Uva - 1152 - 4 Values whose Sum is 0

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虽然时间给了9秒，但显然四重循环直接就会超时，太慢了。。。先枚举前a和b，然后在后两个中找a+b的形式。这样最后的时间也用了5秒多

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cctype>#include <cstring>#include <string>#include <sstream>#include <vector>#include <set>#include <map>#include <algorithm>#include <stack>#include <queue>#include <bitset> #include <cassert> #include <cmath>using namespace std;const int maxn = 4005;int n, c, A[maxn], B[maxn], C[maxn], D[maxn], sums[maxn * maxn];int main(){ios::sync_with_stdio(false);int T;cin >> T;while (T--) {cin >> n;for (int i = 0; i < n; i++) {cin >> A[i] >> B[i] >> C[i] >> D[i];}c = 0;// 枚举a和bfor (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {sums[c++] = A[i] + B[j];}}// 排序sums，需要一个有序数组sort(sums, sums + c);long long cnt = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {cnt += upper_bound(sums, sums + c, -C[i] - D[j]) - lower_bound(sums, sums + c, -C[i] - D[j]);}}cout << cnt << endl;if (T) {cout << endl;}}return 0;}

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