//练习一、计算若干个浮点数的平均值，以 -1 作输入结束。
void doubles() {
?????? float a = 0, b = 0;
?????? int index = 0;
?????? cout << "输入若干个浮点数，以 -1 作输入结束" << endl;
?????? while (true)
?????? {
????????????? cin >> b;
????????????? if (b == -1) {
???????????????????? break;
????????????? }
????????????? a += b;
????????????? index++;
?????? }
?????? cout << "平均值为：" << a / index << endl;
}//练习二、写一个程序读入10个整数，输出所有正数之和，负数之和与所有整数之和。用户可一次性输入
//?????????? 这10个整数，而且可以采用任何的顺序。程序不应该要求用户单独输入正数和负数。
void intArr() {
?????? int arr[10];
?????? int a = 0, ua = 0, sum = 0;
?????? for (int i = 0; i < 10; i++)
?????? {
????????????? cin >> arr[i];
????????????? if (arr[i] > 0) {
???????????????????? a += arr[i];
????????????? }
????????????? else if (arr[i] < 0) {
???????????????????? ua += arr[i];
????????????? }
????????????? sum += arr[i];
?????? }
?????? cout << "所有正数之和为" << a << endl;
?????? cout << "所有负数之和为" << ua << endl;
?????? cout << "所有整数之和为" << sum << endl;
}/*练习三、用于计算数字n的平方根的巴比伦算法如下：
a.先猜一个答案guess（可以将n / 2作为第一个答案）
b.计算r = n / guess
c.令guess = (guess + r) / 2
d.如有必要返回第2步重复多次。步骤2和步骤3重复的次数越多，guess就越接近n的平方根。
写一个程序，输入整数作为n的值，重复执行巴比伦算法，直到guess与前一个guess的误差在1 % 范围
内，将答案作为一个double输出。*/
void maths() {
?????? double a = 0, r = 0;
?????? cout << "请输入一个数字" << endl;
?????? cin >> a;
?????? cout << "请猜一个数字" << endl;
?????? double guess = 0;
?????? cin >> guess;
?????? double pre = 0;
?????? while (true) {
????????????? r = a / guess;
????????????? guess = (guess + r) / 2;
????????????? if (abs(pre - guess) <= 0.01) {
???????????????????? cout << "答案是" << endl;
???????????????????? cout << guess << endl;
???????????????????? break;
????????????? }
????????????? pre = guess;
?????? }
}/*
练习四、“23”游戏是一个双人游戏，道具是23根牙签，玩家轮流取1，2或3根牙签。拿到最后一根牙签
的是输家。写程序和计算机玩“23”。总是玩家先走，轮到计算机时，它根据以下规则采取行动：
a.如果剩余牙签多于4根，计算机就取走4-x根，x为玩家上次取走的牙签数
b.如果剩余2-4根牙签，计算机取走足够多的牙签，确保只剩下1根
c.如果剩余1根牙签，计算机只能取走它并且认输
玩家输入要取走的牙签数量时，程序应对输入的有效性进行检查。要确定玩家输入的数在1到3之间，而
且试图取走的不能超过当前剩余的。
程序不允许输出当前剩余牙签数量。
*/
void ends() {
?????? int sum = 23;
?????? int a = 0, b = 0;
?????? while (true) {
????????????? cout << "还剩下" << sum << "根牙签" << endl;
????????????? cout << "请输入取走的牙签数:取1，2或3根牙签" << endl;
????????????? while (true) {
???????????????????? cin >> a;
???????????????????? if (a > 0 && a < 4 && a <= sum) {
??????????????????????????? break;
???????????????????? }
???????????????????? else
???????????????????? {
??????????????????????????? cout << "一次只能拿走1，2或3根牙签" << endl;
???????????????????? }
????????????? }
????????????? sum -= a;
????????????? if (sum == 0) {
???????????????????? cout << "完家输了" << endl;
????????????? }
????????????? //system("cls");
????????????? //计算机；
????????????? if (sum > 4) {
???????????????????? b = 4 - a;
???????????????????? sum -= b;
???????????????????? cout << "计算机取走" << b << "根牙签" << endl;
????????????? }
????????????? else if (sum >= 2 && sum <= 4) {
???????????????????? sum = sum - (sum - 1);
???????????????????? cout << "计算机取走" << sum - 1 << "根牙签" << endl;
????????????? }
????????????? else {
???????????????????? sum = sum - 1;
