Add More Zero
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2245 Accepted Submission(s): 1053
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.
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
题目大意:给你m,m满足有2^m-1>=10^k,询问最大k值为多少。
解题思路:有题,我们只需要求得最大k值,则可以列出下列公式:
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因此代码就非常好写了。
#include <cstdio> #include <iostream> #include <cmath> using namespace std; int main() { int m, k; int cas = 0; while(~scanf("%d", &m)) { k = log10(2)*m; printf("Case #%d: %d\n", ++cas, k); } return 0; }
题目链接:点击打开链接http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=759