1、 i + j; /* 输出结果 */return 0;改正:int main()int i; / 给i赋值i + j = i;您输入一个数字是 i endl;5您输入一个数字是52-20 打印ASCII码为32127的字符。for (int i = 32; i128; i+) (char) i;!#$%G()*+,./0123456789:?ABCDEFGHIJKLMNOP_QRSTUVWXYZabcdefghijklmnopqrstuvwxyzs2-21 运行下面的程序,观察其输出,与你的设想是否相同?unsigned int x;unsigned int y = 100;unsigned
2、int z = 50;x= y - z;Difference is: x;x = z - y;nNow difference is: x endl; 50Now difference is: 4294967246注意,第二行的输出并非 -50,注意x、y、z的数据类型。2-22 运行下面的程序,观察其输出,体会i+与+i的差别。int myAge = 39; / initialize two integersint yourAge = 39;I am: myAge years old.nYou are: yourAge years oldnmyAge+; / postfix increment
3、+yourAge; / prefix incrementOne year passes.nAnother year passesn myAge+ +yourAge char flag;while(1)(Yes or No): flag;if ( toupper(flag) = Y)break;Nxlqn或:y2-29 编写一个完整的程序,运行时向用户提问你考试考了多少分?(0100),接收输入后判断其等级,显示出来。规则如下:int i,score;你考试考了多少分?(0100): score;if (score100 | scoreint i,j,k,flag;i = 2;while(i =
4、 100)flag = 1;k = sqrt(i);j = 2;while (j = k)if(i%j = 0)flag = 0;j+;if (flag)是质数.i+;使用dowhile循环语句:dowhile (j = k);while(i = 100);使用for循环语句:for(i = 2;= 100;for (j = 2; j m)你猜的值太小了!else if (n 你猜的值太大了!你猜对了!/使用dowhile语句while(n != m);5025101518第三章 函数3-2 观察下面程序的运行输出,与你设想的有何不同?仔细体会引用的用法。int intOne;int &rSo
5、meRef = intOne;intOne = 5;intOne:tt intOne rSomeRef:t rSomeRef int intTwo = 8;rSomeRef = intTwo; / not what you think!nintOne:intTwo: intTwo -1)Answer: answer;Error, cant divide by zero!Enter two numbers.Number one:82 43-8 编写函数把华氏温度转换为摄氏温度,公式为:C = (F - 32) * 5/9; 在主程序中提示用户输入一个华氏温度,转化后输出相应的摄氏温度。部分实验三
6、3-10 编写函数求两个整数的最大公约数和最小公倍数。int fn1(int i,int j); /求最大公约数的函数int i,j,x,y;请输入一个正整数: i ;请输入另一个正整数: j ;x = fn1(i,j);y = i * j / x;和的最大公约数是:的最小公倍数是: y int fn1(int i, int j)int temp;if (i j)temp = i;i = j;j = i;while(j != 0)temp = i % j;j = temp;return i;12072120和72的最大公约数是:24120和72的最小公倍数是:3603-12 在主程序中提示输入
7、整数n,编写函数用递归的方法求1 + 2 + + n的值。int fn1(int i);从1累加到i 的和为: fn1(i) 2;fib(1) = fib(2) = 1;观察递归调用的过程。3-15 用递归的方法编写函数求n阶勒让德多项式的值,在主程序中实现输入、输出;float p(int n, int x);int n,x;请输入正整数n: n;请输入正整数x:n = x = P() = p(n,x) float p(int n, int x)if (n = 0)else if (n = 1)return x;return (2*n-1)*x*p(n-1,x) - (n-1)*p(n-2,
8、x) /n ;1n = 1x = 2P1(2) = 234n = 3x = 4P3(4) = 154 第 四 章 类4-9 设计并测试一个名为Rectangle的矩形类,其属性为矩形的左下角与右上角两个点的坐标,能计算矩形的面积。class Rectanglepublic:Rectangle (int top, int left, int bottom, int right);Rectangle () int GetTop() const return itsTop; int GetLeft() const return itsLeft;int GetBottom() const return
9、 itsBottom;int GetRight() const return itsRight;void SetTop(int top) itsTop = top;void SetLeft (int left) itsLeft = left;void SetBottom (int bottom) itsBottom = bottom;void SetRight (int right) itsRight = right;int GetArea() const;private:int itsTop;int itsLeft;int itsBottom;int itsRight;Rectangle:R
10、ectangle(int top, int left, int bottom, int right)itsTop = top;itsLeft = left;itsBottom = bottom;itsRight = right;int Rectangle:GetArea() constint Width = itsRight-itsLeft;int Height = itsTop - itsBottom;return (Width * Height);Rectangle MyRectangle (100, 20, 50, 80 );int Area = MyRectangle.GetArea(
11、);Area: Area n 3000Upper Left X Coordinate: 204-11 定义一个矩形类,有长、宽两个属性,有成员函数计算矩形的面积 Rectangle(float len, float width)Length = len;Width = width;Rectangle();float GetArea() return Length * Width;float GetLength() return Length;float GetWidth() return Width;float Length;float Width;float length, width;请输入矩形的长度: length;请输入矩形的宽度: width;Rectangle r(length, width);长为 length 宽为 width 的矩形的面积为: r.GetArea () 长为5
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