2017 Chinese Multi-University Training, BeihangU Contest
思路：log10(2^m) = m*log10(2)
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb emplace_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<int, pii> #define puu pair<ULL, ULL> #define pdd pair<long double, long double> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head int n, cs = 0; int main() { while(~scanf("%d", &n)) { printf("Case #%d: %d\n", ++cs, (int)(n*log10(2.0))); } return 0; }