poj2262Goldbach's Conjecture(素数打表）

原题链接

Goldbach's Conjecture

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44278		Accepted: 16922

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

8
20
42
0

Sample Output

8 = 3 + 5
20 = 3 + 17
42 = 5 + 37
解题思路：题意比较简单，就是输入一个偶数输出它由哪两个素数相加得到。我是打了个素数表然后判断i和n-i是不是同为素数。


 

  
#include
   
    
#include
    
     
#include
     
      
using namespace std;
#define MAX 1000000
bool isprime[MAX+1];
void prime(){//打印一个素数表，这个算法很有用
	memset(isprime,true,sizeof(isprime));
	int i;
	for(int i=2;i*i<=MAX;i++){
		if(isprime[i]){
			for(int j=i+i;j<=MAX;j=j+i){
				isprime[j]=false;
			}
		}
	}
}
int main(){
	prime();
	int n;
	while(scanf("%d",&n)&&n!=0){
		for(int i=3;i