fzu-2268

E - Cutting Game

 

Fat brother and Maze are playing a kind of special (hentai) game with a piece of gold of length N where N is an integer. Although, Fat Brother and Maze can buy almost everything in this world (except love) with this gold, but they still think that is not convenient enough. Just like if they would like to buy the moon with M lengths of the gold, the man who sold the moon need to pay back Fat Brother and Maze N-M lengths of the gold, they hope that they could buy everything they can afford without any change. So they decide to cut this gold into pieces. Now Fat Brother and Maze would like to know the number of the pieces they need to cut in order to make them fulfill the requirement. The number of the gold pieces should be as small as possible. The length of each piece of the gold should be an integer after cutting.

Input

The first line of the data is an integer T (1 <= T <= 100), which is the number of the text cases.

Then T cases follow, each case contains an integer N (1 <= N <= 10^9) indicated the length of the gold.

Output

For each case, output the case number first, and then output the number of the gold pieces they need to cut.

Sample Input

13

Sample Output

Case 1: 2

题意：求解一个数可以分成几个连续的数，比如3，分成1  2，如4 分成1 2 1这里题目要求求解切几刀

思路：找规律，写一些数观察到切几下相当于求这个数的二进制有几位。

代码：

#include<cstdio>using namespace std;int main(){    int t,cas;    scanf("%d",&t);    for(int cas=1; cas<=t; cas++)    {        int n,sum=0;        scanf("%d",&n);        while(n)        {            sum++;            n>>=1;        }        printf("Case %d: %d\n",cas,sum);    }    return 0;}