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】给出一个n*4的矩阵，每列上选一个数使得最后加起来为0，问有多少种取法
【思路】先用ab数组存a+b的所有组合，同理，存储cd数组，然后对cd数组进行排序，然后用upper_bound，lower_bound查找是否存在-ab[i],正好两者只差为1，即多了一种组合方式
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int N=+;
int n;
int a[N],b[N],c[N],d[N];
int ab[N*N],cd[N*N];
int main()
{
scanf("%d",&n);
for(int i=;i<=n;i++)
{
scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
}
int k=;
for(int i=;i<=n;i++)
{
for(int j=;j<=n;j++)
{
ab[k]=a[i]+b[j];
cd[k]=c[i]+d[j];
k++;
}
}
sort(cd,cd+k);
long long ans=;
for(int i=;i<k;i++)
{
int tmp=-ab[i];
ans+=(long long )(upper_bound(cd,cd+k,tmp)-lower_bound(cd,cd+k,tmp));
}
printf("%I64d\n",ans);
return ;
}