POJ2262,Goldbach's Conjecture,哥德巴赫猜想水过

本文介绍了一个关于哥德巴赫猜想的经典问题及其简单的验证算法。该算法通过判断素数来找出所有小于一百万的偶数可以如何表示为两个奇素数之和。
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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


分析:

= =

code:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
bool IsPrime(int n)
{
    for(int i=2;i<=sqrt(n);i++)
        if(n%i==0) return false;
    return true;
}
int main()
{
    int num,p;
    while(scanf("%d",&num),num)
        for(p=3;p<=num/2;p+=2)
            if(IsPrime(p)&&IsPrime(num-p))
            {
                printf("%d = %d + %d\n",num,p,num-p);
                break;
            }
    return 0;
}


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