C++程序设计（第3版）谭浩强 第3章 习题

1.怎样区分表达式和表达式语句？C语言为什么要设表达式语句？什么时候用表达式？什么时候用表达式语句？

【解】略。

2.设圆半径r=1.5,圆柱高h=3,求圆周长、圆面积、圆球表面积、圆球体积、圆柱体积。用cin输入数据,输出计算结果,输出时要求有文字说明,取小数点后两位数字。请编程序。

【解】

#include 
#include 
using namespace std;
int main()
{
	float h, r, l, s, sq, vq, vz;
	const float pi = 3.1415926;
	cout << "please enter r,h:";	//要求输入圆半径r和圆柱高h 
	cin >> r >> h;	//输入r和h
	l = 2 * pi*r;	//计算圆周长l
	s = r * r*pi;	//计算圆面积s
	sq = 4 * pi*r*r;	//计算圆球表面积sq
	vq = 4.0 / 3.0*pi*r*r*r;	//计算圆球体积vq
	vz = pi * r*r*h;	//计算圆柱体积vz
	cout << setiosflags(ios::fixed) << setiosflags(ios::right) << setprecision(2);
	//设置输出格式
	cout << " l= " << setw(10) << l << endl;	//输出圆周长l, 指定字段宽度为10
	cout << " s= " << setw(10) << s << endl;	//输出圆面积s
	cout << "sq=" << setw(10) << sq << endl;	//输出圆球表面积sq
	cout << " vq=" << setw(10) << vq << endl;	//输出圆球体积vq
	cout << " vz=" << setw(10) << vz << endl;	//输出圆柱体积vz
	return 0;
}

运行结果：

//please enter r, h: 1.5 3↙
//l = 9.42
//s = 7.07
//sq = 28.27
//vq = 14.14
//vz = 21.20

3.输入一个华氏温度,要求输出摄氏温度。公式为c=5/9(F-32)，输出要有文字说明,取两位小数。

【解】

#include 
using namespace std;
int main()
{
	float c, f;
	cout << "请输入一个华氏温度: ";
	cin >> f;
	c = (5.0 / 9.0) * (f - 32); //注意5和9要用实型表示, 否则5/9值为0
	cout << "摄氏温度为: " << c << endl;
	return 0;
}

运行结果：

//请输入一个华氏温度: 56↙
//摄氏温度为 : 13.33

4.编程序,用getchar函数读入两个字符给c1,c2,然后分别用putchar函数和cout语句输出这两个字符。并思考以下问题:

(1) 变量c1,c2应定义为字符型还是整型？抑或二者皆可？

(2) 要求输出c1和c2值的ASCII码,应如何处理？

【解】

#include 
using namespace std;
int main()
{
	char c1, c2;
	cout << "请输入两个字符c1,c2:";
	c1 = getchar();	//将输入的第1个字符赋给c1
	c2 = getchar();	//将输入的第2个字符赋给c2
	cout << "用putchar函数输出结果为:";
	putchar(c1);
	putchar(c2);
	cout << endl;
	cout << "用cout语句输出结果为:";
	cout << c1 << c2 << endl;
	return 0;
}

运行结果：

//请输入两个字符c1, c2:ab↙
//用putchar函数输出结果为 : ab
//用cout语句输出结果为 : ab

5.整型变量与字符变量是否在任何情况下都可以互相代替？如

char c1, c2;与int c1, c2;

【解】略。

6.什么是算术运算？什么是关系运算？什么是逻辑运算？

【解】略。

7.C++如何表示“真”和“假”？系统如何判断一个量的“真”和“假”？

【解】略。

8.写出下面各逻辑表达式的值。设a=3,b=4,c=5。

【解】略。

9.有3个整数a,b,c,由键盘输入,输出其中最大的数。

【解】

//（1）方法一：用if语句。
#include 
using namespace std;
int main()
{
	int a, b, c;
	cout << "please enter three integer numbers:";
	cin >> a >> b >> c;
	if (a < b)
		if (b < c)
			cout << "max=" << c;
		else
			cout << "max=" << b;
	else if (a < c)
		cout << "max=" << c;
	else
		cout << "max=" << a;
	cout << endl;
	return 0;
}

