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[500010];

void prim(int n);

int main()
{
    prim(500000);
    
    vector<int> prim_vec;//对于未知大小的数据来说非常方便 
    int i;
    for(i=2;i<=500000;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=0;i<prim_vec.size();i++){
        for(j=0;j<prim_vec.size();j++){
            if( (p = prim_vec[i]*prim_vec[j]) <=1000000)
                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
");
        else
            printf("No
");
    }
    return 0;    
} 

void prim(int n)
{
    int i,j,k;
    
    for(i=0;i<=n;i++)
        isPrim[i]=1;//全部初始化为1，假定全是素数 
    isPrim[0]=isPrim[1]=0;//0和1非素数
     
    k=(int)sqrt(n);
    for(i=2;i<=k;i++){
        if(isPrim[i]){
            for(j=2*i;j<=n;j+=i)
                isPrim[j]=0;
        }
    }
    /*
    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
",count);
    */
}