POJ 2785 4 Values whose Sum is 0

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探讨了在四个列表中选取元素使其和为零的问题，采用分组求和结合二分查找的方法解决，并附带了一个易错案例及AC代码。

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4 Values whose Sum is 0

Time Limit:15000MS Memory Limit:228000KB 64bit IO Format:%I64d & %I64u

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 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

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).

题意
给 n * 4 (0 < n <= 4000)的数表，在每列中取一个数，使四个数相加和为 0 。

分析
很明显直接枚举会爆掉，所以两两分组求和之后二分查找。

本题有个坑，当数据为
0 0 0 0

0 0 0 0

时，总数应为256

AC代码如下

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 4000+10;
int arr[N][4];
int sumab[N*N],sumcd[N*N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < 4; j++)
            scanf("%d",&arr[i][j]);
    int t = 0;
    for(int i = 0; i < n; i++)<span style="white-space:pre">				</span>//分组求和
        for(int j = 0; j < n; j++)
            sumab[t++] = arr[i][0] + arr[j][1];
    int tt = 0;
    for(int i = 0; i < n; i++)<span style="white-space:pre">				</span>//分组求和
        for(int j = 0; j < n; j++)
            sumcd[tt++] = arr[i][2] + arr[j][3];
    int sum = 0;
    sort(sumcd,sumcd+tt);
    for(int i = 0; i < t; i++)<span style="white-space:pre">				</span>//二分查找
    {
        int x = lower_bound(sumcd,sumcd+tt,0-sumab[i]) - sumcd;
        for(int j = x; j < tt; j++)
        {
            if(sumcd[j] + sumab[i] > 0) break;
            if(sumcd[j] + sumab[i] == 0) sum++;<span style="white-space:pre">						</span>//跳坑
        }
    }
    printf("%d\n",sum);
    return 0;
}