poj-2262 Goldbach's Conjecture

最新推荐文章于 2021-02-02 10:09:10 发布

原创最新推荐文章于 2021-02-02 10:09:10 发布 · 678 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#numbers #pair #output #integer #input #less

数论专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个简单的程序，用于验证哥德巴赫猜想。该程序接收一个整数n作为输入，并尝试找到一对质数a和b，使得n=a+b。如果找到了这样的一对质数，则输出这对质数；如果没有找到，则输出错误信息。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：http://poj.org/problem?id=2262

Goldbach's Conjecture

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28046		Accepted: 10797

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

题意：好吧，水题又来了。输入一个数n，输出两个数a，b；使得n=a+b，而且a和b都是质数，输出其中a最小的一组。

#include<iostream>
#include<cmath>
using namespace std;
int n,m,p;
bool verify(int n)   //判断函数.判断是否为质数
{
	m=sqrtf(n);
	for(int i=2;i<=m;i++)
	{
		if(n%i==0)
		{return false;break;}
	}
	return true;
}
int main()
{
	while(cin>>n)
	{
		p=0;
		if(n==0) break;
		for(int i=2;i<=n/2;i++)    
		{
			if(verify(i) && verify(n-i))   //使a从2开始,找到一组满足要求的a和b,结束,输出a,b
			{
				p++;
				cout<<n<<" = "<<i<<" + "<<n-i<<endl;
				break;
			}
		}
		if(p==0)
			cout<<"Goldbach's conjecture is wrong."<<endl;   //没有满足要求的a和b
	}
	return 0;
}