K - Goldbach's Conjecture

最新推荐文章于 2021-10-09 16:59:42 发布

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本文介绍了一个程序设计问题：如何验证哥德巴赫猜想，即每个大于4的偶数都可以表示为两个奇质数之和。通过筛选法找出100万以内的所有质数，并使用暴力匹配的方法找到使得两质数之差最大的一对质数组合。

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题目大意:输入一个偶数，要求解出其为两个素数的加法，且两个数素数的差最大，如果没有则输出"Goldbach's conjecture is wrong."

每一个大于4的数都能拆分成两个质数相加
用筛法打表1000000的素数,再依次暴力判断

描述

In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:

Every even number greater than 4 can be
written as the sum of two odd prime numbers.

For example:

8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.

Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.

输入

The input will contain one or more test cases.
Each test case consists of one even integer n with 6 <= n < 1000000.
Input will be terminated by a value of 0 for n.

输出

For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

样例输入

8
20
42
0

样例输出

8 = 3 + 5
20 = 3 + 17
42 = 5 + 37

//题目大意:输入一个偶数，要求解出其为两个素数的加法，且两个数素数的差最大，
//如果没有则输出"Goldbach's conjecture is wrong."
#include<iostream>
#include<cstdio>
#include<cmath>
#define MX 1000005
using namespace std;
bool a[MX];//false代表素数
int fun()
{
    int i,j,k;
    a[0]=a[1]=true;
    for(i=2;i<MX;i++)
    {
        for(j=2;j<=sqrt(i);j++)
            if(i%j==0){a[i]=true;break;}
            if(j>sqrt(i))a[i]=false;
    }
}
int main()
{
    int n,f=0;
    fun();
    while(cin>>n&&n)
    {
        int i;f=0;
        for(int i=2;i<=n;i++)
        {
            if((a[i]==false&&a[n-i]==false)||a[n]==false)
              {
               printf("%d = %d + %d\n",n,i,n-i); f=1;
               break;
              }
        }
        if(!f)
        {
            printf("Goldbach's conjecture is wrong.\n");
        }

    }
    return 0;
}