Goldbach's ConjectureCrawling

最新推荐文章于 2021-01-18 21:37:20 发布

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#Goldbachs Conjecture

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本文介绍了一种验证哥德巴赫猜想的算法实现，该猜想指出每个大于4的偶数都可以表示为两个奇质数之和。通过构建质数表，文章提供了一个C++程序示例来验证这一猜想对于小于一百万的所有偶数的有效性。

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Goldbach's ConjectureCrawling in process...Crawling failed

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

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

解题思路：只要把素数表建好，一切迎刃而解；本题数据我貌似开大了，比较占内存，int类型完全可以存下。。。

代码：

#include<iostream>
#include<cstdio>
using namespace std;

int f[1000005],prime[1000000];
int count=0;

void Select()
{
	for(long long i=2;i<=1000005;i++)
		if(!f[i])
		{
			prime[count++]=i;
			for(long long  j=i*i;j<=1000005;j+=i)
			f[j]=1;
		}
		//cout<<count<<endl;
}

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