zoj 2723 Semi-Prime(set)

最新推荐文章于 2018-03-20 16:55:20 发布

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最新推荐文章于 2018-03-20 16:55:20 发布

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原文链接：http://www.cnblogs.com/qinduanyinghua/p/6530865.html

本文介绍了一种用于判断半素数的有效算法，并提供了完整的C++实现代码。文章首先定义了素数与半素数的概念，然后详细阐述了如何构建素数表及半素数表，最后演示了如何利用这些表来快速判断一个给定的整数是否为半素数。

Semi-Prime

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

建立素数表和半素数表，建立半素数表是通过素数表中的任意两个素数相乘得到保存起来。

 1 #include <iostream>
 2 #include <vector>
 3 #include <set>
 4 #include <cmath>
 5 using namespace std;
 6 //建立全局向量，用来保存素数
 7 vector<int> v;
 8 //在全局内存中定义全局集合容器，用来保存半素数
 9 //集合是平衡二叉检索树，搜索速度最快
10 set<int> s;
11 //建立[a, b]范围内素数表
12 void pt(int a, int b){
13     for(int i = a; i <= b; i++){
14         //2是素数，清除2的倍数
15         if(i != 2 && i % 2 == 0) continue;
16         //消除素数的倍数 
17         for(int j = 3; j * j <= i; j += 2){
18             if(i % j == 0)
19                 goto RL;
20         }
21         v.push_back(i);
22         RL: continue;
23     }
24 }
25 
26 int main(){
27     pt(2, 500000);
28     int i, j, p;
29     for(i = 0; i < v.size(); i++){
30         for(j = 0; j < v.size(); j++){
31             p = v[i] * v[j];
32             if(p < 1000000)
33                 s.insert(p);
34             else
35                 break;
36         }
37     }
38     //读入数据，在半素数表中查找，看是否在该表
39     int n;
40     set<int>::iterator it;
41     while(cin >> n){
42         it = s.find(n);
43         if(it != s.end())
44             cout << "Yes" << endl;
45         else
46             cout << "No" << endl;
47     }
48     return 0;
49 }