/*思路 两个数：a,b 1.求a中有几个b：int n= (int)(a/b) 2.c=a-n*b,余数double */ # include<iostream> # include<iomanip> using namespace std; int main(){ double a,b,c; int n; cin>>a>>b; n=a/b; c=a-n*b; cout<<fixed<<setprecision(2)<<c<<endl; return 0; }