2017 Multi-University Training Contest - Team 1 HDU 6033 Add More Zero【对数】

题目来戳呀

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

1
64

Sample Output

Case #1: 0
Case #2: 19

题意:

用10^k表示出2^m-1,求k的最大整数值。

想法:

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。
问题转化成10^k<=2^m-1,两边同时取对数(-1忽略不计),得k<=m*lg2。

#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int m,cnt;
    cnt=1;
    while(~scanf("%d",&m))
    {
        int ans=(int)m*log10(2.0);//注意对数的表达
        printf("Case #%d: %d\n",cnt++,ans);
    }
    return 0;
}

ps:题意理解很重要!!!

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