//（2）方法二：使用条件表达式，可以使程序更加简明、清晰。
#include 
using namespace std;
int main()
{
	int a, b, c, temp, max;
	cout << "please enter three integer numbers:";
	cin >> a >> b >> c;
	temp = (a > b) ? a : b;	//将a和b中的大者存入temp中
	max = (temp > c) ? temp : c;	//将a和b中的大者与c比较, 最大者存入max中
	cout << "max=" << max << endl;
	return 0;
}

运行结果：

//please enter three integer numbers :: 121 –34 40↙
//max = 121

10.有一函数:

编写一程序，输入x，输出y的值。

【解】

#include < iostream >
using namespace std;
int main()
{
	int x, y;
	cout << "enter x:";
	cin >> x;
	if (x < 1)
	{
		y = x;
		cout << "x=" << x << ", y=x=" << y;
	}
	else if (x < 10)	// 1≤x＜10 
	{
		y = 2 * x - 1;
		cout << "x=" << x << ", y=2*x–1=" << y;
	}
	else	// x≥10
	{
		y = 3 * x - 11;
		cout << "x=" << x << ", y=3*x–11=" << y;
	}
	cout << endl;
	return 0;
}

运行结果：

//① enter x : 4↙
//x = 4, y = 2 * x–1 = 7
//② enter x : –1↙
//x = –1, y = x  = –1
//③ enter x : 20↙
//x = 20, y = 3 * x–11 = 49

11.给出一个百分制的成绩,要求输出成绩等级'A','B','C','D','E'。90分以上为'A',80～89分为'B',70～79分为'C',60～69分为'D',60分以下为'E'。

【解】

#include 
using namespace std;
int main()
{
	float score;
	char grade;
	cout << "please enter score of student:";
	cin >> score;
	while (score > 100 || score < 0)
	{
		cout << "data error,enter data again.";
		cin >> score;
	}
	switch (int(score / 10))
	{
	case 10:
	case 9: grade = 'A'; break;
	case 8: grade = 'B'; break;
	case 7: grade = 'C'; break;
	case 6: grade = 'D'; break;
	default:grade = 'E';
	}
	cout << "score is " << score << ", grade is " << grade << endl;
	return 0;
}

运行结果：

//① please enter score of student : 90.5↙
//score is 90.5, grade is A
//② please enter score of student : 45.5↙
//score is 59.5, grade is E

12.给一个不多于5位的正整数,要求: ①求出它是几位数;②分别打印出每一位数字;③按逆序打印出各位数字,例如原数为321,应输出123。

【解】

#include 
using namespace std;
int main()
{
	long int num;
	int indiv, ten, hundred, thousand, ten_thousand, place;
	//分别代表个位、十位、百位、千位、万位和位数
	cout << "enter an integer(0~99999):";
	cin >> num;
	if (num > 9999)
		place = 5;
	else  if (num > 999)
		place = 4;
	else  if (num > 99)
		place = 3;
	else  if (num > 9)
		place = 2;
	else place = 1;
	cout << "place=" << place << endl;
	//计算各位数字
	ten_thousand = num / 10000;
	thousand = (int)(num - ten_thousand * 10000) / 1000;
	hundred = (int)(num - ten_thousand * 10000 - thousand * 1000) / 100;
	ten = (int)(num - ten_thousand * 10000 - thousand * 1000 - hundred * 100) / 10;
	indiv = (int)(num - ten_thousand * 10000 - thousand * 1000 - hundred * 100 - ten * 10);
	cout << "original order:";
	switch (place)
	{
	case 5:	cout << ten_thousand << "," << thousand << "," << hundred << "," << ten << "," << indiv << endl;
		cout << "reverse order:";
		cout << indiv << ten << hundred << thousand << ten_thousand << endl;
		break;
	case 4:	cout << thousand << "," << hundred << "," << ten << "," << indiv << endl;
		cout << "reverse order:";
		cout << indiv << ten << hundred << thousand << endl;
		break;
	case 3:	cout << hundred << "," << ten << "," << indiv << endl;
		cout << "reverse order:";
		cout << indiv << ten << hundred << endl;
		break;
	case 2:	cout << ten << "," << indiv << endl;
		cout << "reverse order:";
		cout << indiv << ten << endl;
		break;
	case 1:	cout << indiv << endl;
		cout << "vreverse order : ";
		cout << indiv << endl;
		break;
	}
	return 0;
}

