# Conject-It !

Problem:-

You are given a number N. You can perform the following operations on N any number of times:

> If N is evendivide N by 2.

> If N is odd, replace N with 3N+1.

Your task is to find out, for a given Nif it is possible to reach the number 1 after performing the above two valid operations on N any number of times.

INPUT:

• First-line contains T denoting the number of test cases.
• Next T lines contain just one integer, N.

OUTPUT:

Print “YES” (without the quotes) if it is possible to reach 1 for the given N. Else print “NO”(without quotes).

Constraints:

1<= T <=100

2<= N <= 10100,000

Sample Input
`121`
Sample Output
`YES`
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation

The steps are:

21 –> 64 –> 32 –> 16 –> 8 –> 4 –> 2 –> 1.

Code:-

Here if we take any even number and start dividing by 2 then at last we will get 1 for all number . similarly if we take a odd then we have to change it in (3*n+1) according to question and after this process it become a even number and when we start dividing it by 2 then we will get 1.

So we have to  print “YES” in all cases.

#include<bits/stdc++.h>
using namespace std;
int main()
{
int t,i;
long long int n;
cin>>t;
while(t–)
{
cin>>n;
cout<<“YES”<<endl;

}
return 0;
}