SDUT　2809 Goldbach

最新推荐文章于 2021-05-23 16:47:19 发布

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本文通过编程实现验证高斯猜想，即每个大于4的偶数都可以表示为两个奇质数之和，对小于百万的偶数进行求解并输出结果。

Goldbach

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

题目描述

The Goldbach’s Conjecture is stated like the following:

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

Your task is now to verify Goldbach\'s Conjecture for all even numbers less than a million.

输入

The input contains multiple test cases. Each test case consisted of an even integer n with 6 < n < 1000000. Input will be terminated by 0.

输出

For every test case, print one line of the form “n = a + b”, a < b, without quotes. Where a and b are odd prime numbers. If there is more than one pair of odd primes, you are to print the pair a, b with the maximized difference b - a. If there are no such pair, print a line saying “Goldbach\'s conjecture is wrong.”.

示例输入

8
20
0

示例输出

8 = 3 + 5
20 = 3 + 17

#include <bits/stdc++.h>

using namespace std;

bool prim(int n)
{
    for(int i=2;i*i<=n;i++)
    {
        if(n%i == 0)
            return false;
    }
    return true;
}

int main()
{
    int n;
    while(cin>>n && n)
    {
        for(int i=2;i<=n/2;i++)
        {
            if(prim(i))
            {
                int t = n - i;
                if(prim(t))
                {
                    cout<<n<<" = "<<i<<" + "<<t<<endl;
                    break;
                }
            }
        }
    }
    return 0;