运行结果：

//enter an integer(0~99999) :98765↙
//place = 5
//original order : 9, 8, 7, 6, 5
//reverse order : 56789

13.企业发放的奖金来自利润提成。利润i低于或等于10万元的,可提成10%为奖金;利润i高于10万元,低于或等于20万元(100000

要求：（1）用if语句编写程序；

（2）用switch语句编写程序。

【解】

//（1）用if语句
#include 
using namespace std;
int main()
{
	long i;	//i为利润
	float  bonus, bon1, bon2, bon4, bon6, bon10;
	bon1 = 100000 * 0.1;	//利润为10万元时的奖金
	bon2 = bon1 + 100000 * 0.075;	//利润为20万元时的奖金
	bon4 = bon2 + 200000 * 0.05;	//利润为40万元时的奖金
	bon6 = bon4 + 200000 * 0.03;	//利润为60万元时的奖金
	bon10 = bon6 + 400000 * 0.015;	//利润为100万元时的奖金
	cout << "enter i:";
	cin >> i;
	if (i <= 100000)
		bonus = i * 0.1;	//利润在10万元以内按10%提成奖金
	else if (i <= 200000)
		bonus = bon1 + (i - 100000)*0.075;	//利润在10万元至20万元时的奖金
	else if (i <= 400000)
		bonus = bon2 + (i - 200000)*0.05;	//利润在20万元至40万元时的奖金
	else if (i <= 600000)
		bonus = bon4 + (i - 400000)*0.03;	//利润在40万元至60万元时的奖金
	else if (i <= 1000000)
		bonus = bon6 + (i - 600000)*0.015;	//利润在60万元至100万元时的奖金
	else
		bonus = bon10 + (i - 1000000)*0.01;	//利润在100万元以上时的奖金
	cout << "bonus=" << bonus << endl;
	return 0;
}


//（2）用switch语句
#include 
using namespace std;
int main()
{
	long i;
	float bonus, bon1, bon2, bon4, bon6, bon10;
	int c;
	bon1 = 100000 * 0.1;
	bon2 = bon1 + 100000 * 0.075;
	bon4 = bon2 + 200000 * 0.05;
	bon6 = bon4 + 200000 * 0.03;
	bon10 = bon6 + 400000 * 0.015;
	cout << "enter i:";
	cin >> i;
	c = i / 100000;
	if (c > 10)  c = 10;
	switch (c)
	{
	case 0: bonus = i * 0.1; break;
	case 1: bonus = bon1 + (i - 100000)*0.075; break;
	case 2:
	case 3: bonus = bon2 + (i - 200000)*0.05; break;
	case 4:
	case 5: bonus = bon4 + (i - 400000)*0.03; break;
	case 6:
	case 7:
	case 8:
	case 9: bonus = bon6 + (i - 600000)*0.015; break;
	case 10: bonus = bon10 + (i - 1000000)*0.01;
	}
	cout << "bonus=" << bonus << endl;
	return 0;
}

