K - Goldbach's Conjecture解题报告（张宇）

K - Goldbach's Conjecture

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

Submit Status Practice POJ 2262

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

820420

Sample Output

8 = 3 + 520 = 3 + 1742 = 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;}