A. Odd Divisor
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an integer n. Check if n has an odd divisor, greater than one (does there exist such a number x (x>1) that n is divisible by x and x is odd).
For example, if n=6, then there is x=3. If n=4, then such a number does not exist.
Input
The first line contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
Each test case contains one integer n (2≤n≤1014).
Please note, that the input for some test cases won’t fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Output
For each test case, output on a separate line:
“YES” if n has an odd divisor, greater than one;
“NO” otherwise.
You can output “YES” and “NO” in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).
当n大于3的时候，让其一直除以2，每次除完后看是否是奇数，注意四舍五入。
#include<bits/stdc++.h>
using namespace std;
int flag=0;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
long long n;
scanf("%lld",&n);
if(n%2)
{
printf("YES\n");
continue;
}
if(n<3)
{
printf("NO\n");
continue;
}
int flag=0;
double f=n*1.0;
while(f>=3)
{
f/=2;
if((int)(f+0.5)%2)
{
flag=1;
break;
}
}
if(flag==1)
printf("YES\n");
else
printf("NO\n");
}
}