因为侵犯慕课网版权而被投诉了可还行……
第五章 编程题
1
参考答案
#include <iostream>
#include <cstring>
using namespace std;
int main(){
char str[80];
int count = 0;
cin >> str;
for (int i = 0;i < strlen(str);i++){
if (str[i] >= 'a' && str[i] <= 'z'){
str[i] += 'A' - 'a';
count++;
}
}
cout << str << endl << count;
}
2
参考答案
#include <iostream>
using namespace std;
int main(){
int array[10];
for (int i = 0;i < 10;i++){
cin >> array[i];
}
int search;
bool thereBe = false;
cin >> search;
for (int i = 0;i < 10;i++){
if (array[i] == search){
thereBe = true;
cout << i + 1 << endl;
}
}
if(!thereBe)
cout << "数组中没有该数";
}
3
参考答案
#include <iostream>
using namespace std;
int main(){
float score[3][5];
for (int i = 0; i < 3;i++){
float sum = 0;
for (int j = 0; j < 3; j++){
cin >> score[i][j];
sum += score[i][j];
}
score[i][3] = sum;
score[i][4] = sum / 3;
}
int rank[3];
if (score[1][3] > score[2][3]){
rank[0] = 1;
rank[1] = 2;
}
else {
rank[0] = 2;
rank[1] = 1;
}
if (score[0][3] > score[ rank[0] ][3]){
rank[1] = rank[0];
rank[0] = 0;
}else if (score[0][3] > score[ rank[1] ][3]){
rank[2] = rank[1];
rank[1] = 0;
}else rank[2] = 0;
for (int i = 0; i < 3; i++){
for (int j = 0; j < 5; j++){
cout << score[ rank[i] ][j] << " ";
}
cout << i + 1 << " " << endl;
}
}
第六章 编程题
1
参考答案
#include <iostream>
using namespace std;
int main(){
char str1[80], str2[40];
int p;
for (int i = 0; i < 80; i++){
str1[i] = 0;
}
cin >> str1 >> str2 >> p;
char *point = &str1[p];
char *pt;
for (int i = 0; str2[i] != 0; i++){
for (pt = str1; *pt; pt++);
for (; pt != point; pt--){
*pt = *(pt - 1);
}
*point = str2[i];
point++;
}
cout << str1;
}
2
参考答案
#include <iostream>
#include <cstring>
using namespace std;
int main(){
char xnum[40];
int tnum = 0, sxtn = 1;
cin >> xnum;
char *point = &xnum[strlen(xnum) - 1];
while (*point != '\0'){
if (*point >= '0' && *point <= '9')
tnum += (*point - '0') * sxtn;
else if (*point >= 'A' && *point <= 'F')
tnum += (*point - 'A' + 10) * sxtn;
else if (*point >= 'a' && *point <= 'f')
tnum += (*point - 'a' + 10) * sxtn;
else if (*point == '-')
tnum = -tnum;
else {
cout << "input data error!" << endl;
return 0;
}
sxtn *= 16;
point--;
}
cout << tnum << endl;
}
3
参考答案
#include <iostream>
using namespace std;
int main(){
char str[40];
int sum[64];
for (int i = 0; i < 64; i++)
sum[i] = 0;
cin >> str;
char *pch = str;
while (*pch != '\0'){
sum[*pch - 'A']++;
pch++;
}
for (int i = 0; i < 64; i++){
if (sum[i])
cout << (char)(i + 'A') << ' ' << sum[i] << endl;
}
}
第七章 编程题
1
参考答案
#include <iostream>
using namespace std;
int exchange(int num){
int exnum = 0;
while (num) {
exnum = exnum * 10 + num % 10;
num /= 10;
}
return exnum;
}
void sort(int *array, const int num){
int brray[num];
int t1, t2;
for (int i = 0; i < num; i++)
brray[i] = exchange(array[i]);
for (int i = 1; i < num; i++){
for (int j = i; brray[j] < brray[j - 1] && j; j--){
