HDU 6033 Add More Zero
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Add More Zero
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)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 between0 and (2m−1) (inclusive).
As a young man born with ten fingers, he loves the powers of10 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 integerm , 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
As a young man born with ten fingers, he loves the powers of
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
Input
The input contains multiple test cases. Each test case in one line contains only one positive integerm , 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
164
Sample Output
Case #1: 0Case #2: 19
Source
2017 Multi-University Training Contest - Team 1
题意:
给定m,求最大的k 满足
10^k ≦ (2^m) -1
思路:
利用log公式,得到
k <= ,即 k =取下限
由于不存在
所以
代码:
#include <iostream>#include <cstdio>#include <cmath>using namespace std;int main(){ int m; double res = log10(2); int i = 1; while(scanf("%d",&m)!=EOF){ int ans = (int)m*res; printf("Case #%d: %d\n",i,ans); i++; } return 0;}
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