#28469: __求解


1080294@dmhs.kh.edu.tw (cocodor)


#include<iostream>
#include<iomanip>
using namespace std;
int main(){
double a,b;
char c;
while(cin>>a>>b>>c){
if(c=='U'){
    if(a/30.9-b>=0)
    cout<<fixed<<setprecision(2)<<"U"<<" "<<a/30.9-b<<endl;
    else
    cout<<"No Money"<<endl;
}
if(c=='E'){
    if(a/34.5-b>=0)
    cout<<fixed<<setprecision(2)<<"E"<<" "<<(double)a/34.5-b<<endl;
    else
    cout<<"No Money"<<endl;
}
if(c=='J'){
    if(a/0.28-b>=0)
    cout<<fixed<<setprecision(2)<<"J"<<" "<<a/0.28-b<<endl;
    else
    cout<<"No Money"<<endl;
}
if(c=='T'){
    if(a-b>=0)
    cout<<fixed<<setprecision(2)<<"T"<<" "<<a-b<<endl;
    else
    cout<<"No Money"<<endl;
}
}
return 0;
}
老師說
少了0.00的呈現 <0 0.005~0之間 >=0.005 三個判斷
#28478: Re:求解


cges30901 (cges30901)


#include
#include
using namespace std;
int main(){
double a,b;
char c;
while(cin>>a>>b>>c){
if(c=='U'){
    if(a/30.9-b>=0)
    cout<<fixed<<setprecision(2)<<"U"<<" "<<a/30.9-b<<endl;
    else
    cout<<"No Money"<<endl;
}
if(c=='E'){
    if(a/34.5-b>=0)
    cout<<fixed<<setprecision(2)<<"E"<<" "<<(double)a/34.5-b<<endl;
    else
    cout<<"No Money"<<endl;
}
if(c=='J'){
    if(a/0.28-b>=0)
    cout<<fixed<<setprecision(2)<<"J"<<" "<<a/0.28-b<<endl;
    else
    cout<<"No Money"<<endl;
}
if(c=='T'){
    if(a-b>=0)
    cout<<fixed<<setprecision(2)<<"T"<<" "<<a-b<<endl;
    else
    cout<<"No Money"<<endl;
}
}
return 0;
}
老師說
少了0.00的呈現 =0.005 三個判斷


題目不就講得很清楚了?

「若餘額為正數且小於 0.05,列印 0.00」