t1 = brray[j];
t2 = array[j];
brray[j] = brray[j - 1];
array[j] = array[j - 1];
brray[j - 1] = t1;
array[j - 1] = t2;
}
}
}
int main()
{
int A, B, i, k;
int yuan[50];
cin >> A >> B;
for (k = 0, i = A; i <= B; i++)
yuan[k++] = i;
sort(yuan, k);
cout << yuan[0];
for (i = 1; i < k; i++)
cout << " " << yuan[i];
cout << endl;
return 0;
}
2
参考答案
#include <iostream>
#include <cstring>
using namespace std;
bool detection(char *num, char *merchant){
const int mbc = 21;
const int ucc = 12;
const int tcc = 10;
int Mobile[21] = {134, 135, 136, 137, 138, 139, 147, 150, 151, 152, 157, 158, 159, 172, 178, 182, 183, 184, 187, 188, 198};
int Unicom[12] = {130, 131, 132, 145, 155, 156, 166, 171, 175, 176, 185, 186};
int Telecom[10] = {133, 149, 153, 173, 177, 180, 181, 189, 191, 199};
int code = 0;
for (int i = 0; i < 3; i++){
code = code * 10 + (num[i] - '0');
}
for (int i = 0; i < mbc + ucc + tcc; i++){
if (i < mbc){
if (code == Mobile[i]){
strcpy(merchant, "中国移动");
return true;
}
} else if (i < mbc + ucc){
if (code == Unicom[i - mbc]){
strcpy(merchant, "中国联通");
return true;
}
} else if (code == Telecom[i - mbc - ucc]){
strcpy(merchant, "中国电信");
return true;
}
}
return false;
}
int main(){
char number[12];
char merchant[9];
cin >> number;
if (detection(number, merchant))
cout << merchant << endl;
else
cout << "不确定" << endl;
return 0;
}
3
参考答案
#include <iostream>
using namespace std;
long C(int num){
int result = 0;
if (num == 0 || num == 1)
return 1;
else {
for (int i = 0; i < num; i++){
result += C(i) * C(num - 1 - i);
}
return result;
}
}
void CTL(long *lngCTL,const int num){
lngCTL = new long[num];
for (int i = 0; i < num; i++){
lngCTL[i] = C(i);
cout << lngCTL[i];
if (i < num - 1)cout << ' '; //注意：此处最后一项后不得有空格，否则会提示格式错误
}
cout << endl;
}
int main(){
long *lngCTL;
int n, i;
cin >> n;
CTL(lngCTL, n);
return 0;
}
第九章 编程题
第九章的题目都是看起来很长，其实要写的也没那么多，主函数干脆就直接给你提供了，要写的只有那几个成员函数而已。
1
参考答案
#include <iostream>
using namespace std;
class fraction
{
private:
int above; //分子
int below; //分母
int gcd(int,int); //求两个数的最大公约数
int lcm(int,int); //求两个数的最小公倍数
public:
fraction();
fraction(int ,int);
void reduction(); //约分
fraction add(fraction); //两分数相加
fraction div (fraction); //两分数相除
void setfrac(int ,int); //重置分数
void display(); //显示分数
};
fraction::fraction(){ above = 0; below = 1; }
fraction::fraction(int a, int b = 1):above(a),below(b){}
int fraction::gcd(int a, int b){
if (a % b == 0)
return b;
else return gcd(b, a % b);
} //辗转相除法
int fraction::lcm(int a, int b){
return a * b / gcd(a, b);
}
void fraction::reduction(){
int gcd_of_a_b = gcd(above, below);
above = above / gcd_of_a_b;
below = below / gcd_of_a_b;
}
fraction fraction::add(fraction b){
fraction result;
result.below = below * b.below;
result.above = above * b.below + b.above * below;
result.reduction();
