HDU 6033 Add More Zero 【数学】【基础】

Add More Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 547    Accepted Submission(s): 390

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

 

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题目链接：

2017多校训练第一场 1001

题目大意：

用10^k表示出2^m-1，求k的整数部分

解题思路：

因为1≤m≤1e5，所以2^m的最后一位只能为2，4，8，6，因此2^m-1中的-1对它的结果不会造成影响。

令10^k = 2^m 两边取对数，得 k = m*log10(2)的整数部分。

Mycode：

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int MAX = 10005;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;int main(){    int m, cas = 0;    while(~scanf("%d",&m))    {        int ans = floor(m*log10(2));        printf("Case #%d: %d\n",++cas, ans);    }    return 0;}