POJ 2785 (二分)

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 18785 Accepted: 5579Case Time Limit: 5000MS

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 -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).

Source

Southwestern Europe 2005

题意是给出4个数列，每个数列n个数，要求从每个数列中取出一个数使得4个数的和为0。

先取两组记录两两的和，然后枚举另外两组两两的和从已知的和里面去找。

可能map常数比较大过不了，二分就可以了。

#include <iostream>#include <cmath>#include <cstdlib>#include <time.h>#include <vector>#include <cstdio>#include <cstring>#include <map>#include <algorithm>using namespace std;#define maxn 4111int n;long long a[maxn][4];long long sum[maxn*maxn];int main () {while (scanf ("%d", &n) == 1) {for (int i = 1; i <= n; i++) {scanf ("%lld%lld%lld%lld", &a[i][0], &a[i][1], &a[i][2], &a[i][3]);}int cnt = 0;for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {long long num = a[i][2]+a[j][3];sum[cnt++] = num;}}long long ans = 0;sort (sum, sum+cnt);for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {long long num = -1 * (a[i][0] + a[j][1]);ans += upper_bound (sum, sum+cnt, num) - lower_bound (sum, sum+cnt, num);}}printf ("%lld\n", ans);}return 0;}