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~??
f s s-1 f ,
.
floor/2^(s-1)+s-1~

floor/2^s 1 s
s floor/2^s 1 s+1 .

s() ...
7, 7 4 6 7 3 .
8, 8 4 6 7 8 4 .
2 .
8 2^3 1 한 4 .
9 9 4회 (16 )

, floor/2^s 1 s+1회 .

~cpp
#include <iostream> 
#include <cmath> 
using namespace std; 

int testcase, nFloor, nSaver; 

int calculate(int f, int s) 
{ 
	if(f/pow(2,s)<1)
	{
		while(s--)
		{
			if(f/pow(2,s)>=1)
				return s+1;
		}
	}
    return f/pow(2,s-1)+s-1;   // f/pow(2,s-1) =>s-1 통해      
} 

int main() 
{ 
    cin>>testcase; 
    while(testcase--) 
    { 
        cin>>nFloor>>nSaver; 
        cout << calculate(nFloor, nSaver) <<endl; 
    } 
    return 0; 
} 
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