1 条题解
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05-何承哲 (938936) LV 9 MOD @ 2020-03-26 21:36:57
这题也是之前做过的改编
注意输出的时候不再输出名字#include<iostream> #include<iomanip> #include<string> #include<vector> using namespace std; class Person { friend class Package; friend istream& operator>>(istream& c, Person& p); friend ostream& operator<<(ostream& c, Person p); private: string name; string addr; string city; string state; int zipc; public: //默认构造函数 Person() :name(""), addr(""), city(""), state(""), zipc(0.0) {} //赋值构造函数 Person(const Person& p) { name = p.name; addr = p.addr; city = p.city; state = p.state; zipc = p.zipc; } }; class Package { protected: Person sed, rec; double weg, cos; public: //初始化构造函数 Package(Person a, Person b, double w, double c) :sed(a), rec(b), weg(w), cos(c) {} //赋值构造函数,调用初始化构造函数进行赋值 Package(const Package& p) :Package(p.sed, p.rec, p.weg, p.cos) {} virtual double calculateCost() { return weg * cos; } void print() { cout << sed << endl; cout << rec << endl; } bool check() { return weg > 0 && cos > 0; } }; class TwoDayPackage : public Package { private: double ff; public: virtual double calculateCost() { return weg * cos + ff; } //调用package的赋值构造函数来初始化,下同 TwoDayPackage(Package p, double f) : Package(p), ff(f) {} }; class OvernightPackage :public Package { public: OvernightPackage(Package p, double f) : Package(p), ff(f) {} virtual double calculateCost() { return weg * (cos + ff); } private: double ff; }; istream& operator>>(istream& c, Person& p) { c >> p.name >> p.addr >> p.city >> p.state >> p.zipc; return c; } ostream& operator<<(ostream& c, Person p) { c << p.addr << ' ' << p.city << ' ' << p.state << ' ' << p.zipc; return c; } int main() { vector<Package*> ps; Person sender; Person recipient; double weight; double price; double flatFee; double addFee; cin >> sender; cin >> recipient; cin >> weight; cin >> price; cin >> flatFee; ps.push_back(new TwoDayPackage(Package(sender, recipient, weight, price), flatFee)); cin >> sender; cin >> recipient; cin >> weight; cin >> price; cin >> addFee; ps.push_back(new OvernightPackage(Package(sender, recipient, weight, price), addFee)); double sumc = 0.0; for (int i = 0; i < ps.size(); i++) { ps[i]->print(); cout << fixed << setprecision(3) << ps[i]->calculateCost() << endl; sumc += ps[i]->calculateCost(); } cout << sumc << endl; return 0; }
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- ID
- 1008
- 难度
- 1
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- 递交数
- 112
- 已通过
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- 通过率
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