C++14(g++5.4) 解法, 执行用时: 20ms, 内存消耗: 1400K, 提交时间: 2020-08-24 17:48:59

#include <bits/stdc++.h>
using namespace std;
#define MAXN 1000000
#define MAXSQRTN 1000
int n;
bitset<MAXN+5> is_not_prime;
int main(){
    //freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
    for(int i=2;i<=MAXSQRTN;i++){
        if(!is_not_prime[i]){
            for(int j=i*2;j<=MAXN;j+=i){
                is_not_prime[j]=true;
            }
        }
    }
    while(scanf("%d",&n)==1&&n!=0){
        for(int i=3;i<=n;i++){
            if(!is_not_prime[i]&&!is_not_prime[n-i]){
                printf("%d = %d + %d\n",n,i,n-i);
                break;
            }
        }
    }
    return 0;
}

C++(clang++11) 解法, 执行用时: 124ms, 内存消耗: 1596K, 提交时间: 2021-05-14 16:18:37

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int n;
bool check(int x){
    if(x <= 1) return false;
    for(int i = 2;i <= x / i;i ++ ){
        if(x % i == 0) return false;
    }
    return true;
}
int main(){
    
    while(scanf("%d", &n), n){
        for(int i = 2;i <= n / 2;i ++ ){
            if(check(i) && check(n - i)){
                printf("%d = %d + %d\n", n, i, n - i);
                break;
            }
        }
    }
    return 0;
}