运行结果：

//enter i : 234000↙
//bonus = 19200

14.输入4个整数,要求按由小到大的顺序输出。

【解】

#include 
using namespace std;
int main()
{
	int t, a, b, c, d;
	cout << "enter four numbers:";
	cin >> a >> b >> c >> d;
	cout << "a=" << a << ", b=" << b << ", c=" << c << ",d=" << d << endl;
	if (a > b)
	{
		t = a; a = b; b = t;
	}
	if (a > c)
	{
		t = a; a = c; c = t;
	}
	if (a > d)
	{
		t = a; a = d; d = t;
	}
	if (b > c)
	{
		t = b; b = c; c = t;
	}
	if (b > d)
	{
		t = b; b = d; d = t;
	}
	if (c > d)
	{
		t = c; c = d; d = t;
	}
	cout << "the sorted sequence:" << endl;
	cout << a << ", " << b << ", " << c << ", " << d << endl;
	return 0;
}

运行结果：

//enter four numbers : 6 8 1 4↙
//a = 6, b = 8, c = 1, d = 4
//the sorted sequence :
//1, 4, 6, 8

15.输入两个正整数m和n,求其最大公约数和最小公倍数。

【解】

#include 
using namespace std;
int main()
{
	int p, r, n, m, temp;
	cout << "please enter two positive integer numbers n,m:";
	cin >> n >> m;
	if (n < m)
	{
		temp = n;
		n = m;
		m = temp;	//把较大数放在n中, 较小数放在m中
	}
	p = n * m;	//先将n和m的乘积保存在p中, 以便求最小公倍数时使用
	while (m != 0)	//求n和m的最大公约数
	{
		r = n % m;
		n = m;
		m = r;
	}
	cout << "HCF=" << n << endl;
	cout << "LCD=" << p / n << endl;	// p是原来两个整数的乘积
	return 0;
}

运行结果：

//please enter two positive integer numbers n, m:15 9 ↙
//HCF = 3
//LCD = 45

16.输入一行字符,分别统计出其中英文字母、空格、数字和其他字符的个数。

【解】

#include 
using namespace std;
int main()
{
	char c;
	int letters = 0, space = 0, digit = 0, other = 0;
	cout << "enter one line:" << endl;
	while ((c = getchar()) != '\n')
	{
		if (c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z')
			letters++;
		else if (c == ' ')
			space++;
		else if (c >= '0' && c <= '9')
			digit++;
		else
			other++;
	}
	cout << " letter:" << letters << ",space:" << space << ",digit:" << digit << ",other:" << other << endl;
	return 0;
}

运行结果：

//enter one line :
//My teacher's address is "#123 Beijing Road,Shanghai".
//letter : 38, space : 6, digit : 3, other : 6

17.求Sn=a+aa+aaa+…+aa…an个a之值,其中a是一个数字。例如: 2+22+222+2222+22222(此时n=5),n由键盘输入。

【解】

#include 
using namespace std;
int main()
{
	int  a, n, i = 1, sn = 0, tn = 0;
	cout << "a,n=:";
	cin >> a >> n;
	while (i <= n)
	{
		tn = tn + a;	//赋值后的tn为i个a组成数的值
		sn = sn + tn;	//赋值后的sn为多项式前i项之和
		a = a * 10;
		++i;
	}
	cout << "a+aa+aaa+…=" << sn << endl;
	return 0;
}

运行结果：

//a, n: 2 5↙
//a + aa + aaa + … = 24690

18.求∑20n=1n! (即求1!+2!+3!+4!+…+20!)。

【解】

#include 
using namespace std;
int main()
{
	float s = 0, t = 1;
	int n;
	for (n = 1; n <= 20; n++)
	{
		t = t * n;	//求n! 
		s = s + t;	//将各项累加 
	}
	cout << "1!+2!+…+20!=" << s << endl;
	return 0;
}

运行结果：

//1!+ 2!+ … + 20 != 2.56133e+018

19.输出所有的“水仙花数”,所谓“水仙花数”是指一个3位数,其各位数字立方和等于该数本身。例如,153是一水仙花数,因为153=13+53+33。

【解】

#include 
using namespace std;
int main()
{
	int i, j, k, n;
	cout << "narcissus numbers are:" << endl;
	for (n = 100; n < 1000; n++)
	{
		i = n / 100;
		j = n / 10 - i * 10;
		k = n % 10;
		if (n == i * i * i + j * j * j + k * k * k)
			cout << n << " ";
	}
	cout << endl;
	return 0;
}

