HDU6033-Add More Zero

Add More Zero

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                               Total Submission(s): 663    Accepted Submission(s): 463

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

 

Source

2017 Multi-University Training Contest - Team 1

 

题意：问2的n次方有几位

解题思路：位数其实求一个log即可，并且减一可以省去，不会影响结果

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int main(){    int  n;    int cas=0;    while(~scanf("%d",&n))    {       int x=n*log10(2);       printf("Case #%d: %d\n",++cas,x);    }    return 0;}