题解：Goldbach's Conjecture（素数表）

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本文介绍了一个基于C++的程序，用于验证哥德巴赫猜想，即每个大于4的偶数都可以表示为两个奇素数之和。通过预处理素数列表并使用算法查找符合条件的素数对，该程序能够处理小于一百万的所有偶数。

每个大于4的整数可以以转换为两个素数和

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 <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <algorithm>
typedef long long ll;
using namespace std; 
#define N 1000010
int n,a[N];
void prime()
{
 memset(a,0,sizeof(a));
 a[0]=a[1]=1;
 for(int i=2;i<=N/2;i++)
  if(a[i]==0)
  {
   for(int j=2*i;j<=N;j+=i)
    a[j]=1;	
  }	
}
int main()
{
    prime();
    int flag;
    while(cin>>n&&n)
    {
     flag=1;
     for(int i=3;i<=N/2;i++)//
      if(a[i]==0&&a[n-i]==0)
      {
      	printf("%d = %d + %d\n",n,i,n-i);
		flag=0;
		break;
	  }
	 if(flag) printf("Goldbach's conjecture is wrong.\n");
	}
    return 0;
}