Alexandra and Prime Numbers(思维)

Alexandra and Prime Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1658    Accepted Submission(s): 565

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

题解：让找最小的正整数M使N/M是素数；

水暴力，但是完全暴力会超，就想着折半找，先找这个最小正整数，如果不行就找素数；都找到sqrt(N)结束，这样就省了好多时间；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
bool is_prim(int x){
    if(x == )return false;
    for(int i = ; i <= sqrt(x); i++){
        if(x % i == )return false;
    }
    return true;
}
int main(){
    int N;
    while(~scanf("%d",&N)){
        if(N ==  || N == ){
            puts("");continue;
        }
        int i;
        int ans = ;
        for(i = ; i <= sqrt(N); i++){
            if(N % i ==  && is_prim(N/i)){
                ans = i;
                break;
            }
        }
        if(ans){
            printf("%d\n",ans);continue;
        }
            for(int j = sqrt(N); j > ; j--){
                if(N % j ==  && is_prim(j)){
                    ans = N / j;
                    break;
                }
        }
        printf("%d\n",ans);
    }
    return ;
}