POJ - 2785：4 Values whose Sum is 0

4 Values whose Sum is 0

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题目

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 .

输入

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 2 28 ) that belong respectively to A, B, C and D .

输出

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

输入样例

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

输出样例

5

样例解释

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).

题目大意

参考代码

#include
#include
#define MAXN 4005
using namespace std;
int A[MAXN],B[MAXN],C[MAXN],D[MAXN];
int AB[MAXN*MAXN];

int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);
	}
	int cnt=0;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			AB[cnt++]=A[i]+B[j];
		}
	}
	sort(AB,AB+cnt);
	int ans=0;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			int t=-C[i]-D[j];
			ans += upper_bound(AB,AB+cnt,t) - lower_bound(AB,AB+cnt,t);//因为存在AB[]中可能存在相同元素，普通的二分查找不再适用 
		}
	}
	printf("%d\n",ans);
	return 0;
}