Hello people , I am stuck at this problem [SITTING ARRANGEMENTS](https://www.hackerrank.com/contests/hour-of-code-round-ii/challenges/sitting-arrangements) , i read the editorial but couldn't figure why the solution works, if anyone can provide an explanation/proof for the solution , it would be great. Thanks in advance :) ↵
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This is the author's solution ↵
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~~~~~↵
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#include<bits/stdc++.h>↵
using namespace std;↵
typedef long long int ll;↵
ll solve(ll n,ll m)↵
{↵
if(n==0||m==0)↵
return 0;↵
else if(n%2==0&&m%2==0)↵
return solve(n/2,m/2); // Halving both dimensions doesn't change the number of tiles↵
else if(n%2==0&&m%2==1)↵
return (n+solve(n/ 2,m/ 2));// Use a row of 1x1 tiles↵
else if(n%2==1&&m%2==0)↵
return (m+solve(n/ 2,m/ 2));// Use a column of 1x1 tiles↵
else↵
return (n+m-1+solve(n/2,m/2)); //ROW OR COLUMN (WHICHEVER OVERLAP)↵
}↵
int main()↵
{↵
ll t;↵
cin>>t;↵
while(t--)↵
{↵
ll n,m,i,j,p,q,r;↵
cin>>n>>m;↵
cout<<solve(n,m)<<endl;↵
}↵
}↵
↵
~~~~~↵
↵
↵
This is the author's solution ↵
↵
~~~~~↵
↵
#include<bits/stdc++.h>↵
using namespace std;↵
typedef long long int ll;↵
ll solve(ll n,ll m)↵
{↵
if(n==0||m==0)↵
return 0;↵
else if(n%2==0&&m%2==0)↵
return solve(n/2,m/2); // Halving both dimensions doesn't change the number of tiles↵
else if(n%2==0&&m%2==1)↵
return (n+solve(n/ 2,m/ 2));// Use a row of 1x1 tiles↵
else if(n%2==1&&m%2==0)↵
return (m+solve(n/ 2,m/ 2));// Use a column of 1x1 tiles↵
else↵
return (n+m-1+solve(n/2,m/2)); //ROW OR COLUMN (WHICHEVER OVERLAP)↵
}↵
int main()↵
{↵
ll t;↵
cin>>t;↵
while(t--)↵
{↵
ll n,m,i,j,p,q,r;↵
cin>>n>>m;↵
cout<<solve(n,m)<<endl;↵
}↵
}↵
↵
~~~~~↵
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