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最新推荐文章于 2021-03-19 09:27:50 发布

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最新推荐文章于 2021-03-19 09:27:50 发布

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分类专栏： 2017杭电多校联赛 ACM 数学

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本文探讨了一道数学计算题目，目标是在特定超算环境下找到10的最高次幂k，使得10^k不超过2^m-1。文章提供了实现这一目标的算法思路与C++代码示例。

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 00 and (2m−1)(2m−1) (inclusive).

As a young man born with ten fingers, he loves the powers of 1010 so much, which results in his eccentricity that he always ranges integers he would like to use from 11 to 10k10k (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 mm, your task is to determine maximum possible integer kkthat is suitable for the specific supercomputer.

Input

The input contains multiple test cases. Each test case in one line contains only one positive integer mm, satisfying 1≤m≤1051≤m≤105.

Output

For each test case, output " Case #xx: yy" in one line (without quotes), where xxindicates the case number starting from 11 and yy 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)
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const double IP=0.30102999566398;
int main()
{
    int n,m,cas=1;
    while(~scanf("%d",&n))
    {
        int m=(int)(n*IP);
        printf("Case #%d: %d\n",cas++,m);
    }
    return 0;
}