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 from1 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 from1 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
題目大意:給你m,m滿足有2^m-1>=10^k,詢問最大k值爲多少。
解題思路:有題,我們只需要求得最大k值,則可以列出下列公式:
因此代碼就非常好寫了。
#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;
}
題目鏈接:點擊打開鏈接http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=759