HDU 5108 最大素数因子

Alexandra and Prime Numbers

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

Total Submission(s): 706 Accepted Submission(s): 247



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]);

}

}