K - Alexandra and Prime Numbers

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≤1,000,000,000N≤1,000,000,000. 
Number of cases with N>1,000,000N>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

题意：给你一个数n(1<=n<=1e9)  要你求满足n/m是素数的最小m的值

分析：p=n/m p是素数，m=n/p 要使m最小，那么p就要最大,且由唯一分解定理的n=p1*p2*p3....*pn(pi为素数），那么就是要求n最大的素数因子就为p,此时m最小，p从sqrt(n)+1开始枚举，因为n不可能有两个大于sqrt(n)的素数因子

Ac code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn=1e5+2;
//int prime[maxn];
bool vis[maxn];
int cnt;
void set_prime()
{
    cnt=0;
    memset(vis,0,sizeof vis);
    for(int i=2;i<=100000;i++)
    {
        if(!vis[i])
        {
            //prime[cnt++]=i;
            for(int j=i+i;j<=100000;j+=i)
                vis[j]=1;
        }
    }
}
int main()
{
    set_prime();
    ll n;
    while(scanf("%lld",&n)!=EOF)
    {
        if(n==1)
        {
            printf("0\n");
            continue;
        }
        ll temp=n;
        ll ans=0;
        ll m=(ll)sqrt(n)+1;
        for(m;m>1;m--)///找小于n的最大的素数因子，且不会有两个素数因子
                      ///都大于sqrt(n),故从sqrt(n)+1开始找
            if(n%m==0)
        {
            while(n%m==0&&!vis[m])
                n/=m;
            if(!vis[m])
                ans=max(ans,m);
        }
        ans=max(ans,n);///n若本来就是个素数，或剩下的因子更大
        printf("%lld\n",temp/ans);
    }
    return 0;
}