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 .

InputThe 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 .OutputFor each input file, your program has to write the number quadruplets whose sum is zero.Sample Input6-45 22 42 -16-41 -27 56 30-36 53 -37 77-36 30 -75 -4626 -38 -10 62-32 -54 -6 45Sample Output5HintSample 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).

#include<iostream>#include<cmath>#include<vector>#include<algorithm>#define maxn 4000using namespace std;int map1[maxn*maxn];int map2[maxn*maxn];int a[maxn],b[maxn],c[maxn],d[maxn];int main(){      int n,i,j,k,sum,p;      scanf("%d",&n);      for(i=0;i<n;i++)      {         scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);       }       for(i=0;i<n;i++)        for(j=0;j<n;j++)         map1[i*n+j]=a[i]+b[j];      for(i=0;i<n;i++)        for(j=0;j<n;j++)          map2[i*n+j]=c[i]+d[j];      sort(map1,map1+n*n);      sort(map2,map2+n*n);      sum=0;      p=n*n-1;      for(i=0;i<n*n;i++)      {         while(p>=0&&map1[i]+map2[p]>0) p--;        if(p<0) break;        int temp=p;        while(temp>=0&&map1[i]+map2[temp]==0)        {            sum++; temp--;        }      }      printf("%d/n",sum);       return 0;}