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ex: n=3
1 2 3
2 3 4
3 4 5
The answer is 1 because only number 3 appear in every row


What I have tried:

#include <iostream>

using namespace std;

void input(int **a,int n)
{
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
            cin>>a[i][j];
    }
}

void output(int **a,int n)
{
    int result=0;
    for(int i=1; i<n; i++)
    {
        bool appear=false;
        for(int j=0; j<n; j++)
        {
            for(int k=0; k<n; k++)
            {
                if(a[i][j]==a[0][k])
                {
                    appear=true;
                    break;
                }
            }
        }
        if(appear)
            result++;
    }
    cout<<result<<endl;
}

int main()
{
    int t, n;
    cin>>t;
    while(t--)
    {
        cin>>n;
        int **a=new int*[n];
        for(int i=0; i<n; i++)
            a[i]=new int[n];
        input(a,n);
        output(a,n);
    }
    return 0;
}
I think my code is really bad because I have no idea
Posted
Updated 10-Mar-22 23:53pm

1 solution

You need to check it by row, and keep a count of each number that you find. So create a second array that is as long as the highest number input + 1 (to allow for zeroes). Then as you traverse each row, for each number that you find use it as an index into the array, and add 1 to the relevant cell. So for your matrix above you would end up with something like:
C++
0, 1, 2, 3, 2, 1 // i.e no zeroes, 1 x 1, 2 x 2, 3 x 3, 2 x 4, 1 x 5.
 
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