NUC1090 Goldbach's Conjecture【哥德巴赫猜想 】

Goldbach's Conjecture

时间限制: 1000ms 内存限制: 65536KB

问题描述

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.

输入描述

The input file 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.

输出描述

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."

样例输入

8
20
42
0

样例输出

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

来源

Ulm Local 1998

问题分析：

这个题与《POJ2262 ZOJ1951 UVa543 Goldbach's Conjecture》完全相同，代码直接拿过来就可以了。

程序说明：

参见参考链接。

参考链接：POJ2262 ZOJ1951 UVa543 Goldbach's Conjecture

题记：

程序写多了，似曾相识的也就多了。

AC的C++程序如下：

/* POJ2262 ZOJ1951 UVa543 Goldbach's Conjecture */  
  
#include <stdio.h>  
#include <math.h>  
  
#define MAXN 1000000  
  
// 试除法判断一个数是否为素数  
int isprime(int n)  
{  
    int end2, i;  
  
    if(n == 3)  
        return 1;  
  
    end2 = sqrt(n);  
    for(i=3; i<=end2; i+=2) {  
        if(n % i == 0)  
            break;  
    }  
  
    return i > end2 ? 1 : 0;  
}  
  
int main(void)  
{  
    int n;  
    int okflag, i;  
  
    while(scanf("%d", &n) != EOF) {  
        // 判定结束条件  
        if(n == 0)  
            break;  
  
        // 寻找素数  
        okflag = 0;  
        for(i=3; i<=n/2; i+=2)  
            if(isprime(i) && isprime(n-i)) {  
                okflag = 1;  
                break;  
            }  
  
        // 输出结果  
        if(okflag)  
            printf("%d = %d + %d\n", n, i, n-i);  
        else  
            printf("Goldbach's conjecture is wrong.\n");  
    }  
  
    return 0;  
}