HACKER RANK Algorithms - Implementation - ACM ICPC Team - Solution.'C++'

                                Algorithms - Implementation - ACM ICPC Team - Solution
Problem Statement

You are given a list of N people who are attending ACM-ICPC World Finals. Each of them are either well versed in a topic or they are not. Find out the maximum number of topics a 2-person team can know. And also find out how many teams can know that maximum number of topics.

Note Suppose a, b, and c are three different people, then (a,b) and (b,c) are counted as two different teams.

Input Format

The first line contains two integers, N and M, separated by a single space, where Nrepresents the number of people, and M represents the number of topics. N lines follow.
Each line contains a binary string of length M. If the ith line's jth character is 1, then the ithperson knows the jth topic; otherwise, he doesn't know the topic.

Constraints 
2≤N≤500 
1≤M≤500

Output Format

On the first line, print the maximum number of topics a 2-person team can know. 
On the second line, print the number of 2-person teams that can know the maximum number of topics.

Sample Input

4 5
10101
11100
11010
00101

Sample Output

5
2

Explanation

(1, 3) and (3, 4) know all the 5 topics. So the maximal topics a 2-person team knows is 5, and only 2 teams can achieve this.

Solution:

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

void myfun(int m,int t,int **mat)
{   int i,j,q=0,maxQ=0,count=0;
    for(i=0;i<m;i++)
        for(j=0;j<m;j++)
        {   q=0;
           if (i<j)
            {  
                for(int k=0;k<t;k++)
                    q+= max(mat[i][k],mat[j][k]);
           
                if ( q > maxQ )
                  {  maxQ=q;
                     count=0;
                  }
                if( q==maxQ)
                 {   count++;
                 }
             }
        
        }
 cout<<maxQ<<"\n"<<count;
   
    
}

int main() {
    long int t,m,i,j;
    cin>>m>>t;
    char c;
    
    int **mat;
    
    mat = (int **)malloc(m*sizeof(int*));
    for(i=0;i<m;i++)
        mat[i] = (int*)malloc(t*sizeof(int));
    
    for(i=0;i<m;i++)
        for(j=0;j<t;)
         {    c=getchar();
              if( c!='\n')
              mat[i][j++] = (int)c-48;            
         }
   myfun(m,t,mat);
    return 0;
}

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