CodeForces - 797A k-Factorization（唯一分解定理）

A. k-Factorization

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

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.

Input

The first line contains two integers n and k (2 ≤ n ≤ 100000, 1 ≤ k ≤ 20).

Output

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.

Examples

input

100000 2

output

2 50000 

input

100000 20

output

-1

input

1024 5

output

2 64 2 2 2 

思路：唯一分解定理，把n分解成k个质因数相乘的形式，如果 k >= m 则情况存在，否则输出"-1".

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define clr(a) memset(a,0,sizeof(a))

const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int N = 1010;

int a[MAXN];
int sum;
int Prime_num(int n){
    clr(a);
    sum=0;
    for(int i=2;i*i<=n;i++){
        if(n % i == 0){
            while(n % i == 0){
                n /= i;
                a[sum++] = i;
            }
        }
    }
    if(n != 1){
        a[sum ++] = n;
    }
    return sum ;
}

int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        int k = Prime_num(n);
        if(k >= m){
            int ans = 1;
            for(int i=0;i<m-1;i++){
                printf("%d ",a[i]);
                ans *= a[i];
            }
            printf("%d\n",n/ans);
        }
        else cout<<"-1"<<endl;
    }
    return 0;
}