Educational Codeforces Round 76 (Rated for Div. 2) B. Magic Stick 水题

B. Magic Stick

Recently Petya walked in the forest and found a magic stick.

Since Petya really likes numbers, the first thing he learned was spells for changing numbers. So far, he knows only two spells that can be applied to a positive integer:

If the chosen number 𝑎 is even, then the spell will turn it into 3𝑎2;
If the chosen number 𝑎 is greater than one, then the spell will turn it into 𝑎−1.
Note that if the number is even and greater than one, then Petya can choose which spell to apply.

Petya now has only one number 𝑥. He wants to know if his favorite number 𝑦 can be obtained from 𝑥 using the spells he knows. The spells can be used any number of times in any order. It is not required to use spells, Petya can leave 𝑥 as it is.

Input

The first line contains single integer 𝑇 (1≤𝑇≤104) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains two integers 𝑥 and 𝑦 (1≤𝑥,𝑦≤109) — the current number and the number that Petya wants to get.

Output

For the 𝑖-th test case print the answer on it — YES if Petya can get the number 𝑦 from the number 𝑥 using known spells, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

input
7
2 3
1 1
3 6
6 8
1 2
4 1
31235 6578234
output
YES
YES
NO
YES
NO
YES
YES

题意

现在给你一个数x。
如果这个数是偶数，你可以让这个数变成x/2*3。
你也可以让这个数变成x-1

问你x经过若干次变换之后，能否变成y，能输出YES，不能输出NO

题解

其实，当x大于等于4的时候，这个x就可以变成无限大了，然后让x不断减1，就可以得到y了。
其他情况我们暴力就可以。

代码

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

void solve(){
	long long x,y;
	map<long long,int>H;
	cin>>x>>y;
	while(H[x]==0){
		if(x>=y){
			cout<<"YES"<<endl;
			return;
		}
		H[x]=1;
		if(x%2==1)x--;
		x=x/2*3;
	}
	cout<<"NO"<<endl;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--)solve();
}