这不是楼上的做法吗，这个方法首先就排除了在oj上马上就TLE了

给个网址 看看什么题目撒 呵呵

Description

Your task is simple.
Give you a number N, you should judge whether N is a prime number or not.
Input

There are multiple test cases. For each test case, there is an integer N(2<=N<=10^18).
Output

For each test case, you should output whether N is a prime number or not.
If N is a prime number , you should output "It is a prime number."; otherwise you should output "It is not a prime number.";
Sample Input

2
4
Sample Output

It is a prime number.
It is not a prime number.

给个网址 好去ACM

unsigned long int i,a;int flag=1;
  scanf("%ld",&a);
  for(i=2;i<=sqrt(a);i++)
     if(a%i==0)
      { flag=0;printf("%d不是素数"，a);break;}
  if(flag) printf("%d是素数"，a);
  

我改了下输入输出格式  你提交下看看

TLE了

能给个网址不？

http://acm.fzu.这个oj上

我给你船长的代码：
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int main(void)
{
    int a, b=0, n;
    printf("请输入一个自然数：");
    scanf("%d",&n);
    if(n==1 || n==0){
        printf("输入错误，0 和 1 既不是质数也不是合数\n");
        exit (0);
    }
    for (a=2;a<=sqrt(n); a++) {
        if ( n%a==0 ) {
            ++b;
        }   
    }
    if ( b >0 ) {
           printf("%d:是一个合数\n",n);
           printf("约数为:");
           for(a=2;a<n;a++){
                   if(n%a==0) printf("%4d", a);
               }
    }
    else if (b==0){
        printf(" %d:是一个质数",n);
    }
    putchar(10);
    return 0;
}