70分3RE求调（调对必关）

#include <bits/stdc++.h>
using namespace std;
int n, sum;
int a[105];
bool sf[105];
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1; i < n; i++) {
		for (int j = i + 1; j <= n; j++) {
			for (int k = 1; k <= n; k++) {
				if (!sf[a[k]] && k != i && k != j && a[i] + a[j] == a[k]) {
					sum ++;
					sf[a[k]] = true;
				}
			}
		}
	}
	cout << sum;
	return 0;
}