POJ2785-4 Values whose Sum is 0

传送门：http://poj.org/problem?id=2785

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 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 2²⁸ ) 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). 

 

 

给你一个数字n，然后有n组每组4个数，求每一列取出一个数，使最终四个数的和为0，求有多少种组合方式

解题思路：先求出来前两列和后两列的和，然后二分就可以了，对于这道题来说结果没有去重

#include<vector>
#include<set>
#include<stdio.h>
#include<algorithm>
using namespace std;

vector <int > a;
vector <int > b;
vector <int > c;
vector <int > d;
vector <int > sum1;
vector <int > sum2;
int main()
{
    int n, i, j, a1, b1, c1, d1, sum;
    while(scanf("%d", &n)!=EOF) {
        sum = 0;
        for(i = 0; i < n; i++)
        {
            scanf("%d%d%d%d", &a1, &b1, &c1, &d1);
            a.push_back(a1);
            b.push_back(b1);
            c.push_back(c1);
            d.push_back(d1);
        }
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++) {
                sum1.push_back(a[i] + b[j]);
                sum2.push_back(c[i] + d[j]);
            }
        }
            
    /*    for(i = 0; i < sum1.size(); i++) {
            for(j = 0; j < sum2.size(); j++) {
                if(sum1[i]+sum2[j] == 0) {
                    sum++;
                }
            }
        }*/
        n = sum2.size();
        sort(sum2.begin(),sum2.end());
        
        for(i = 0; i < n; i++) {
            int l = 0, r = n-1, mid;
            while(l <= r) {
                mid = (l + r) / 2;
                if(sum1[i] + sum2[mid] == 0) {
                    sum++;
                    for(j = mid + 1; j < n; j++) {
                        if(sum1[i] + sum2[j] != 0) 
                            break;
                        else 
                            sum++;
                    }
                    for(j = mid - 1; j >= 0; j--) {
                        if(sum1[i] + sum2[j] != 0) 
                            break;
                        else 
                            sum++;
                    }
                    break;
                }
                if(sum1[i] + sum2[mid] < 0) 
                    l = mid + 1;
                else 
                    r = mid - 1;
            }
        }
        printf("%d
", sum);
    }
    return 0;
}