2017 Multi-University Training Contest - Team 1 簽到題 ...
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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2042 Accepted Submission(s): 1278
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 1to 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
Source 2017 Multi-University Training Contest - Team 1
- 對於2^m-1>=10^k
- 解得k<=lb(2^m-1)/lb(10)
- 即k<=m/lb(10)
1 #include <iostream> 2 #include <string> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <climits> 7 #include <cmath> 8 #include <vector> 9 #include <queue> 10 #include <stack> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL ; 15 typedef unsigned long long ULL ; 16 const int maxn = 1e5 + 10 ; 17 const int inf = 0x3f3f3f3f ; 18 const int npos = -1 ; 19 const int mod = 1e9 + 7 ; 20 const int mxx = 100 + 5 ; 21 const double eps = 1e-6 ; 22 const double PI = acos(-1.0) ; 23 24 int main(){ 25 // freopen("in.txt","r",stdin); 26 // freopen("out.txt","w",stdout); 27 int T=0; 28 double m; 29 while(~scanf("%lf",&m)){ 30 printf("Case #%d: %d\n",++T,(int)(m/log2((double)10))); 31 } 32 return 0; 33 }
