【2017 Multi-University Training Contest - Team 1 1001】Add More Zero

:

让你求最大的k;
使得
10^k<=2^m-1

求出2^m-1的位数就好;
[lg(2^m-1)] = lg(2^m) = m*lg2


1(没加case)

#include <bits/stdc++.h>
using namespace std;

int n;

int main(){
    int kk = 0;
    while (~scanf("%d",&n)){
        kk++;
        int temp = n*log10(2);
        printf("Case #%d: %d\n",kk,temp);
    }
    return 0;
}