Add More Zero

Problem Description

There is a youngster known for amateur propositions concerning several mathematical hard problems.

Nowadays, he is preparing a thought-provoking problem on a specific type of supercomputer which has ability to support calculations of integers between0  and (2 m −1)  (inclusive).

As a young man born with ten fingers, he loves the powers of 10  so much, which results in his eccentricity that he always ranges integers he would like to use from1  to 10 k   (inclusive).

For the sake of processing, all integers he would use possibly in this interesting problem ought to be as computable as this supercomputer could.

Given the positive integer m , your task is to determine maximum possible integer k  that is suitable for the specific supercomputer.

Input

The input contains multiple test cases. Each test case in one line contains only one positive integerm , satisfying 1≤m≤10 5  .

Output

For each test case, output "Case #x :y " in one line (without quotes), where x  indicates the case number starting from 1  and y  denotes the answer of corresponding case.

Sample Input

164

Sample Output

Case #1: 0Case #2: 19

题目的大体意思是吧2^m用10进制表示出来有多少位。

刚开始的时候傻傻的吧2^m的10进制用数组表示出来了，数数组存了多少个就是位数。（然后就超时了）

然后看他们的博客就知道了直接对2^m取对数就可以…

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>//http://acm.split.hdu.edu.cn/diy/contest_showproblem.php?pid=1001&cid=32494using namespace std;int main(){int m,count,ans;count=1;while(scanf("%d",&m)!=EOF){ans=log10(2)*m;printf("Case #%d: %d\n",count++,ans);}return 0; }