HW 12

àì夳堔傛蜴生んèń 2021-09-30 04:06 241阅读 0赞

知道自己究竟想做什么，知道自己究竟能做什么是成功的两大关键

—— 比尔盖茨

## Problem A: 杯子 ##

题面：  
![1618425-20190627182846838-931266516.png][]

## main函数： ##

int main()
    {
        Cup c1;
        int i, j;
        cin>>i>>j;
        Cup c2(i), c3(c2);
        c3.setVolume(j);
        return 0;
    }

考点：类的使用

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    class Cup
    {
    public:
        int vol;
        Cup(int _v = 0):vol(_v)
        {
            printf("A cup of %d ml is created.\n",vol);
        }
        Cup(const Cup &b):vol(b.vol)
        {
            printf("A cup of %d ml is copied.\n",vol);
        }
        ~Cup()
        {
            printf("A cup of %d ml is erased.\n",vol);
        }
        void setVolume(int _v){vol=_v;}
    };

## Problem B: QQ好友 ##

题面：

![1618425-20190627183354694-1669761320.png][]

![1618425-20190627183424956-704814017.png][]

## main函数： ##

int main()
    {
        int m, n, i, j, id;
        string str;
        Friends friends;
        cin>>m;
        for (i = 0; i < m; i++)
        {
            cin>>str>>n;
            Group group(str);
            for (j = 0; j < n; j++)
            {
                cin>>id>>str;
                QQ qq(id, str);
                group.addQQ(qq);
            }
            friends.addGroup(group);
        }
     
        cin>>m;
        for (i = 0; i < m; i++)
        {
            cin>>str;
            friends.findGroup(str);
        }
     
        cin>>n;
        for (i = 0; i < n; i++)
        {
            cin>>j;
            friends.findQq(j);
        }
        return 0;
    }

考点：类的组合，vector容器的使用，注意，要是对传进来的值进行排序，则该参数不能加const，注意格式控制，有点坑。

## AC代码： ##

#include <algorithm>
    #include <iostream>
    #include <vector>
    #include <cstdio>
    using namespace std;
    class QQ
    {
    public:
        int val;
        string name;
        QQ(int _val,string _name):val(_val),name(_name){}
        bool operator <(const QQ&b)
        {
            return val<b.val;
        }
    };
    class Group
    {
    public:
        string n;
        vector<QQ> QQlist;
        Group(string _n):n(_n){}
        void addQQ(const QQ&b)
        {
            QQlist.push_back(b);
        }
    };
    class Friends
    {
    public:
        vector<Group> Glist;
        void addGroup( Group&b)
        {
            sort(b.QQlist.begin(),b.QQlist.end());
            Glist.push_back(b);
        }
        void findGroup(string name)
        {
            for(int i = 0;i<Glist.size();i++)
            {
                if(Glist[i].n ==name)
                {
                    if(Glist[i].QQlist.size()==0)
                    {
                        cout << "Group " << name << " : empty." << endl;
                    }
                    else
                    {
                        cout << "Group " << name << " :";
                        int len = Glist[i].QQlist.size();
                        for(int j= 0;j<len-1;j++)
                        {
                            cout << " " << Glist[i].QQlist[j].val << " " << Glist[i].QQlist[j].name << ";";
                        }
                        cout << " " << Glist[i].QQlist[len-1].val << " " << Glist[i].QQlist[len-1].name << "." << endl;
                    }
                    return ;
                }
            }
             cout << "Group " << name << " : not existing." << endl;
             return ;
        }
        void findQq(int v)
        {
            //sort(Glist.begin(),Glist.end());
            //bool flag = false;
            int len = Glist.size();
            vector <string> f;
            cout << "QQ " << v << " in :";
            for(int i = 0;i<len;i++)
            {
                int len1 = Glist[i].QQlist.size();
                for(int j = 0;j<len1;j++)
                {
                    if(Glist[i].QQlist[j].val==v)
                    {
                        f.push_back(Glist[i].n);
                    }
                }
            }
            sort(f.begin(),f.end());
            if(f.size()==0)
                cout << " empty." << endl;
            else
            {
                for(int i = 0;i<f.size()-1;i++)
                    cout << " " << f[i] << ";";
                cout << " " << f[f.size()-1] << "." << endl;
            }
            return ;
        }
    };

