Alexandra and Prime Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Alexandra has a little brother. He is new to programming. One day he is solving the following problem: Given an positive integer N, judge whether N is prime.
The problem above is quite easy, so Alexandra gave him a new task: Given a positive integer N, find the minimal positive integer M, such that N/M is prime. If such M doesn‘t exist, output 0.
Help him!

Input

There are multiple test cases (no more than 1,000). Each case contains only one positive integer N.
N≤1,000,000,000.
Number of cases with N>1,000,000 is no more than 100.

Output

For each case, output the requested M, or output 0 if no solution exists.

Sample Input

3
4
5
6

Sample Output

1
2
1
2

【题目简述】：本题让我们输入一个正整数N，然后求出一最小的正整数M使得 N/M 是一个素数。

【分析】：刚刚拿到这题的第一感觉就是用素数的相关操作去处理这道题，然后超时但还是不死心，不停的优化。但本题和其他的素数不一样的地方就是这道题它求的是N/M，所以既然涉及到除法，其实我们还是用另外一种方法去处理这道题，那就是质因数分解。

质因数分解：就是把一个合数分解成几个素数相乘的形式。例如：

48 = 2*2*2*3

58 = 2*3*3*3

由此我们可以将N一直除以一个比它自己本身小的质数，按照质因数分解的方法，不停的进行分解，我们要找到一个分解出的最大的素数P,因为只有这样，我们才能使得M最小。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#include <cassert>
using namespace std;
typedef long long LL;
const int N = 50;
int n;

void work() {
    int i , m , x = 0;
    m = n;
    for (i = 2 ; i * i <= n ; ++ i) {
        if (n % i == 0) {
            while (n % i == 0)
                n /= i;
            x = max(x , i);
        }
    }
    if (n > 1)
        x = max(x , n);
    printf("%d\n" , x ? m / x : 0);
}

int main() {
    while (~scanf("%d",&n))
        work();
    return 0;
}