K - Goldbach's Conjecture解题报告

最新推荐文章于 2021-02-02 10:09:10 发布

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本文探讨了哥德巴赫猜想，并提供了一种通过筛选法求解素数的方法来验证该猜想对于小于一百万的所有偶数的有效性。通过具体实例展示了如何找出两个奇素数之和等于给定偶数。

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K - Goldbach's Conjecture

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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

这题超时了N遍，就是不记得筛选法求素数了，哎~~~这次可要牢记了，其他的就不多说了。。。

#include<iostream>
#include<time.h>
short prime[1000000];
using namespace std;
int main()
{
	int i,j,s=0,n,num;
	memset(prime,0,sizeof(prime));
for(i=2;i<1000000;i++)
for(j=2;i*j<1000000;j++)
{
	if(prime[i*j]==0)
		prime[i*j]=1;
}
	while(scanf("%d",&n)&&n!=0)
	{
			for(i=3;i*2<=n;i+=2)
				if(prime[i]==0&&prime[n-i]==0)
				{
					printf("%d = %d + %d\n",n,i,n-i);
					goto end;
				}
				printf("Goldbach's conjecture is wrong.\n");
end:;
	}

	return 0;
}