题目
原题链接
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 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
1
64
Sample Output
Case #1: 0
Case #2: 19
Statistic | Submit | Clarifications | Back
题意：问2^m-1有几位…
思路：你会发现-1只有在 10的倍数时候位数才会发生变化 10-1 = 9， 100-1， 1000-1…2^m不可能出现这种数这里写链接内容，所以只考虑2^m有几位就好了…就是10的几次方， 取个log10， 答案就是m×log2；
代码
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<math.h>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int maxn=10e5+10;
int main()
{
ll m;
ll las=0;
while(cin>>m)
{
ll k=(ll)m*log10(2);
cout<<"Case #"<<++las<<": "<<k<<endl;
}
return 0;
}