17 多校

Add More Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 311    Accepted Submission(s): 228

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 between 0 and (2m−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 from 1 to 10k (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 integer m, satisfying 1≤m≤105.

 

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

 

题目链接：点击打开链接

题意就是给你一个m求使 10 ^ k <= ( 2 ^ m - 1 )成立的k的最大值

就是求2 ^ m - 1这个数的位数，k为位数-1

k = log10(2^m - 1)，因为最后一位只有可能为2，4，8，6，所以减一不影响位数，即k = log10（2^m）= m * log10 (2) 。向下取整。

#include<string.h>#include<stdio.h>#include<math.h>int m,cas=0;int main(){    while(~scanf("%d",&m))        printf("Case #%d: %d\n",++cas,int(m*log10(2)));    return 0;}

当时写这题时完全没往这方面想，就想着暴力算出来