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
NO
題意
現在給你一個數x。
如果這個數是偶數，你可以讓這個數變成x/2*3。
你也可以讓這個數變成x-1
問你x經過若幹次變換之後，能否變成y，能輸出YES，不能輸出NO
題解
其實，當x大于等于4的時候，這個x就可以變成無限大了，然後讓x不斷減1，就可以得到y了。
其他情況我們暴力就可以。
代碼
#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();
}