zoukankan      html  css  js  c++  java
  • hdu1397(Goldbach`s Conjecture )

    Goldbach's Conjecture

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 7550    Accepted Submission(s): 2877


    Problem Description
    Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2.
    This conjecture has not been proved nor refused yet. No one is sure whether this conjecture actually holds. However, one can find such a pair of prime numbers, if any, for a given even number. The problem here is to write a program that reports the number of all the pairs of prime numbers satisfying the condition in the conjecture for a given even number.

    A sequence of even numbers is given as input. Corresponding to each number, the program should output the number of pairs mentioned above. Notice that we are interested in the number of essentially different pairs and therefore you should not count (p1, p2) and (p2, p1) separately as two different pairs.
     
    Input
    An integer is given in each input line. You may assume that each integer is even, and is greater than or equal to 4 and less than 2^15. The end of the input is indicated by a number 0.
     
    Output
    Each output line should contain an integer number. No other characters should appear in the output.
     
    Sample Input
    6 10 12 0
    Sample Output
    1 2 1
     
    分析:用筛法求素数并用数组s记录下来就好了。
    #include<cstdio>
    bool a[10000010];
    int s[1000000];
    int cnt=0;
    void prime()
    {
        for(int i=4;i<10000001;i+=2) a[i]=1;
        s[cnt++]=2;
        for(int i=3;i<10000001;i++)
        {
            if(!a[i])
            {
                s[cnt++]=i;
                for(int j=i+i;j<10000001;j+=i)
                a[j]=1;
            }
        } 
    }
    
    int main()
    {
        int n;
        prime();
        while(scanf("%d",&n)&&n)
        {
            int N=n/2,ans=0;
            int i=0;
            while(s[i]<=N)
            {
                if(!a[n-s[i]]) ans++;
                i++;
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
    View Code
     
  • 相关阅读:
    MYSQL数据库设计规范与原则
    PHP-CI框架数据库连接默认是长连接,需要注意应用场景
    mysql性能调优与架构设计笔记
    JavaScript学习笔记
    PHP一维数组和二维数字排序整理
    MYSQL常用的Show命令笔记
    设计模式之建造者模式
    设计模式之工厂模式
    设计模式前提篇二(C++编程原则)
    设计模式前提篇一(C++/基础)
  • 原文地址:https://www.cnblogs.com/ACRykl/p/8576041.html
Copyright © 2011-2022 走看看