return result;
}
fraction fraction::div(fraction b){
fraction result;
result.above = above * b.below;
result.below = below * b.above;
result.reduction();
return result;
}
void fraction::setfrac(int a, int b){ above = a; below = b; }
void fraction::display()
{
if (below < 0){
above = -above;
below = -below;
}
if (above == 0) cout << 0 << endl;
else if (below == 1) cout << above << endl;
else cout << above << '/' << below << endl;
}
int main()
{
cout << "--------测试构造函数-----------" << endl;
fraction f1(20, -9), f2(-7, -9), f3(5, -9);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3="; f3.display();
cout << "--------测试加法函数-----------" << endl;
f3 = f1.add(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1+f2="; f3.display();
f1.setfrac(-3, -15);
f2.setfrac(-10,12);
f3 = f1.add(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1+f2="; f3.display();
f1.setfrac(12, 45);
f2.setfrac(19, -30);
f3 = f1.add(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1+f2="; f3.display();
f1.setfrac(3, 5);
f2.setfrac(-9, 15);
f3 = f1.add(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1+f2="; f3.display();
cout << "--------测试除法函数-----------" << endl;
f1.setfrac(-24, 7);
f2.setfrac(16, 35);
f3 = f1.div(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1/f2=";f3.display();
f1.setfrac(33, -45);
f2.setfrac(-11, 20);
f3 = f1.div(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1/f2="; f3.display();
f1.setfrac(12, 35);
f2.setfrac(-24, 7);
f3 = f1.div(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1/f2="; f3.display();
f1.setfrac(-11, 20);
f2.setfrac(33, -50);
f3 = f1.div(f2);
cout << "f1="; f1.display();
cout << "f2="; f2.display();
cout << "f3=f1/f2="; f3.display();
cout << "------------测试结束------------" << endl;
return 0;
}
要说有什么要注意的地方的话，大概就是要注意正负号了。我这里是在display是规范了分数的正负号形式，因为这样写最省事。
还有就是在写约分函数时有个要注意的，不能直接写above = above / gcd(above, below0; below = below / gcd(above, below); 因为第一句修改了above，第二句算below时就会用修改后的above算，刚开始我这里就写错了……
gcd函数我用了辗转相除法，lcm函数则直接用了最小公倍数=两数之积/最大公约数
2
参考答案
#include <iostream>
using namespace std;
class timeC{
int hour;
int minute;
int sec;
public:
timeC(int h = 0, int m = 0, int s = 0):hour(h), minute(m), sec(s){}
timeC add(timeC &);
timeC minus(timeC &);
void showtimeC();
};
timeC timeC::add(timeC &t)
{
timeC k;
k.hour = hour + t.hour;
k.minute = minute + t.minute;
k.sec = sec + t.sec;
int i;
i = k.sec / 60;
k.sec = k.sec % 60;
k.minute += i;
i = k.minute / 60;
k.minute = k.minute % 60;
k.hour += i;
k.hour = k.hour % 24;
return k;
}
timeC timeC::minus(timeC &t){
timeC result;
result.hour = hour - t.hour;
result.minute = minute - t.minute;
result.sec = sec - t.sec;
if (result.sec < 0){
result.minute--;
result.sec += 60;
}
if (result.minute < 0){
result.hour--;
result.minute += 60;
}
if (result.hour < 0)
result.hour += 24;
return result;
}
