POJ - 2785 4 Values whose Sum is 0

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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 2 ²⁸ ) 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).

思路：枚举所有的话是4000^4，数据太大，肯定要拆分，然后二分查找；分别分成letf[] 和 right[]两个数组；

代码如下：

#include <stdio.h>
#include <algorithm>
using namespace std;
int left[16000005] , right[16000005];
struct node{
	int a, b, c, d;
}s[4005];
int main() {
	int n;
	scanf("%d",&n);
	for(int i = 0; i < n; i++) {
		scanf("%d %d %d %d",&s[i].a, &s[i].b, &s[i].c, &s[i].d);
	}
	int t = 0;
	for(int i = 0; i < n; i++) {
		for(int j = 0; j < n; j++) {
			left[t] = s[i].a + s[j].b;
			right[t] = s[i].c + s[j].d;
			t++;
		}
	}
	int sum = 0;
	sort(left , left + t);
	for(int i = 0; i < t; i++) {
		sum += upper_bound(left,left+t,-right[i])-lower_bound(left,left+t,-right[i]);
	}
	printf("%d\n",sum);
	return 0;
}

upper_bound(a , a+n , m)返回第一个大于m的位置；

lower_bound(a , a+n , m)返回第一个大于等于m的位置；

以上两者类似于二分找到答案；

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