## Problem C: 玩家PK ##

题面：

![1618425-20190627184335595-325222034.png][]

## main函数： ##

int main()
    {
        int a, b, c, d;
        string name;
        cin>>name>>a>>b>>c>>d;
        Role one(name, a, b, c, d);
        cin>>name>>a>>b>>c>>d;
        Role two(name, a, b, c, d);
        one.combat(two);
        return 0;
    }

考点：小模拟，注意一下当防御最多只能减到0，含有攻击发起者和被攻击者在输出语句的顺序。

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    class Role
    {
    public:
        int hp,ce,de,fa;
        string name;
        Role(string _name,int _hp,int _ce,int _de,int _fa):hp(_hp),ce(_ce),de(_de),fa(_fa),name(_name) {}
        void combat(Role &b)
        {
            bool flag = false;
            if(fa>b.fa)
                flag = true;
            while(hp>0&&b.hp>0)
            {
                if(flag)
                {
                    if(ce<=b.de)
                    {
                        b.hp--;
                    }
                    else
                        b.hp-=(ce-b.de);
                    b.de--;
                    if(b.hp<0)
                        b.hp=0;
                    if(b.de<0)
                        b.de = 0;
                    cout << name << " attacks " << b.name << ":" << b.name;
                    printf(" hp=%d,de=%d\n",b.hp,b.de);
                }
                else
                {
                    if(b.ce<=de)
                    {
                        hp--;
                    }
                    else
                        hp-=(b.ce-de);
                    de--;
                    if(hp<0)
                        hp=0;
                    if(de<0)
                        de = 0;
                    cout << b.name << " attacks " << name << ":" << name;
                    printf(" hp=%d,de=%d\n",hp,de);
                }
                flag = !flag;
            }
            if(hp==0)
                cout << b.name << " wins." << endl;
            else
                cout << name << " wins." << endl;
        }
    };

## Problem D: 时间之差 ##

题面：

![1618425-20190627184950209-1232335081.png][]

## main函数： ##

int main()
    {
        int a, b, c;
        cin>>a>>b>>c;
        Time t1(a, b, c);
        cin>>a>>b>>c;
        Time t2(a, b, c);
        cout<<"Deference is "<<(t2 - t1)<<" seconds."<<endl;
        return 0;
    }

考点：运算符重载，注意要取秒之差的绝对值

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    class Time
    {
    public:
        int h, m, s;
        Time(int _h,int _m,int _s):h(_h),m(_m),s(_s){};
        int operator -(const Time& b)
        {
            int d  = h*3600+m*60+s;
            int d1 = b.h*3600+b.m*60+b.s;
            return abs(d-d1);
        }
    };

## Problem E: 家禽和家畜 ##

题面：  
![1618425-20190627185306683-735111807.png][]

## Main函数： ##

int main()
    {
        Animal *animal;
        char ch;
        while(cin>>ch)
        {
            switch(ch)
            {
            case 'c':
                animal = new Cat();
                break;
            case 'd':
                animal = new Dog();
                break;
            case 'r':
                animal = new Rooster();
                break;
            }
            animal->eat();
            animal->fun();
            delete animal;
        }
        return 0;
    }

