zoj 2723 Semi-Prime（素筛打表+搜索优化）

题目链接：

　　http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2723

题目描述：

Prime Number Definition
An integer greater than one is called a prime number if its only
positive divisors (factors) are one and itself. For instance, 2, 11, 67,
89 are prime numbers but 8, 20, 27 are not.

Semi-Prime Number Definition
An integer greater than one is called a semi-prime number if it can be
decompounded to TWO prime numbers. For example, 6 is a semi-prime number
but 12 is not.

Your task is just to determinate whether a given number is a semi-prime number.

Input

There are several test cases in the input. Each case contains a single integer N (2 <= N <= 1,000,000)

Output

One line with a single integer for each case. If the number is a semi-prime number, then output "Yes", otherwise "No".

Sample Input

3
4
6
12

Sample Output

No
Yes
Yes
No

 /*问题 判断输入的n（n的范围是2到100 0000）是否为半素数
 解题思路 先用素筛法建立一张[2,50 0000]的素数表放在数组里，再将素数存进向量里，然后计算出2到100 0000的半素
 数表放在set集合里后查表即可*/
 #include <cstdio>
 #include <set>
 #include <cmath>
 #include <vector>
 using namespace std;

 int isPrim[];

 void prim(int n);

 int main()
 {
     prim();

     vector<int> prim_vec;//对于未知大小的数据来说非常方便
     int i;
     for(i=;i<=;i++){
         if(isPrim[i])
             prim_vec.push_back(i);
     }

     set<int> prim_set;//查询起来非常快速、方便
     int j,p;
     for(i=;i<prim_vec.size();i++){
         for(j=;j<prim_vec.size();j++){
             if( (p = prim_vec[i]*prim_vec[j]) <=)
                 prim_set.insert(p);
             else
                 break;//一种优化方法，对于有顺序的数据
         }
     }

     int n;
     set<int>::iterator it;
     while(scanf("%d",&n) != EOF)
     {
         it=prim_set.find(n);
         if(it != prim_set.end())
             printf("Yes\n");
         else
             printf("No\n");
     }
     return ;
 } 

 void prim(int n)
 {
     int i,j,k;

     for(i=;i<=n;i++)
         isPrim[i]=;//全部初始化为1，假定全是素数
     isPrim[]=isPrim[]=;//0和1非素数

     k=(int)sqrt(n);
     for(i=;i<=k;i++){
         if(isPrim[i]){
             for(j=*i;j<=n;j+=i)
                 isPrim[j]=;
         }
     }
     /*
     int count=0;
     for(i=2;i<=n;i++){
         if(isPrim[i])
             count++;//记录50 0000以内的素数有41538个，验证素筛写的对不对
     }
     printf("%d\n",count);
     */
 }