POJ2785 4 Values whose Sum is 0(暴力二分查找)

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 -4626 -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).

思路：

题目是要求四个数相加之和为零， 我们可以两 两相加， 这样的话就变成是求两个数之和为0了，再进一步想就是在另一个数组中找相反数，利用low_bound

()和up_bound可以快速的二分查找。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int mx = 4050;int a[mx], b[mx], c[mx], d[mx];int n, sum1[mx * mx], sum2[ mx * mx]; void slove(){  int cnt =0, nn = n * n;  sort(sum1, sum1 + nn);   for(int i = 0; i < nn; i++){ cnt += upper_bound(sum1, sum1+ nn, -sum2[i]) - lower_bound(sum1, sum1+ nn, -sum2[i]); //记得负号  } cout<<cnt<<endl; }int main(){while(scanf("%d", &n) != EOF){for(int i = 0; i < n; i++)scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);int c1 = 0, c2 = 0; for(int i =0; i < n; i++)for(int j =0; j < n; j++){sum1[c1++] = a[i] + b[j];sum2[c2++] = c[i] + d[j];}slove();}return 0;}