二分-G

G - 4 Values whose Sum is 0

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

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).

　　题目大意：每行会给出四个整数，要求在每列整数中找出一个整数，使四个整数之和为0，求有几种不同组合

#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 4010;
const int maxm = 4010*4010;
int a[maxn],b[maxn],c[maxn],d[maxn];
int ab[maxm],cd[maxm];

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++)//前两列整数求和 
        for(int j=0;j<n;j++)
            ab[cnt++] = a[i] + b[j];
    cnt = 0;    
    for(int i=0;i<n;i++)//后两列整数求和 
        for(int j=0;j<n;j++)
            cd[cnt++] = c[i] + d[j];
    
    sort(ab,ab+n*n);
    sort(cd,cd+n*n);
    
    int p = n*n-1;
    cnt = 0;
    for(int i=0;i<n*n;i++){
        for(;p>=0;p--)
            if(ab[i]+cd[p]<=0)    break;
        if(p<0)                    break;
        for(int j=p;j>=0;j--){
            if(ab[i]+cd[j]==0)    cnt++;
            if(ab[i]+cd[j]<0)    break;
        } 
    }
    printf("%d",cnt);
}

　　面对四列整数，一一组合将会造成极其庞大的数据，可将其中两两组合，组合以后再相加寻找和为0的情况。