考点：虚基类的使用，记得写虚析构函数

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    class Animal
    {
    public:
        Animal(){};
        virtual void eat()=0;
        virtual void fun()=0;
        virtual ~Animal(){};
    };
    class Cat:public Animal
    {
    public:
        Cat():Animal(){}
        void eat()
        {
            printf("Cat eats fishes.\n");
        }
        void fun()
        {
            printf("Cat catches mouses.\n");
        }
        ~Cat(){};
    };
    class Dog:public Animal
    {
    public:
        Dog():Animal(){};
        void eat()
        {
            printf("Dog eats bones.\n");
        }
        void fun()
        {
            printf("Dog can be housekeeping.\n");
        }
        ~Dog(){};
    };
    class Rooster:public Animal
    {
    public:
        Rooster():Animal(){};
        void eat()
        {
            printf("Rooster eats corns.\n");
        }
        void fun()
        {
            printf("Rooster crows.\n");
        }
        ~Rooster(){}
    };

## Problem F: 复数类模板 ##

题面：

![1618425-20190627190050588-1041810383.png][]

## main函数： ##

int main()
    {
        int a, b;
        double c, d;
        cin>>a>>b;
        Complex<int> c1(a, b);
        cout<<setiosflags(ios::fixed)<<setprecision(2)<<c1.getModulus()<<endl;
        cin>>c>>d;
        Complex<double> c2(c, d);
        cout<<setiosflags(ios::fixed)<<setprecision(2)<<c2.getModulus()<<endl;
        return 0;
    }

考点：类模板

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    template<class T>
    class Complex
    {
    public:
        T r,i;
        Complex(T _r,T _i):r(_r),i(_i){}
        double getModulus()
        {
            return sqrt(r*r+i*i);
        }
    };

## Problem G: 相邻的素数 ##

题面：

![1618425-20190627190335045-1509835829.png][]

## main函数： ##

int main()
    {
        int a, b;
        cin>>a>>b;
        Compute compute(a, b);
        compute.showResult();
        return 0;
    }

考点：毒瘤题，注意判断一下当要求输出数目大于满足条件素数的数目时，输出应该停止。

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    bool isPrime(LL x)
    {
        if(x<=1)
            return false;
        if(x==2)
            return true;
        for(LL i= 2; i*i<=x; i++)
        {
            if(x%i==0)
                return false;
        }
        return true;
    }
    class Compute
    {
     
    public:
        int m,n;
        Compute(int _m,int _n):m(_m),n(_n) {};
        void showResult()
        {
            if(n>0)
            {
                LL num = m;
                int k = 0;
                while(k<n)
                {
                    if(isPrime(num))
                    {
                        if(k==0)
                            cout << num ;
                        else
                            cout << " " << num;
                        k++;
                    }
                    num++;
                }
            }
            else if(n<0)
            {
                n = abs(n);
                LL num = m;
                int k = 0;
                while(k<n&&num>1)
                {
                    if(isPrime(num))
                    {
                        if(k==0)
                            cout << num ;
                        else
                            cout << " " << num;
                        k++;
                    }
                    num--;
                }
            }
            cout << endl;
        }
    };

## Problem H: 数组的平滑 ##

题面：  
![1618425-20190627190833938-504352386.png][]

考点：vector的使用，以及负数的四舍五入等于其绝对值四舍五入结果取负。

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    vector<LL> vec,vec1;
    LL check = 0;
    void op()
    {
        for(int i = 1;i<vec.size()-1;i++)
        {
            LL num = vec[i-1]+vec[i+1];
            LL n = abs(num);
            if(num<0)
                vec[i] = -(n+1)/2;
            else
                vec[i] = (n+1)/2;
        }
        while(vec1!=vec)
        {
            vec1 = vec;
             for(int i = 1;i<vec.size()-1;i++)
        {
            LL num = vec[i-1]+vec[i+1];
            LL n = abs(num);
            if(num<0)
                vec[i] = -(n+1)/2;
            else
                vec[i] = (n+1)/2;
        }
     
        }
        return ;
    }
    int main()
    {
        LL num;
        while(cin >> num)
        {
            vec.push_back(num);
            vec1.push_back(num);
        }
        if(vec.size()<=2&&vec.size()>0)
        {
            cout << vec[0] ;
            for(int i = 1;i<vec.size();i++)
                cout << " " << vec[i];
            cout << endl;
        }
        else if(vec.size()==0)
            return 0;
        else
        {
            check = vec[1];
            op();
            cout << vec[0];
            for(int i = 1;i<vec.size();i++)
                cout << " " << vec[i];
            cout << endl;
        }
        return 0;
    }

