B - 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. $N \leq 1,000,000,000$. 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

 

这个题呢，我很纠结，因为理解错题了，首题目让找一个 最小的数使n/m，必须能被整除，而且使n/m的值是个素数。

当时我呢，没有找必须被整除的数，一直，时间超限，而且答案是错误的。

 

结果我问了我的师傅，才知道那儿错了。首先任何一个大于1的数x，都能写成x=a1^（n1）×a2^（n2）×a3^（n3）×……×an^（nn）；

这里a1，a2，a3，……，an，都是奇数。

比如当n=450；那么n=2×3×3×5×5；只能保留到最后一个，最大的，素数，

即m=n/5；

 

所以，

#include<stdio.h>

#include<math.h>

#include<algorithm>

using namespace std;

int fun(int n)

{

int i,j=0,k;

k=n;

for(i=2;i<=sqrt(n);i++)   //不能用i<=n因为会超时

{

while(k%i==0)

{

k=k/i;

j=max(j,i);

}

}

j=max(j,k);  //   比如当n=10时

if(n==1)printf("0\n");

else

{

printf("%d\n",j==0?1:n/j);

}

 

}

 

 

int main()

{

int n;

while(scanf("%d",&n)!=EOF)

{

fun(n);

 

}

return 0;

}