[exercise -11] 4 values why sum is 0 (and 4 values of 0)

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).

The question ：

Enter one per line a,b,c,d, Ask the input a[n],b[n],c[n],d[n] How many combinations can make a+b+c+d=0 establish .

The question is very simple .
First of all, we can definitely think of the enumeration of violence , four layers for loop , But it will obviously time out （4000^4）.
This time , You can think of divide and conquer （ Turn the big problem into several small problems ）. I can put a+b recorded , Then see if there is -c-d（a+b+c+d=0 ==> a+b = -c-d）.

AC Code ：

#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;
}

use scanf To prevent timeout ..
The most important thing here is to understand ：

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

upper_bound Yes, return greater than ,lower_bound Is to return greater than or equal to . If this time is not equal to , Then it must be 0,（lower = upper = The first one is greater than sum Number of numbers ）. But if there is equal to sum The number of will return its number .
You can use this to find The number of a certain number .