Goldbach's Conjecture - POJ 2262 水题

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本文探讨哥德巴赫猜想的实现，即验证每个大于4的偶数都能被表示为两个奇质数之和。通过输入偶数，输出最优解的奇质数组合，展示了一种高效且简洁的解决方案。

Goldbach's Conjecture

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 37157		Accepted: 14253

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

Sample Output

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

题意：输入一个偶数，输出两个奇质数的和。并且让b-a最大。

思路：首先哥德巴赫猜想是对的，其次，打表会超时……好久好久……

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int vis[2000010];
int main()
{ int i,j,k,num=0,n;
  for(i=2;i<=1000000;i++)
   if(vis[i]==0)
    for(j=i*2;j<=1000000;j+=i)
     vis[j]=1;
  while(~scanf("%d",&n) && n>0)
  { for(i=3;i<=n;i++)
     if(vis[i]==0 && vis[n-i]==0)
     { printf("%d = %d + %d\n",n,i,n-i);
       break;
     }
  }
}