Goldbach's Conjecture

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

题意描述：
验证是否一个大于4的偶数都能写成两个奇素数的和，若存在多组只输出两奇素数差值最大的一组。

解题思路：

利用筛法将1000000以内的素数标记出来，利用循环从3开始判断若i与n-i都为素数输出，并跳出循环。

#include<stdio.h>
#include<math.h>
int isprime[10000010];
int main()
{
	int i,j,m;//素数为0 
	isprime[0]=1;
	isprime[1]=1;
	isprime[2]=0;
	m=sqrt(1000010);
	for(i=2;i<=m;i++)
	{
		if(isprime[i]==0)
			for(j=2*i;j<=1000010;j=j+i)
				isprime[j]=1;
	}
	int n,f;
	while(scanf("%d",&n))
	{
		if(n==0)
			break;
		f=0;
		for(i=3;i<=n-3;i++)
			if(isprime[i]==0&&isprime[n-i]==0)
			{
				f=1;
				printf("%d = %d + %d\n",n,i,n-i);
				break;
			}
		if(f==0)
			printf("Goldbach's conjecture is wrong.\n");
	}
	return 0;
}