Codeforces Round #635 (Div. 2) B.Kana and Dragon Quest game

Codeforces Round #635 (Div. 2) B.Kana and Dragon Quest game

题目链接
Kana was just an ordinary high school girl before a talent scout discovered her. Then, she became an idol. But different from the stereotype, she is also a gameholic.

One day Kana gets interested in a new adventure game called Dragon Quest. In this game, her quest is to beat a dragon.

The dragon has a hit point of x initially. When its hit point goes to 0 or under 0, it will be defeated. In order to defeat the dragon, Kana can cast the two following types of spells.

• Void Absorption
  Assume that the dragon’s current hit point is h, after casting this spell its hit point will become ⌊h2⌋+10. Here ⌊h2⌋ denotes h divided by two, rounded down.

• Lightning Strike
  This spell will decrease the dragon’s hit point by 10. Assume that the dragon’s current hit point is h, after casting this spell its hit point will be lowered to h−10.

Due to some reasons Kana can only cast no more than n Void Absorptions and m Lightning Strikes. She can cast the spells in any order and doesn’t have to cast all the spells. Kana isn’t good at math, so you are going to help her to find out whether it is possible to defeat the dragon.

Input

The first line contains a single integer t (1≤t≤1000) — the number of test cases.

The next t lines describe test cases. For each test case the only line contains three integers x, n, m (1≤x≤1e5, 0≤n,m≤30) — the dragon’s intitial hit point, the maximum number of Void Absorptions and Lightning Strikes Kana can cast respectively.

Output

If it is possible to defeat the dragon, print “YES” (without quotes). Otherwise, print “NO” (without quotes).

You can print each letter in any case (upper or lower).

Example

input

7
100 3 4
189 3 4
64 2 3
63 2 3
30 27 7
10 9 1
69117 21 2

output

YES
NO
NO
YES
YES
YES
YES

对第一种操作我们不难发现，当 $x>20$ 时，该操作会一直减小 $x$
所以当 $x>20$ 时，我们可以一直执行第一种操作，然后再执行第二种操作，这样能尽可能的减少 $x$，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    int t,x,n,m;
    cin>>t;
    while(t--){
        cin>>x>>n>>m;
        while(x>20 && n){
            x=x/2+10;
            n--;
        }
        while(m){
            x-=10;
            m--;
        }
        if(x<=0) puts("YES");
        else puts("NO");
    }
}