Add More Zero
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 0 and (2m−1) (2m−1) (inclusive).
As a young man born with ten fingers, he loves the powers of10 10 so
much, which results in his eccentricity that he always ranges integers he would like to use from 1 1 to 10k 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 m,
your task is to determine maximum possible integer k kthat
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
1 64
Case #1: 0 Case #2: 19
题意:
(1)
(2)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
简单的说就是,在0到(2^m-1)之间,求最大的10^k,求k的值。(英语好重要啊,想哭,真的!)
思路如下:
10^k<=2^m-1 ==> 10^k<2^m ==> k<m*log10(2)
代码如下:
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <map> #include <math.h> using namespace std; int main() { double mi = log10(2); int n,num=1; while(scanf("%d",&n)!=EOF) { double ans = n*mi; printf("Case #%d: %lld\n",num++,(long long)ans); } return 0; }