Goldbach’s Conjecture（哥德巴赫猜想）（素数筛）

For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2.

给你一个n >=4, 至少一对 p1 p2， n = p1+p2
This conjecture has not been proved nor refused yet. No one is sure whether this conjecture actually holds.

没被证伪

However, one can find such a pair of prime numbers, if any, for a given even number.

给你一个n，你可以找到那么一对prime

The problem here is to write a program that reports the number of all the pairs of prime numbers satisfying the condition in the conjecture for a given even number.

程序目的：cout有多少对prime满足p1+p2 = n

A sequence of even numbers is given as input. Corresponding to each number, the program should output the number of pairs mentioned above. Notice that we are interested in the number of essentially different pairs and therefore you should not count (p1, p2) and (p2, p1) separately as two different pairs.

Input
An integer is given in each input line. You may assume that each integer is even, and is greater than or equal to 4 and less than 2^15. The end of the input is indicated by a number 0.

Output
Each output line should contain an integer number. No other characters should appear in the output.

Sample Input
6
10
12
0

Sample Output
1
2
1

Source
Asia 1998, Tokyo (Japan)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <map>
#include <stack>
using namespace std;


bool judge[50001];// 0-5W 是否素数
vector<int> prime;

void initial(int bound) {
    judge[0] = 0;
    judge[1] = 0;
    for(int i = 2; i <= bound ; i ++) {
        judge[i] = 1;
    }

    for (int i = 2 ; i <= bound; i ++) {
        if(judge[i]) {
            prime.push_back(i);
            //倍数不是
            for ( int j = 2 * i; j <= bound; j += i) {
                judge[j] = 0;
            }
        }
    }
}




int getResult(int n) {
    int cnt = 0;
    int maxIndex = prime.size() - 1 ;
    for(int i = 0; i <= maxIndex; i++) {
        if(prime[i] > n / 2) {
            return cnt;
        }
        int res = n - prime[i];
        if (res <= 0) {
            return cnt;
        }
        if (judge[res]) {
            cout << "prime:" << prime[i] << " + " << res << endl;
            cnt ++;
        }

    }

    return cnt;

}


int main() {
    initial(50000);
    int n;
    while(cin >> n && n != 0) {
        cout << getResult(n) << endl;

    }

}