FZU

/*
  这题主要是想到突破口--每次除以2(我感觉还是思维题，就像我当时做时，就是想不到 T^T)
  然后，用1个奇数和1个偶数举一下例子
  f(7) = f(3) + 1, f(3) = f(1) + 1, f(1) = 1
  f(8) = f(4) + 1, f(4) = f(2) + 1, f(2) = f(1) + 1, f(1) = 1
*/

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
	int t, n;
	scanf("%d", &t);
	for (int c = 1; c <= t; c++)
	{
		scanf("%d", &n);
		int res = 0;
		while (n)
		{
			res++;
			n /= 2;
		}
		printf("Case %d: %d
", c, res);
	}
	return 0;
}

//以及这题还想明白除以2以后，就可以用对数去做了，注意每次除以2，除以log2n次，最后还要加上f(1)，如下
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

int main()
{
	int t, n;
	scanf("%d", &t);
	for (int c = 1; c <= t; c++)
	{
		scanf("%d", &n);
		int res = (int)(log(n) / log(2) + 1);
		printf("Case %d: %d
", c, res);
	}
	return 0;
}