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).
题意：
给定各有n个整数的四个数列A,B,C,D。要从每个数列中各取出一个数，使四个数的和为0.求出这样的组合的个数，当一个数列中有多个相同的数字时，把它们作为不同的数字看待。
分析：
从四个数列中选择的话总共有n^4种情况，所以全部判断一遍不可行。不过将它们对半分成AB和CD再考虑，就可以快速解决了。从两个序列中选择的话只有n^2种组合，所以可以进行枚举。先从A、B中选择a，b后，为了使总和为0，则需要从C，D中取出c+d=-a-b。因此先将从A，B中取数字的n^2种方法全部枚举出来，将这些和排好序，这样就可以运用二分搜索了。
有时候，问题的规模较大，无法枚举所有元素的组合，但能够枚举一半元素的组合。此时，将问题拆成两半后分别枚举，再合并它们的结果，这一方法往往非常有效。
代码：
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int n,a[4009],b[4009],c[4009],d[4009];
int num[16000000];
int main()
{
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
num[i*n+j]=a[i]+b[j];
sort(num,num+n*n);
long long int ans=0;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
int cd=-(c[i]+d[j]);
ans+=upper_bound(num,num+n*n,cd)-lower_bound(num,num+n*n,cd);
}
printf("%lld
",ans);
return 0;
}