运行结果：

//narcissus numbers are : 153 370 371 407

20.一个数如果恰好等于它的因子之和,这个数就称为“完数”。例如,6的因子为1,2,3,而6=1+2+3,因此6是“完数”。编程序找出1000之内的所有完数,并按下面格式输出其因子:

6,its factors are 1,2,3

【解】

//方法一：
#include < iostream >
using namespace std;
int main()
{
	const int m = 1000;	//定义寻找范围 
	int k1, k2, k3, k4, k5, k6, k7, k8, k9, k10;
	int i, a, n, s;
	for (a = 2; a <= m; a++)	// a是2～1000的整数, 检查它是否为完数
	{
		n = 0;	//n用来累计a的因子的个数 
		s = a;	//s用来存放尚未求出的因子之和, 开始时等于a 
		for (i = 1; i < a; i++)	//检查i是否为a的因子 
			if (a%i == 0)	//如果i是a的因子 
			{
				n++;	//n加1，表示新找到一个因子 
				s = s - i;	//s减去已找到的因子, s的新值是尚未求出的因子之和 
				switch (n)	//将找到的因子赋给k1,…, k10
				{
				case 1:
					k1 = i;  break;	//找出的第1个因子赋给k1 
				case 2:
					k2 = i;  break;	//找出的第2个因子赋给k2 
				case 3:
					k3 = i;  break;	//找出的第3个因子赋给k3 
				case 4:
					k4 = i;  break;	//找出的第4个因子赋给k4 
				case 5:
					k5 = i;  break;	//找出的第5个因子赋给k5 
				case 6:
					k6 = i;  break;	//找出的第6个因子赋给k6 
				case 7:
					k7 = i;  break;	//找出的第7个因子赋给k7 
				case 8:
					k8 = i;  break;	//找出的第8个因子赋给k8 
				case 9:
					k9 = i;  break;	//找出的第9个因子赋给k9 
				case 10:
					k10 = i;  break;	//找出的第10个因子赋给k10 
				}
			}
		if (s == 0)	//s=0表示全部因子都已找到了
		{
			cout << a << " is a 完数" << endl;
			cout << "its factors are:";
			if (n > 1)  cout << k1 << "," << k2;	//n>1表示a至少有2个因子 
			if (n > 2)  cout << "," << k3;	//n>2表示至少有3个因子, 故应再输出一个因子 
			if (n > 3)  cout << "," << k4;	//n>3表示至少有4个因子, 故应再输出一个因子 
			if (n > 4)  cout << "," << k5;	//以下类似 
			if (n > 5)  cout << "," << k6;
			if (n > 6)  cout << "," << k7;
			if (n > 7)  cout << "," << k8;
			if (n > 8)  cout << "," << k9;
			if (n > 9)  cout << "," << k10;
			cout << endl << endl;
		}
	}
	return 0;
}


//方法二：
#include 
using namespace std;
int main()
{
	int m, s, i;
	for (m = 2; m < 1000; m++)
	{
		s = 0;
		for (i = 1; i < m; i++)
			if ((m % i) == 0) s = s + i;
		if (s == m)
		{
			cout << m << "is a完数" << endl;
			cout << "its factors are:";
			for (i = 1; i < m; i++)
				if (m % i == 0)  cout << i << " ";
			cout << endl;
		}
	}
	return 0;
}


//方法三：此题用数组方法更为简单。
#include 
using namespace std;
int main()
{
	int k[11];
	int i, a, n, s;
	for (a = 2; a <= 1000; a++)
	{
		n = 0;
		s = a;
		for (i = 1; i < a; i++)
			if ((a % i) == 0)
			{
				n++;
				s = s - i;
				k[n] = i;  //将找到的因子赋给k[1],…, k[10] 
			}
		if (s == 0)
		{
			cout << a << " is a 完数" << endl;
			cout << "its factors are: ";
			for (i = 1; i < n; i++)
				cout << k[i] << " ";
			cout << k[n] << endl;
		}
	}
	return 0;
}

运行结果：
//6 is a 完数
//its factors are : 1, 2, 3

//28 is a 完数
//its factors are : 1, 2, 4, 7, 14

//496 is a 完数
//its factors are : 1, 2, 4, 8, 16, 31, 62, 124, 248

21.有一分数序列

2/1,3/2,5/3,8/5,13/8,21/13,...

