2017多校第一场 Add More Zero(hdu6033)

最新推荐文章于 2018-09-10 20:54:13 发布

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本文介绍了一个关于特定超级计算机能够处理的整数范围的问题——AddMoreZero。该问题旨在找到最大的整数k，使得从1到10^k的所有整数能够在(2^m-1)的范围内被表示。文章提供了问题背景、输入输出样例及解决方案，并附带了实现代码。

Add More Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1795 Accepted Submission(s): 1147

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

题解：求2进制的10进制表达的指数；

log10(2)*m,取下界

即ans=(int)floor(m*log10(2));

注意：log10(2)要写成log10((long double)2)

比赛时将log10(2)提前计算出来，但是不知道要取几位，导致wa

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int solve(int m)
{
	int ret=(int)floor(m*log10((long double)2));
	return ret;
}
int main()
{
    for(int t=1,m;scanf("%d",&m)==1;t++)
		printf("Case #%d: %d\n",t,solve(m));
	return 0;
}