Goldbach‘s Conjecture

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本文探讨了哥德巴赫猜想的内容，即每个大于4的偶数都可以表示为两个奇素数之和，并通过两种不同的C语言程序实现来验证该猜想对于小于一百万的所有偶数的有效性。

Description

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.

Input

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.

Output

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."

Sample Input

8

20

42

0

Sample Output

8 = 3 + 5

20 = 3 + 17

42 = 5 + 37

#include<stdio.h>
#include<math.h>
int isprime(int n)
{
    int i;
    for(i=2;i<=sqrt(n)+1;i++)      //基本思路 穷举
    {
        if(n%i==0)
        {
            return 0;
        }
    }
    return 1;
}
int main()
{
    int n,i;
    while(scanf("%d",&n)&&n)
    {
    for(i=3;i<=n/2;i+=2)
    {
        if(isprime(i)&&isprime(n-i))
        {
            printf("%d = %d + %d",n,i,n-i);
            break;
        }
    }
    }
    return 0;
}

打表改进一下，以空间换时间

#include<stdio.h>
int main()
{
    char flag[1000000]={0};
    int i,j;
    for(i=2;i<1000000;i++)
    {
        if(flag[i]==0)
        {
            for(j=2*i;j<1000000;j+=i)         //Eratosthenes筛法打质数表 
            {
                flag[j]=1;
            }
        }
    }
    int n;
    while(scanf("%d",&n)&&n)
    {
        for(i=3;i<=n/2;i+=2)
        {
            if(flag[i]==0&&flag[n-i]==0)        
            {
                printf("%d = %d + %d",n,i,n-i);
                break;
            }
        }
    }
    return 0;
}