[解题报告]543 - Goldbach's Conjecture

最新推荐文章于 2021-10-09 16:59:42 发布

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最新推荐文章于 2021-10-09 16:59:42 发布

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原文链接：http://www.cnblogs.com/TheLaughingMan/archive/2013/02/24/2924716.html

本文旨在通过编程方式验证哥德巴赫猜想，即所有大于2的偶数都可以表示为两个奇质数的和。对于每个输入的偶数n（小于1百万），输出其分解形式，或确认猜想在该数值范围内成立。

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Goldbach's Conjecture

In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:

Every number greater than 2 can be written as the sum of three prime numbers.

Goldbach cwas considering 1 as a primer number, a convention that is no longer followed. Later on, Euler re-expressed the conjecture as:

Every even number greater than or equal to 4 can be expressed as the sum of two 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 as expressed by Euler for all even numbers less than a million.

Input

The input file will contain one or more test cases.

Each test case consists of one even integer n with $6 \le n < 1000000$ .

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

Miguel A. Revilla
1999-01-11

哥德巴赫猜想，建质数表，前面有类似的题，速秒

#include <stdio.h>
#include <string.h>
int used[1000005];
int prim[1000005];
int main()
{
    int n,count=0;
    int i,j;
    memset(used,0,sizeof(used));
    for (i=2;i<=1000000;i++)
        if (!used[i])
        {
            prim[count++] = i;
            for(j=2*i;j<=1000000;j+=i)
                used[j]=1;
        }
    while(scanf("%d",&n)&&n)
    {
        int flag=0;
        for(i=1;prim[i]<=n/2;i++)
            if (!used[n-prim[i]])
            {
                printf("%d = %d + %d\n",n,prim[i],n-prim[i]);
                flag=1; break;
            }
        if (!flag) printf("Goldbach's conjecture is wrong.\n");
    }
    return 0;
}