题目链接

Add More Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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 denotes the answer of corresponding case.

 

Sample Input

164

 

Sample Output

Case #1: 0Case #2: 19

 

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思路：
题目是求[0,2^m-1]上的十进制数最大有多少位，即10^k中k的最大值。
显然k=log10(2^m-1),要是2^m就好了，我们算的是位数，如果+1影响到它的位数，说明2^m-1末位是9，显然不可能，则k=log10(2^m)=m*log10(2)。

ac代码：

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