#36350: C++認真3行解


tommyyang1025@gmail.com (Tommy Yang)


程式碼:

int n,m;
cin>>n>>m;
cout<<log(m/n+1)/log(2)<<endl;

因為是比例,因此n : m可以改寫成1 : (m/n)

day1是1:1

day2是1:3

day3是1:7

day4是1:15

因此可以找到規律:

m/n = 2^day -1
=>m/n + 1 = 2^day

=>day = log(m/n + 1)/log(2)