1483.Problem_A

Time Limit: 1000 MS    Memory Limit: 32768 KB
Total Submission(s): 153    Accepted Submission(s): 27

Description

Check whether an integer n is a power of 2.

Input

First line contains a single integer T (T<=20000000) which denotes the number of test cases. For each test case, there is an big integer N(0 < N < 2^1000)

Output

For each case, output the "Yes" or "No" in a single line.

Sample Input

313 8

Sample Output

YesNoYes

Hint


Source

Unknown

               现在~~题目给出的是一个大整数~求这个是不是2的次方~~这种情况下~我们可以选择模2取余法的模拟；

               将这个大整数变成一个2进制~~如果除了最高位还有1的出现~那么这就不是2的次方；

#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int m[10005];
#define mm 1000
int main()
{
	int te;
	cin >> te;
	while (te--)
	{
		memset(m, 0, sizeof(m));
		string n;
		cin >> n;
		int len = n.length();
		for (int s = len - 1; s >= 0; s--)
		{
			m[s] = n[s]-'0';
		}
		int lenn;
		int out[10005];
		int sum=1, d = 0, k = 0;
		while (sum)
		{
			sum = 0;
			for (int s = 0; s < len; s++)
			{
				d = m[s] / 2;
				sum += d;
				if (s == len - 1)
				{
					out[k++] = m[s] % 2;
				}
				else
				{
					m[s + 1] += (m[s] % 2) * 10;
				}
				m[s] = d;
			}
		}
		int spot = 1;
		for (int s =0; s <k-1; s++)
		{
			if (out[s] != 0)
			{
				spot = 0;
				break;
			}
		}
		if (spot)
		{
			printf("Yes\n");
		}
		else
		{
			printf("No\n");
		}
	}
	return 0;
}