求出这个数列的前20项之和。

【解】

#include 
using namespace std;
int main()
{
	int i, t, n = 20;
	double a = 2, b = 1, s = 0;
	for (i = 1; i <= n; i++)
	{
		s = s + a / b;
		t = a;
		a = a + b;	//将前一项分子与分母之和作为下一项的分子 
		b = t;	//将前一项的分子作为下一项的分母 
	}
	cout << "sum=" << s << endl;
	return 0;
}

运行结果：

//sum = 32.6603

22.猴子吃桃问题。猴子第1天摘下若干个桃子,当即吃了一半,还不过瘾,又多吃了一个。第2天早上又将剩下的桃子吃掉一半,又多吃了一个。以后每天早上都吃了前一天剩下的一半另加一个。到第10天早上想再吃时,就只剩一个桃子了。求第1天共摘了多少个桃子。

【解】

#include 
using namespace std;
int main()
{
	int day, x1, x2;
	day = 9;
	x2 = 1;
	while (day > 0)
	{
		x1 = (x2 + 1) * 2; //第1天的桃子数是第2天桃子数加1后的2倍 
		x2 = x1;
		day--;
	}
	cout << "total=" << x1 << endl;
	return 0;
}

运行结果：

//total = 1534

23.用迭代法求x=a。求平方根的迭代公式为xn+1=12xn+axn要求前后两次求出的x的差的绝对值小于10-5。

【解】

#include 
#include 
using namespace std;
int main()
{
	float a, x0, x1;
	cout << "enter a positive number:";
	cin >> a;            // 输入a的值 
	x0 = a / 2;
	x1 = (x0 + a / x0) / 2;
	do
	{
		x0 = x1;
		x1 = (x0 + a / x0) / 2;
	} while (fabs(x0 - x1) >= 1e-5);
	cout << "The square root of " << a << " is " << x1 << endl;
	return 0;
}

运行结果：

//enter a positive number : 2↙
//The square root of  2 is 1.41421

24.输出以下图案:

【解】

#include 
using namespace std;
int main()
{
	int i, k;
	for (i = 0; i <= 3; i++)	//输出上面4行*号 
	{
		for (k = 0; k <= 2 * i; k++)
			cout << "*";	//输出*号 
		cout << endl;	//输出一行*号后换行 
	}
	for (i = 0; i <= 2; i++)	//输出下面3行*号 
	{
		for (k = 0; k <= 4 - 2 * i; k++)
			cout << "*";	//输出*号 
		cout << endl;	//输出一行*号后换行
	}
	return 0;
}

运行结果：

//*
//***
//*****
//*******
//*****
//***
//*

25.两个乒乓球队进行比赛,各出3人。甲队为A,B,C 3人,乙队为X,Y,Z 3人。已抽签决定比赛名单。有人向队员打听比赛的名单,A说他不和X比,C说他不和X,Z比,请编程序找出3对赛手的名单。

【解】

#include 
using namespace std;
int main()
{
	char i, j, k; // i是a的对手; j是b的对手; k是c的对手
	for (i = 'X'; i <= 'Z'; i++)
		for (j = 'X'; j <= 'Z'; j++)
			if (i != j)
				for (k = 'X'; k <= 'Z'; k++)
					if (i != k && j != k)
						if (i != 'X' && k != 'X' && k != 'Z')
							cout << "A– –" << i << "  B– –" << j << "  C– –" << k << endl;
	return 0;
}

运行结果：

//A– – –Z    B– – –X    C– – –Y