???????????????????? cout << "还剩下" << sum << "跟牙签" << endl;
???????????????????? cout << "计算机输了" << endl;
???????????????????? break;
????????????? }
????????????? cout << endl;
?????? }
}/*
练习五、任意三角形的面积计算公式为：
面积= 根号下s(s-a)(s-b)(s-c)
其中，a,b,c为三角形的3个边长，s是半周长 s=（a+b+c）/2
编写函数获取5个参数，其中3个是传值参数，指定了3个边长，另两个是传引用参数，他们根据前3个参
数来计算面积和周长。
注意：构成三角形的三边需要两个满足条件：较小的两边之和大于第三边，较大两边之差小于第三边。
要求：需自定义异常类，当不满足条件时抛出该异常，然后提示用户重新输入。
*/
//自定义异常类
class erorr1 {
public:
?????? erorr1(string a) :a(a) {}
?????? ~erorr1() {}
?????? void getMessage() {
????????????? cout << a << endl;
?????? }
private:
?????? string a;
};
void exer1(double a, double b, double c, double& area, double& per) {
?????? if (!(a + b > c && b + c > a && a + c > b)) {
????????????? throw erorr1("三角形较小两边之和要大于第三边");
?????? }
?????? if (!(abs(a - b) < c && abs(a - c) < b && abs(b - c) < a)) {
????????????? throw erorr1("三角形较大两边之差要小于第三边");
?????? }
?????? per = a + b + c;
?????? double s = (a + b + c) / 2;
?????? double arr = s * (s - a) * (s - b) * (s - c);
?????? area = sqrt(arr);
}
void maino1() {
?????? try
?????? {
????????????? //面积
????????????? double area = 0, perimeter = 0, a, b, c;
????????????? //周长：
????????????? cout << "输入三角形三边的长度" << endl;
????????????? cin >> a;
????????????? cin >> b;
????????????? cin >> c;
????????????? exer1(a, b, c, area, perimeter);
????????????? cout << "三角形的面积为：" << area << endl;
????????????? cout << "三角形的周长为：" << perimeter << endl;
?????? }
?????? catch (erorr1 e)
?????? {
????????????? e.getMessage();
?????? }
}//练习六：5X5的螺旋数组
void maino() {
?????? int arr[5][5];
?????? int index = 0;
?????? for (int i = 0; i < 3; i++) {
????????????? for (int h = i;h < 5 - i;h++) {
???????????????????? index++;
???????????????????? arr[i][h] = index;
????????????? }
????????????? for (int s = i + 1;s < 5 - i;s++) {
???????????????????? index++;
???????????????????? arr[s][5 - i - 1] = index;
????????????? }
????????????? for (int d = 5 - i - 2;d >= i;d--) {
???????????????????? index++;
???????????????????? arr[5 - i - 1][d] = index;
????????????? }
????????????? for (int j = 5 - i - 2; j >= i + 1;j--) {
???????????????????? index++;
???????????????????? arr[j][i] = index;
????????????? }
?????? }
?????? for (int i = 0;i < 5;i++) {
????????????? for (int j = 0;j < 5;j++) {
???????????????????? cout << arr[i][j] << "\t";
????????????? }
????????????? cout << endl;
?????? }
}//练习七、某商店经销一种货物。货物购进和卖出时以箱为单位，各箱的重量不一样，因此，商店需要记
//录目前库存的总重量（使用静态成员变量）。现在用C++模拟商店货物购进和卖出的情况。
//提示：使用链表结构和静态成员变量
//需要完善的内容包括：
//Goods类，包括类的构造函数，析构函数，成员函数，静态成员函数以及私有的成员变量。
//两个单独的函数：购进（ purchase ）和售出（ sale ）
//程序要求：
//1.main.cpp中只包含main函数，且不可变更。
//2.程序结构清晰，.h文件中放置声明代码，.cpp文件中放置定义代码
//3.保证程序可以运行且结果正确。
class Goods {
public:
?????? Goods() {
????????????? next = NULL;
?????? }
?????? ~Goods() {}
?????? static int TotalWeight() {
????????????? return weight;
?????? }
?????? //设置重量
?????? void setWeight(int a) {
????????????? weight += a;
????????????? b = a;
?????? }
?????? void delWeight() {
????????????? weight -= this->b;
?????????????
