poj-2785-4 Values whose Sum is 0（折半枚举）

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本文探讨了一个经典的计算机科学问题——四数求和。给定四个整数列表A、B、C和D，目标是找出所有可能的四元组(a, b, c, d)，使得a + b + c + d = 0。文章详细介绍了如何通过两层循环构建辅助数组ab，并使用二分查找算法来计算满足条件的四元组数量。

Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 228 ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a[5000],b[5000],c[5000],d[5000],ab[16000010];
int main(){
	int n;
	scanf("%d",&n);
	for(int i = 0;i < n;i++){
		scanf("%lld%lld%lld%lld",&a[i],&b[i],&c[i],&d[i]);
	}
	int k = 0;
	for(int i = 0;i < n;i++){
		for(int j = 0;j < n;j++){
			ab[k++] = a[i]+b[j];
		}
	}
	sort(ab,ab+k);
	ll ans = 0;
	for(int i = 0;i < n;i++){
		for(int j = 0;j < n;j++){
			ll te = -(c[i]+d[j]);
			ans+= upper_bound(ab,ab+k,te)-lower_bound(ab,ab+k,te);
		}
	}
	printf("%lld\n",ans);
	return 0;
}