2017 Multi-University Training Contest - Team 1 - 1001

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Add More Zero

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 761    Accepted Submission(s): 523
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[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
The input contains multiple test cases. Each test case in one line contains only one positive integerm,
satisfying 1≤m≤105.
 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

1
64

 

[align=left]Sample Output[/align]

Case #1: 0
Case #2: 19

 题目大意：给你一个数m求2的m次方-1等于10的k次方求出这个k。

题目做法：k=log10（2的m次方）即k=m*log10（2）。

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