Problem Code -Double Cola

I have understood the pattern that we have there. But, I am having some difficulties in implementing it. Please help me in devising the solution to this problem.

THANKS IN ADVANCE

Problem Code -Double Cola

I have understood the pattern that we have there. But, I am having some difficulties in implementing it. Please help me in devising the solution to this problem.

THANKS IN ADVANCE

There is probably a more efficient way, but in this case you can simply simulate it. Start by subtracting 5*1, then 5*2, 5*4, 5*2^3 … while n does not become negative. Divide the result by the current power of two and you have the position in the initial query (0-based)

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Actually I was able to figure out the solution to this question. I felt like I should be sharing it here, so here it is.

```
//JUST LIKE ANIMALS !!!!
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
ll n,i;cin>>n;
ll c=1;
while(c*5<n){
n-=c*5;
c*=2;
}
//cout<<--n/c;
ll tmp=(n-1)/c;
switch(tmp){
case 0:cout<<"Sheldon";break;
case 1:cout<<"Leonard";break;
case 2:cout<<"Penny";break;
case 3:cout<<"Rajesh";break;
case 4:cout<<"Howard";break;
}
return 0;
}
```

#include <stdio.h>

```
int main()
{
int N,a,i,j;
char SName[5][10]={"Sheldon","Leonard","Penny","Rajesh","Howard"};
while(scanf("%d",&N)!=EOF)
{
for(a=1,i=0;i<30;++i)
{
if(N>5*a)
{
N-=5*a;
}
else
{
for(j=0;j<5;++j)
{
if(N>a)
{
N-=a;
}
else
{
puts(SName[j]);
break;
}
}
break;
}
a*=2;
}
}
return 0;
}
```

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Thanks but I figured it out and now I have done it