hdu6033--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 (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 from1
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 integerm,
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

1
64

 

Sample Output

Case #1: 0
Case #2: 19

 

题意

给定一个数m，求2m−1所表示的10进制数的位数k

题解

[align=left]开始没敢往数学上想，自己暴力瞎做....[/align]
[align=left]其实只要两边同时取log就好了....[/align]
2m−1
= 10k，因为2m一定是偶数，所以减一不会改变10进制位数
可以化简为log10(2m)
= log10(10k)2m2m
[align=left]然后k = m*log10(2)[/align]
[align=left]-----QAQ------[/align]
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代码

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
int main()
{
double m;
int num=1;
while(scanf("%lf",&m)!=EOF)
{
int k;
k=(int)(1.0*m*log10(2.0));
printf("Case #%d: %d\n",num++,k);
}
return 0;
}

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2m−1