 # CHEFWARS - Editorial

Author: Aryan Agarwala
Tester: Данило Мочернюк
Editorialist: Ritesh Gupta

CakeWalk

NONE

# PROBLEM:

You are given two numbers A and B. You can perform several operations on it. In one operation:

1. Update A = A - B
2. Update B = B/2

If at any point of time A becomes non-positive before B then print 1 else 0.

# QUICK EXPLANATION:

Since B becomes half in one operation, that implies in log B operations, it will become zero. Hence, we can optimally iterate over the value of B to find the answer.

# EXPLANATION:

We just need to implement the process. Iterate over the number of operations and at each operation, first, A is decreased by B and then B becomes half of the current value.

• A = A - B
• B = B/2

Keep on iterating until B is greater than or equal to the A. So, if after this A becomes zero then answer is 1 else answer is 0.

TIME: O(Log N)
SPACE: O(1)

# SOLUTIONS:

Setter's Solution
Tester's Solution
``````#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define x first
#define y second
#define sz(a) ((int)(a.size()))
using namespace std;
const int mod = 1000 * 1000 * 1000 + 7;

int main()
{
int T;
cin >> T;
while(T--)
{
int p , h;
cin >> h >> p;
int ans = 0;
while(p != 0)
{
ans += p;
p >>= 1;
}
cout << (ans >= h) << endl;
}
}
``````
Editorialist's Solution
``````#include <bits/stdc++.h>

using namespace std;

int main()
{
int t;
cin >> t;

while(t--)
{
int a,b;
cin >> a >> b;

while(a > 0 && b > 0)
{
a -= b;
b /= 2;
}

cout << (a <= 0 ? 1 : 0) << endl;
}
}
``````

### Video Editorial

if any one wants video editorail in hindi

1 Like

My AC code was literally: print(h<=2*p)

1 Like

lol! This should fail for p = 7, h = 12.