2017 Multi-University Training Contest - Team 1 之Add More Zero

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本文介绍了一种计算2^m-1位数的方法，通过使用log10(2)*m来确定特定超级计算机能够处理的最大整数范围，提供了一个简洁的C++实现。

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Add More Zero
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 338 Accepted Submission(s): 246

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

题意就是求2^m-1的位数

思路因为2^m次方中没有10，实际上就是求2^m的位数。

我们用log10（2）*m就好了

#include<iostream>
#include<cmath>
#include<cstring>
#include<vector>
#include<stdlib.h>
#include<stdio.h>
#include<algorithm>
#include<map>
#include <set>
#include <list>
#include <deque>
#include<sstream>
#include<time.h>
#define pi  3.1415926
#define INF 100000000
typedef long long ll;
using namespace std;

int main()
{
    int n,k=0;
    while(~scanf("%d",&n))
    {
        k++;
        int d=log10(2)*n;
        printf("Case #%d: %d\n",k,d);
    }
    return 0;
}