HDU 6033 Add More Zero【】

Add More Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1082    Accepted Submission(s): 725

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 1to 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

 

Source

2017 Multi-University Training Contest - Team 1

 

答案就是 \left \lfloor \log_{10}(2^m - 1) \right \rfloor⌊log​10​​(2​m​​−1)⌋，注意到不存在 10^k = 2^m10​k​​=2​m​​ ，所以\left \lfloor \log_{10}(2^m - 1) \right \rfloor = \left \lfloor \log_{10}{2^m} \right \rfloor = \left \lfloor m \log_{10}{2} \right \rfloor⌊log​10​​(2​m​​−1)⌋=⌊log​10​​2​m​​⌋=⌊mlog​10​​2⌋，这样做的时间复杂度是 O(1) 。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))#define maxn 510const int M=1e6+10;const int inf=0x3f3f3f3f;const int mod=1e9+7;const double eps=1e-10;ll n,m;int main(){    int cas=1;    while(~scanf("%d",&m))    {        int ans=m*(log10(2.0));        printf("Case #%d: %d\n",cas++,ans);    }    return 0;}