Alexandra and Prime Numbers

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

 

Source

BestCoder Round #19

 

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  题意 ： 给你一个N   寻找一个最小M ，使得N/M=素数

  不解释了，简单代码

   #include<stdio.h>
#include<math.h>
#include<iostream>
#include<cmath>
using namespace std;
int f(int x)
{
  int  flag = 0;
  for(int i = 2; (i*i) <= x; i ++)
  {
    if(x % i == 0)
    {
      flag = 1;
      break;     
    }        
  }   
  if(flag)  return 0;
  return 1; 
}
int main()
{
    int n;
    int mark[200005];
    while(scanf("%d",&n)!=EOF)
    {
        if( n == 1 )  {printf("0\n");  continue;}
        int  count = 0, flag = 0;
        for(int i = 1; (i * i) <= n; i++)
        {
          if( n%i == 0)   {mark[count++] = i; mark[count++] = n / i;}
        }            
        sort(mark, mark + count);      
        int i;
        for(i = count - 1 ; i >= 0; i --)
        {
          if(f(mark[i])) {  break;}        
        }
        printf("%d\n",n / mark[i]); 
    }    
}