SGU113 水题 Easy Problem

问题：判断一个数是否是两个质数的乘积。

Problem: Nearly prime number is an integer positive number for which it is possible to find such primes P1 and P2 that given number is equal to P1*P2. There is given a sequence on N integer positive numbers, you are to write a program that prints “Yes” if given number is nearly prime and “No” otherwise.

解法：筛法求质数，再枚举。枚举时要用O(Sqrt(N))的复杂度判断另一个质数。

Solution: Obtain prime numbers by sieving method, then enumerate them. Please note that you need to use the method of complexity of O(Sqrt(N)) to judge the other number whether it is a prime number in case it is very large.

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <cstring>
#include <math.h>
using namespace std;
int n,m,pnum,p[100100];
bool f[1000100];
bool isprime(int x) {
     if (x<2) return 0;
     for (int i=2;i*i<=x;i++)
         if (x%i==0) return 0;
     return 1;
}
int main() {
    pnum = 0;
    memset(f,0,sizeof(f));
    f[1] = true;
    for (int i=2;i<=1000010;i++)
        if (!f[i]) {
           pnum++;
           p[pnum] = i;
           for (int j=i+i;j<=1000010;j+=i) f[j] = true;
        }
        
    scanf("%d",&n);
    while (n--) {
          bool ans = false;
          scanf("%d",&m);
          for (int i=1;i<=pnum;i++) {
              if (p[i]*p[i]>m) break;
              if ((m%p[i]==0) && isprime(m/p[i])) {
                    ans = true;
                    break;
              }
          }
          if (ans) printf("Yes\n");
          else printf("No\n");
    }
    return 0;
}

SGU113 水题 Easy Problem

原文：http://blog.csdn.net/lotus_land/article/details/18844237