4 Values whose Sum is 0
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).
题目大意：
给出4个数列，从每个数列中选取一个元素，有多少种组合使得和为0？
思路：
首先将4个数列合并为两个数列的部分和，然后比较合并后的两个数列是否有和为0的情况。
将合并后的第二个数组进行排序，用二分查找找到相反数的下限，然后统计下限到上限的距离。
AC代码：
#include <iostream>
#include <algorithm>
using namespace std;
int num[4][4000];//原数组
int nu1[4000*4000];//合并数组1
int nu2[4000*4000];//合并数组2
int main(){
int n;
cin>>n;//原数组长度
int m=n*n;//合并数组长度
for(int i=0;i<n;i++){
for(int j=0;j<4;j++){
cin>>num[j][i];//输入
}
}
int p=0;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
nu1[p]=num[0][i]+num[1][j];//数组1合并
nu2[p++]=num[2][i]+num[3][j];//数组2合并
}
}
sort(nu2,nu2+m);//排序
int time=0;//条件符合计量
for(int i=0;i<m;i++){//二分查找下限
int l=0,r=m,mid;
int _t=0;
while(r>l){
mid=(l+r)/2;
if(nu1[i]+nu2[mid]<0) l=mid+1;//如果小于0，则下限+1
else r=mid;
while(nu1[i]+nu2[l]==0&&l<m){
time++;//条件符合计次
l++;//下限右移
}
}
}
cout<<time;//输出
}