ZOJ.1951 Goldbach's Conjecture【素数】 2015/09/22

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本文介绍了一个程序设计问题：验证哥德巴赫猜想对于所有小于一百万的偶数是否成立。通过预先筛选出素数并使用这些素数尝试组合成目标偶数，程序能够找出符合条件的素数对。

Goldbach's Conjecture

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

Source: University of Ulm Local Contest 1998

将一个数分解成两个素数相加，若有多解，则输出两解差值最大的一组

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

bool prime[1000010];

void init(){
    memset(prime,false,sizeof(prime));
    for( int i = 2 ; i <= 1000000 ; ++i ){
        if( !prime[i] ){
            for( int j = i+i ; j <= 1000000 ; j+=i )
                prime[j] = true;
        }
    }
}

int main(){
    init();
    int n;
    while( cin>>n ){
        if( !n ) break;
        for( int i = 2 ; i <= n/2 ; ++i ){
            if( !prime[i] && !prime[n-i] ){
                printf("%d = %d + %d\n",n,i,n-i);
                break;
            }
        }
    }
    return 0;
}