2020程设期中解题报告 by Jesse

01:编程填空：第i字节

这题考察基本位运算。

(x >> (8*i)) & ((1<<8) - 1);

02:编程填空：又双叒叕是Fun和Do！

稍微排列组合一下试试即可。

virtual void Fun() {
    cout << "A::Fun()" << n+1 << endl;
}

03:编程填空：变来变去的数

这题主要坑点在于如何输出 6 2 2，只要设置一个条件不减就可以了。

static int counter;
void print_avaliable() {
    cout << counter << " ";
}
virtual void work() {
    if(counter - 4 >= 0) counter -= 4;
    print_avaliable();
}

04:编程填空：惊呆！分数竟然也能这样输入输出和运算！

重载练习题。

friend Fraction operator * (const Fraction &x, const Fraction &y) {
    Fraction ret;
    ret.p = x.p * y.p;
    ret.q = x.q * y.q;
    int d = gcd(ret.p, ret.q);
    ret.p /= d;
    ret.q /= d;
    return ret;
}
friend Fraction operator * (const Fraction &x, const int &y) {
    Fraction ret;
    ret.p = x.p * y;
    ret.q = x.q;
    int d = gcd(ret.p, ret.q);
    ret.p /= d;
    ret.q /= d;
    return ret;
}
friend istream& operator >> (istream &is, Fraction &f) {
    int x,y;
    is >> x >> y;
    f.p = x; f.q = y;
    return is;
}
friend ostream& operator << (ostream &os, Fraction f) {
    int d = gcd(f.p, f.q);
    f.p /= d; f.q /= d;
    os << f.p;
    if (f.q != 1) os << "/" << f.q;
    return os;
}
Fraction(int x) : p(x), q(1) {}
Fraction() {}

05:编程填空：Lambda函数可以这样用！

考察高阶 lambda 表达式的使用。

auto f = [](int d) {
    return [d](int x) {
        return x % d == 0;
    };
};

06:编程填空：序列累加与字符串复制

这题难点在于是把加了之后的结果复制num次，所以每次都需要复制好。

template<typename T>
class MyFunc {
public:
    int num;
    T &result;
    T temp;
    MyFunc(int num, T &result): num(num), result(result) {temp = T();}
    void operator () (T &x) {
        temp += x;
        result = T();
        for (int i = 1;i <= num ;i += 1) result += temp;
    }
};

07:石油交易

以下解法用到了 map 用迭代器访问是按照 key 从小到大排序的。

#include <bits/stdc++.h>
using namespace std;
int main () {
    int n;
    cin >> n;
    string stmp;
    map<int,int> all;
    for (int i = 1; i <= n; i += 1) {
        cin >> stmp;
        if (stmp[0] == 'S') {
            int x,y;
            cin >> x >> y;
            if (all.count(y)) all[y] += x;
            else all[y] = x;
        } else {
            int x;
            cin >> x;
            int cost = 0;
            vector<map<int,int> ::iterator> remove;
            for (map<int,int>::iterator it = all.begin(); it != all.end() && x; ++it) {
                if(it->second <= x) {
                    cost += (it->first) * (it->second);
                    x -= it->second;
                    it->second = 0;
                    remove.push_back(it);
                } else {
                    cost += x * it->first;
                    it->second -= x;
                    x = 0;
                }
            }
            for (auto &it: remove) {
                all.erase(it);
            }
            if (x) cost += 40 * x;
            cout << cost << endl;
        }
    }
    return 0;
}

08:编程填空：a+b+c问题

难点在于需要重载从未重载过的 -> 号，试一下之后发现可以这样操作。

class A {
public:
    int x;
    A (int x) : x(x) {}
    int get_value () { return x; }
    A* operator -> () { return this; }
};
A operator + (const A x, const A *y) {
    return A(x.x + y->x);
}
class B : public A {
public:
    B (int x) : A(x) {}
};
class C : public A {
public:
    C (int x) : A(x) {}
};