哥德巴赫猜想（1e5）

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
bool notprime[MAXN];
int n;
void init() {
	memset(notprime, false, sizeof(notprime));
	notprime[0] = notprime[1] = true;
	for (int i = 2; i < MAXN; i++) {
		if (!notprime[i]) {
			if (i > MAXN / i) continue;
			for (int j = i * i; j < MAXN; j += i) {
				notprime[j] = true;
			}
		}
	}
}
int main() {
	init();
	while (scanf("%d", &n) != EOF) {
		for (int i = 2; i <= n / 2; i++) {
			if (!notprime[i] && !notprime[n - i]) {
				printf("%d %d\n", i, n - i);
			}
		}
		printf("\n");
	}
	return 0;
}

 

来源：CSDN

作者：Leonardoss

链接：https://blog.csdn.net/weixin_42769575/article/details/104065969