greedy implementation math *900

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Python Code:

for i in range(int(input())):
    x,v,l=map(int,input().split())
    while x>20 and v>0 :
        x=(x//2)+10
        v-=1
    x-=(l*10)
    if x>0:
        print("NO")
    else:
        print("YES")

C++ Code:

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define pb push_back;
#define mod 1000000007
 
// SUBMITTED BY PARVEJ AALAM 7077

ll fact(ll n)
{
    if(n==1||n==0)
    return 1;
    else{
      return n*fact(n-1)%mod;
    }
}
int main()
{
  ll tc;
  cin>>tc;
  while(tc--)
  {
     ll h,m,n;
     cin>>h>>n>>m;
     while(h>0&&n>0&&h/2+10<h){
        n--;
        h=h/2+10;
     }
     if(h<=m*10){
     cout<<"YES" <<endl;
     }
     else{
      cout<<"NO"<<endl;
     }
  }
  return 0;

}


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