CF - 797A. k-Factorization - 思维+贪心+数学

本文介绍了一种解决k-因子分解问题的算法实现，即如何将一个正整数n分解为k个大于1的整数的乘积，并提供了解题思路及AC代码示例。

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题目描述：
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
题意概述：给你一个数字N要你把它分解成k个数相乘的形式。
解题思路：我们知道一个数分解质因数的上限是sqrt(N)，那么直接将n分到sqrt部分，如果小于k则输出-1，否则贪心地把前sz - k个数合并即可。

AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define lson root << 1
#define rson root << 1 | 1
#define lent (t[root].r - t[root].l + 1)
#define lenl (t[lson].r - t[lson].l + 1)
#define lenr (t[rson].r - t[rson].l + 1)
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
vector<int> v;
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int n, k;
	while (~scanf("%d%d", &n, &k))
	{
		v.clear();
		for (int i = 2; i * i <= n; i++)
			while (n % i == 0)
			{
				v.push_back(i);
				n /= i;
			}
		if (n > 1)
			v.push_back(n);
		if (v.size() < k)
		{
			puts("-1");
			continue;
		}
		int sz = v.size();
		int cur = 1;
		for (int i = 0; i + k < sz + 1; i++)
			cur *= v[i];
		printf("%d", cur);
		for (int i = sz - k + 1; i < sz; i++)
			printf(" %d", v[i]);
		puts("");
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}