HDU 6033 Add More Zero (数学)

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 $\mathrm{00 and (2m-1)(2m-1) (inclusive). As\; a\; young\; man\; born\; with\; ten\; fingers,\; he\; loves\; the\; powers\; of 1010 so\; much,\; which\; results\; in\; his\; eccentricity\; that\; he\; always\; ranges\; integers\; he\; would\; like\; to\; use\; from 11 to 10k10k (inclusive). For\; the\; sake\; of\; processing,\; all\; integers\; he\; would\; use\; possibly\; in\; this\; interesting\; problem\; ought\; to\; be\; as\; computable\; as\; this\; supercomputer\; could. Given\; the\; positive\; integer m,\; your\; task\; is\; to\; determine\; maximum\; possible\; integer k that\; is\; suitable\; for\; the\; specific\; supercomputer.}$

 

Input

The input contains multiple test cases. Each test case in one line contains only one positive integer $\mathrm{mm,\; satisfying 1\le m\le 1051\le m\le 105.}$

 

Output

For each test case, output " Case #$\mathrm{xx: yy}$" in one line (without quotes), where $\mathrm{xx indicates\; the\; case\; number\; starting\; from 11 and yy denotes\; the\; answer\; of\; corresponding\; case.}$

 

Sample

Sample Input
1
64 
 
Sample Output
Case #1: 0
Case #2: 19 

题意：

　　给出10^k ≥ 2^m -1,求k的最大整数

思路：

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

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<math.h>
using namespace std;
int main()
{
    int flag = 1,n;
    while(scanf("%d",&n)!=EOF)
    {
        int ans= (n*log10(2));
        printf("Case #%d: %d
",flag++,ans);
    }
    return 0;
}