?????? }
?????? void linGoods(Goods* nest) {
????????????? this->next = nest;
?????? }
?????? Goods* getGoods() {
????????????? return next;
?????? }
private:
?????? //计算总重量
?????? static int weight;
?????? //
?????? int b = 0;
?????? Goods* next;
};
int Goods::weight = 0;
//从表尾插入
void purchase(Goods*& front, Goods*& rear, int w) {
?????? Goods* abb = new Goods();
?????? abb->setWeight(w);
?????? abb->linGoods(NULL);
?????? if (!front) {
????????????? front = abb;
????????????? rear = abb;
?????? }
?????? else
?????? {
????????????? rear->linGoods(abb);
????????????? rear = abb;
?????? }
}
//从表头删除
void sale(Goods*& front, Goods*& rear) {
?????? if (front==NULL) {
????????????? return;
?????? }
?????? else if (front == rear) {
????????????? delete front;
?????? }
?????? else
?????? {
????????????? Goods* temp = front;
????????????? front = front->getGoods();
????????????? //temp->delWeight();
????????????? delete temp;
?????? }
}
void maino2()
{
?????? Goods* front = NULL, * rear = NULL;
?????? int? w;? int? choice;
?????? do
?????? {
????????????? cout << "请选择:\n";
????????????? cout << "1：进货, \n2：售出, \n0：退出.\n";
????????????? cin >> choice;
????????????? switch (choice)????????????? // 操作选择
????????????? {
????????????? case 1:?????????????????????????????????????????????? // 键入1，购进1箱
???????????????????? //货物
????????????? { cout << "请输入重量: ";
????????????? cin >> w;
????????????? purchase(front, rear, w);????????? // 从表尾插入1个结点
????????????? break;
????????????? }
????????????? case 2:?????????????? // 键入2，售出1箱货物
????????????? {
???????????????????? sale(front, rear);
???????????????????? break;
????????????? }?????? // 从表头删除1个结点
????????????? case 0:? break;?????????????? // 键入0，结束
????????????? }
????????????? cout << "现在的总重量为:" << Goods::TotalWeight() << endl;//count
?????? } while (choice);
}//第八题：
// 练习八、完成复数类的运算符重载函数，包括：
//基本算术运算（ + 、 - 、 * 、 / 、 = ）
//自增自减运算（前置++、后置++、前置--、后置--）
//流运算符（ << 、 >> ）
//
//复数的概念：
//复数的形式： a + bi 代表一个复数，其中a为实部，b为虚部，i为复数符号（i2 = -1）。
//运算法则：
//加法：(a + bi) + (c + di) = (a + c) + (b + d)i
//减法：(a + bi) - (c + di) = (a - c) + (b - d)i
//乘法：(a + bi)(c + di) = ac + adi + bci + bdi2 = (ac - bd) + (bc + ad)i
//除法：(a + bi) / (c + di) = (ac + bd) / (c2 + d2) + ((bc - ad) / (c2 + d2))i
//clas
s complex {
private:
?????? double real;
?????? double imag;
public:
?????? friend ostream& operator<<(ostream& out, const complex& c2);
?????? friend istream& operator>>(istream& in,complex& c2);
?????? complex() {
????????????? real = 0;
????????????? imag = 0;
?????? }
?????? complex(double real, double imag):real(real),imag(imag) {
?????? }
?????? //成员函数重载
?????? complex operator+(const complex& c2) {
????????????? this->real = this->real + c2.real;
????????????? this->imag = this->real + c2.imag;
????????????? return *this;
?????? }
?????? complex operator-(const complex& c2) {
????????????? this->real = this->real - c2.real;
????????????? this->imag = this->real - c2.imag;
????????????? return *this;
?????? }
?????? complex operator*(const complex& c2) {
????????????? this->real = this->real * c2.real-this->imag*c2.imag;
????????????? this->imag = this->real * c2.imag+this->imag*c2.real;
????????????? return *this;
?????? }
?????? complex operator/(const complex& c2) {
????????????? this->real = (this->real * c2.real + this->imag * c2.imag) / (c2.real * c2.real + c2.imag * c2.imag);
????????????? this->imag = (this->imag * c2.real - this->real * this->imag) / ((c2.real * c2.real + c2.imag * c2.imag));
????????????? return *this;
?????? }
?????? complex operator=(const complex& c2) {
????????????? this->real =? c2.real;
????????????? this->imag = c2.imag;
????????????? return *this;
?????? }
?????? complex& operator++() {
????????????? this->real++;
????????????? this->imag++;
????????????? return *this;
?????? }
?????? complex operator++(int) {
????????????? this->real++;
????????????? this->imag++;
????????????? return *this;
?????? }
?????? complex& operator--() {
????????????? this->real--;
????????????? this->imag--;
????????????? return *this;
?????? }
?????? complex operator--(int) {
????????????? this->real--;
????????????? this->imag--;
????????????? return *this;
?????? }
};
ostream& operator<<(ostream& out, const complex& c2) {
?????? out << "real=" << c2.real << " imag=" << c2.imag << endl;
?????? return out;
}
istream& operator>>(istream& in, complex& c2) {
?????? cout << "请输入实数" << endl;
?????? in >> c2.real;
?????? cout << "请输入虚数" << endl;
?????? in >> c2.imag;
?????? return in;
}//练习十、编写程序统计一个文件中的字符总数、非空白字符总数、字母总数和平均单词长度，并输出到
//屏幕和输出文件output.txt。
//要求：统计的文件为当前工程的main.cpp。输出文件要求和main.cpp在同一级目录，且程序中要求使
//用相对路径表示。
void put() {
?????? ifstream opens("output.txt", ios::in);
?????? if (!opens.is_open())
?????? {
????????????? cout << "文件打开失败" << endl;
????????????? return;
?????? }
?????? int a = 0, b = 0, c = 0, d = 0;
?????? char abb;
?????? while ((abb = opens.get()) != EOF)
?????? {
????????????? a++;
????????????? if (abb != 32) {
???????????????????? b++;
????????????? }
????????????? if ((abb >= 65 && abb <= 90) && (abb >= 97 && abb <= 122)) {
???????????????????? c++;
????????????? }
?????? }
?????? d = a / (a - b);
?????? cout << "字符总数=" << a << endl;
?????? cout << "非空白字符总数=" << b << endl;
?????? cout << "字母总数=" << c << endl;
?????? cout << "平均单词长度=" << d << endl;
}
