Codeforces 797A - k-Factorization
2 seconds
256 megabytes
standard input
standard output
Given a positive integer n, find k integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to n.
The first line contains two integers n and k (2 ≤ n ≤ 100000, 1 ≤ k ≤ 20).
If it's impossible to find the representation of n as a product of k numbers, print -1.
Otherwise, print k integers in any order. Their product must be equal to n. If there are multiple answers, print any of them.
5 1
5
5 2
-1
1024 5
2 64 2 2 2
题目大意:输入两个数n和k,找k个大于1的整数,使他们的乘积为n,输出这k个数(如果有多个,输出任意一组),如果没有,输出-1。
方法:把n分解成质因数的乘积,并把这些质因数保存在数组里。如果质因数的个数c小于k,输出-1;否则,输出前k-1个质因数和k到c个质因数的乘积(这样就刚好k个)。
代码:
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=1e5;
int main()
{
int n,k;
cin>>n>>k;
int f[N];
int c=;
for(int i=;i<=n;i++)
{
while(n%i==)
{
n/=i;
f[c++]=i;
}
}
if(c<k)cout<<-<<endl;
else
{
for(int i=;i<k-;i++)
cout<<f[i]<<' ';
int sum=;
for(int i=k-;i<c;i++)
sum*=f[i];
cout<<sum<<endl;
}
return ;
}
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