题解

只有十二行。

#include <iostream>
using namespace std;

int n, sum, a[10000], s[10000]={0};
long long all = 0;
int main() {
    cin >> n >> sum;
    for (int i = 0; i < n; i++) cin >> a[i];
	for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) s[a[i] + a[j]]++;
	for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (sum - a[i] - a[j] > 0) all += s[sum - a[i] - a[j]];
	cout << all; 
	return 0;
}

一行流算是被我玩明白了