【练习-11】4 Values whose Sum is 0（和为0的4个值）

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×B×C×D are such that a+b+c+d=0. In the following, we assume that all lists have the same size n.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

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 test case, your program has to write the number quadruplets whose sum is zero.

The outputs of two consecutive cases will be separated by a blank line.

Samples

Input
1
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
Output
5

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).

题意：

每行输入一个a,b,c,d，问输入的这些a[n],b[n],c[n],d[n]有多少种组合可以使a+b+c+d=0成立。

题意很简单。
首先我们肯定能想到的暴力枚举，四层for循环，但是很显然会超时（4000^4）。
这时间，可以想到分治（把大问题化成若干个小问题）。我可以先把a+b记录下来，然后看是否有-c-d（a+b+c+d=0 ==> a+b = -c-d）。

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define CLEAR(a) memset(a,0,sizeof a)
#define Clear(a) memset(a,-1,sizeof a)
typedef long long ll;
const int N = 1e5+5;
const ll mod = 1e9+7;
ll a[4005],b[4005],c[4005],d[4005];
ll tmp[16000005];
int main()
{
    
	int n,t;
    ll res;
    cin>>t;
    while(t--)
    {
    
    	while (~scanf("%d",&n)) 
		{
    
	        res = 0;
	        for (int i = 0; i < n; i++) 
	            scanf ("%lld%lld%lld%lld", &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++) 
	                tmp[cnt++] = c[i] + d[j];
	        sort(tmp, tmp + cnt);
	        for (int i = 0; i < n; i++) 
	            for (int j = 0; j < n; j++) 
				{
    
	                ll sum =-a[i]-b[j];
	                res += upper_bound(tmp,tmp+cnt,sum) - lower_bound(tmp,tmp+cnt,sum);
	            }
	        printf ("%lld\n\n",res);
	    }
	}
    return 0;
}

用scanf是为了防止超时。。
这里最重要的是理解：

res += upper_bound(tmp,tmp+cnt,sum) - lower_bound(tmp,tmp+cnt,sum);

upper_bound是返回大于，lower_bound是返回大于等于。这时间如果没有等于的话，那么返回的一定是0,（lower = upper = 第一个大于sum的数）。但是如果有等于sum的数就会返回它的个数。
可以用这个来求某个数的个数。