x_n in such a way that the chosen integers form an ideal sequence, print 0.
Otherwise, print k pairwise distinct integers — the elements of the chosen ideal sequence. If there are multiple answers, print any of them.
Examples
Input
3 3
2 4 6
Output
0
Input
3 3
2 4 16
Output
2 4 16
Input
4 3
2 4 6 16
Output
2 4 16
Submitted Solution:
```
from math import gcd
def t_prime(n):
if n == 1:
return False
for p in [2,3,5,7,11,13,17,19,23]:
if p < n and pow(p,n-1,n) != 1:
return False
return True
primes = [i for i in range(2,10**5) if t_prime(i)]
pset = set(primes)
n, k = map(int, input().split())
l = list(map(int, input().split()))
if k == 1:
print(0)
exit()
for i in range(n):
for j in range(i):
u, v = l[i], l[j]
poss = gcd(u,v)
poss2 = max(u,v)//poss
smol = min(poss,poss2)
if t_prime(smol) and smol not in pset:
primes.append(smol)
powers = set()
count = 0
outLs = []
pgood = []
for p in primes:
curr = []
fp = [v for v in l if v % p == 0]
for v in fp:
v2 = v
while v2 % p == 0:
v2 //= p
if v2 == 1:
curr.append(v)
powers.add(v)
if len(curr) > 1:
count += len(curr)
outLs.append(curr)
pgood.append(p)
order = [(len(lis), lis) for lis in outLs]
order.sort(key = lambda x: x[0])
if order[-1][0] == 2 and k % 2 and count > k:
extra = -1
need = -1
last = []
for v in l:
if v in powers:
continue
v2 = v
primesn = []
for p in pgood:
add = 1
while v2 % p == 0:
v2 //= p
if add:
primesn.append(p)
add = 0
if v2 == 1 and (need == -1 or need > len(primesn)):
extra = v
last = primesn
need = len(last)
assert need >= 2
if need == -1 or 2 * need + 1 > k:
print(0)
exit()
other = []
out = [extra]
for a,b in outLs:
works = False
for p in last:
if a % p == 0:
works = True
break
if works:
out.append(a)
out.append(b)
else:
other.append(a)
other.append(b)
assert len(out) == 2 * need + 1
ret = out + other[:(k - 2*need - 1)]
assert len(ret) == k
print(' '.join(map(str,ret)))
exit()
out = []
need = k
for i in range(len(order)):
assert need != 1
lis = order[i][1]
if len(lis) < need - 1 or len(lis) == need or (len(lis) == need - 1 and i == len(order) - 1):
out += lis
need -= len(lis)
elif len(lis) == need - 1:
if len(lis) > 2:
out += lis[:-1]
need -= (len(lis) - 1)
assert need == 2
else:
out += lis[:need]
need = 0
assert need + len(out) == k
for v in l:
if v in powers:
continue
v2 = v
for p in pgood:
while v2 % p == 0:
v2 //= p
if v2 == 1:
out.append(v)
need -= 1
if need == 0:
break
if need == 0:
print(' '.join(map(str,out)))
exit()
else:
print(0)
``` | instruction | 0 | 50,717 | 22 | 101,434 |
No | output | 1 | 50,717 | 22 | 101,435 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.