杭电 Add More Zero

最新推荐文章于 2021-11-03 00:45:41 发布

原创最新推荐文章于 2021-11-03 00:45:41 发布 · 1.3k 阅读

0 ·

CC 4.0 BY-SA版权

水题同时被 2 个专栏收录

75 篇文章

订阅专栏

数学思维题

35 篇文章

订阅专栏

本文介绍了一个算法问题，即给定一个整数m，如何确定10^k的最大值，使得10^k在0到2^m-1的范围内。通过使用对数转换，文章提供了一种有效的方法来解决这个问题，并给出了AC代码实现。

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 #

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：输入一个数m，求0到2^m-1范围内，使得10^k最大，求k

思路：10^k<=2^m-1   10^k<2^m     取对数   log10^k<log2^m     k<m*log2/log10    k<m*log10(2)

AC代码如下：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const double maxi=0.30102999566398;//log10(2)

int main()
{
    int num=1;
    int m;
    while(scanf("%d",&m)!=EOF)
    {
        printf("Case #%d: ",num);
        num++;
        double ans=m*maxi;
        cout<<(int)ans<<endl;
    }
    return 0;
}