## Problem I: 需要重载吗？ ##

题面：

![1618425-20190627191149033-1881005947.png][]

## main函数： ##

int main()
    {
        int i;
        char ch;
        cin>>i>>ch;
        Overload t1, t2(i), t3(ch), t4(i, ch);
        return 0;
    }

考点：类的重载。

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    class Overload
    {
    public:
        int a;
        char c;
        Overload(){printf("Default constructor is called to make a = 0, c = '0'.\n"); }
        Overload(int i):a(i){printf("First constructor is called to make a = %d, c = '0'.\n",i); }
        Overload(char _c):c(_c){printf("Second constructor is called to make a = 0, c = '%c'.\n",c); }
        Overload(int i,char _c):a(i),c(_c){printf("Third constructor is called to make a = %d, c = '%c'.\n",a,c); }
    };

## Problem J: 很难的题 ##

题面：  
![1618425-20190627191408290-1603774128.png][]

## main函数： ##

int main()
    {
        int a;
        cin>>a;
        Difficult difficult(a);
        difficult.show();
        return 0;
    }

考点：类的使用

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    class Difficult
    {
    public:
        int num;
        Difficult(int _n):num(_n){};
        void show(){printf("%d\n",abs(num)); }
    };

## Problem K: 数组的归一化 ##

题面：

![1618425-20190627191606278-794738701.png][]

考点：毒瘤题，全为0时应该全输出1。

## AC代码： ##

#include <bits/stdc++.h>
    using namespace std;
    vector<int> vec;
    int main()
    {
        int num = 0;
        double sum = 0;
        while(cin >>num)
        {
            sum+=num;
            vec.push_back(num);
        }
        if(sum==0)
        {
            printf("%.2f",1.00);
            for(int i = 1;i<vec.size();i++)
                printf(" %.2f",1.00);
            printf("\n");
        }
        else
        {
            printf("%.2f",vec[0]/sum);
            for(int i =  1;i<vec.size();i++)
                printf(" %.2f",vec[i]/sum);
            printf("\n");
        }
        return 0;
    }

转载于:https://www.cnblogs.com/baihualiaoluan/p/11099212.html

[1618425-20190627182846838-931266516.png]: /images/20210725/bea27468b8064ce7882ec4478f16c9cf.png
[1618425-20190627183354694-1669761320.png]: /images/20210725/afa398d11c8f412a84e1db0d6a1f8fac.png
[1618425-20190627183424956-704814017.png]: /images/20210725/060fcdd6e6aa450e9bea63f7e5e1e940.png
[1618425-20190627184335595-325222034.png]: /images/20210725/e493f6b6b8bb4739b5e4534d8405d77f.png
[1618425-20190627184950209-1232335081.png]: /images/20210725/39df5de0a8bf41508fe2a17bf28c36a2.png
[1618425-20190627185306683-735111807.png]: /images/20210725/5548358641d1408788b688cccdf58399.png
[1618425-20190627190050588-1041810383.png]: /images/20210725/0cc7f1e02be7475b934c1c5e768a3850.png
[1618425-20190627190335045-1509835829.png]: /images/20210725/e2cebf26ca3c48968b7b9d8dc6d0a811.png
[1618425-20190627190833938-504352386.png]: /images/20210725/51339e5480cb4591a5d1c8e62155fa26.png
[1618425-20190627191149033-1881005947.png]: /images/20210725/fe81008a7b5240a79900eb57cb2b519e.png
[1618425-20190627191408290-1603774128.png]: /images/20210725/dcee332ed08840138677c0e7aa2c2467.png
[1618425-20190627191606278-794738701.png]: /images/20210725/c21302706bcd46bf960d07537fd9e79f.png