HDU 6033 （2017 多校训练赛1 1001） Add More Zero

最新推荐文章于 2021-03-19 09:27:50 发布

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本文介绍了一个算法问题，即给定一个正整数m，求解2^m-1用10进制表示的最大长度k。通过简单的数学转换，将问题转化为k<m*log10(2)，并提供了一个C语言实现的示例程序。

2017 Multi-University Training Contest - Team 1 1001

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

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 #

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

题意： 给出一个m 求 2^m-1 用10进制表示最大长度
分析：
10^k<=2^m-1 log 10^k<=log 2^m-1 log10^k<log2^m k<mlog2/log10 k<m*log10(2)

int main()
{
    long long n,cas=0;
    while(scanf("%lld",&n)==1)
    {
    printf("Case #%lld: %lld\n",++cas, (long long)(n*log10(2)));
    }
}