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 = ;
         for(i = ; 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 = ; i < n; i++) {
             for(j = ; 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 = ; i < n; i++) {
             int l = , r = n-, mid;
             while(l <= r) {
                 mid = (l + r) / ;
                 if(sum1[i] + sum2[mid] == ) {
                     sum++;
                     for(j = mid + ; j < n; j++) {
                         if(sum1[i] + sum2[j] != )
                             break;
                         else
                             sum++;
                     }
                     for(j = mid - ; j >= ; j--) {
                         if(sum1[i] + sum2[j] != )
                             break;
                         else
                             sum++;
                     }
                     break;
                 }
                 if(sum1[i] + sum2[mid] < )
                     l = mid + ;
                 else
                     r = mid - ;
             }
         }
         printf("%d\n", sum);
     }
     return ;
 }