void timeC::showtimeC()
{
cout << hour << " : " << minute << " : " << sec << endl;
}
int main()
{
timeC t1(21, 12, 34), t2(12, 56, 45);
timeC t3(9,23);
t3.showtimeC();
t3 = t3.add(t2);
t3.showtimeC();
timeC t4;
t4.showtimeC();
t4 = t1.minus(t2);
t4.showtimeC();
return 0;
}
这道题要写的也就只有构造函数和minus函数，不算难
3
参考答案
#include <iostream>
#include <cstring>
using namespace std;
class opera
{
char name[30]; //歌剧名称
int ts[4]; // ts[0]：包厢的总数，ts[1]：一等座的总座位数
// ts[2]：二等座的总座位数，ts[3]：三等座的总座位数
int es[4]; // es[0]：空包厢的数量，es[1]：一等座的空闲座位数
// es[2]：二等座的空闲座位数，es[3]：三等座的空闲座位数
int ps[4]; // ps[0]：包厢的票价，ps[1]：一等座的票价
// ps[2]：二等座的票价，，ps[3]：三等座的票价
int income; //总收入
public:
opera(); //默认构造函数，数据成员的初始值请参看输出样例
void set_name(char *); //更改剧名
void set_ps(int []); //更改座位价格
void booking(int,int); //售票，如果余票小于订票数，
//终止该次售票，并输出购票失败提示信息
void refund(int,int); //退票，不收手续费
void print();
~opera(); //析构函数，完成票房统计，内容与格式请参看输出样例
};
opera::opera(){
strcpy(name, "未命名");
ts[0] = es[0] = 20; ts[1] = es[1] = 100;
ts[2] = es[2] = 240; ts[3] = es[3] = 300;
ps[0] = ps[1] = ps[2] = ps[3] = 0;
income = 0;
}
void opera::set_name(char *n){ strcpy(name, n); }
void opera::set_ps(int *p){
for (int i = 0; i < 4; i++)
ps[i] = p[i];
}
void opera::booking(int rank, int num){
if (num > es[rank])
cout << "余票(" << es[rank] << ")不足，购买失败\n";
else {
es[rank] -= num;
income += num * ps[rank];
}
}
void opera::refund(int rank, int num){
es[rank] += num;
income -= ps[rank] * num;
}
void opera::print()
{
char line[20]="------------------";
cout<<line<<"戏曲名："<<name<<line<<endl;
char st[4][20]={ "包厢","一等座","二等座","三等座"};
for(int i=0;i<4;i++)
cout<<st[i]<<"<"<<"票价："<<ps[i]<<" /总数："<<ts[i]<<" /可售："<<es[i]<<">"<<endl;
cout<<line<<"总收入："<<income<<line<<endl<<endl;
}
opera::~opera(){
cout << name << "的总收入：" << income << endl;
cout << name << "的票房如下：\n";
cout << "包厢售出" << ts[0] - es[0] << endl;
cout << "一等座售出" << ts[1] - es[1] << endl;
cout << "二等座售出" << ts[2] - es[2] << endl;
cout << "三等座售出" << ts[3] - es[3] << endl;
}
int main()
{
opera p1;
p1.print();
p1.set_name("二进宫");//void set_name(char *),更改剧名
int p[]={700,380,180,80};
p1.set_ps(p);//void set_ps(int []),更改座位价格
p1.booking(0,9);//购买9张包厢票
p1.booking(1,15);//购买15张一等票
p1.booking(3,20);//购买20张三等票
p1.print();
p1.booking(0,13);//购买13张包厢票
p1.booking(2,14);//购买14张二等票
p1.refund(3,6);//退6张三等票
p1.print();
return 0;
}
4
参考答案
#include <iostream>
#include <cstring>
using namespace std;
class phone{
char number[12]; //11位本机号码
char city[5]; //本机归属地编号
double fee; //本机话费余额
public:
phone(); //默认构造函数，初始值请参看输出样例第一行
void recharge(double m); //给手机充值
void insertcard(const char*,const char*); //设置手机号码和归属地
double getfee(); //返回本机话费
char *getno(); //返回本机号码
char *getcity(); //返回归属地
int call(phone &other,int m); //和手机other通话m分钟，
/* 1.归属地相同，本机支付话费，话费为每分钟0.7元。
2.归属地不同，双方都要支付话费，本机话费为每分钟1.2元，other话费为每分钟0.7元。
3.通话过程中，如果因为某方话费不足，通话中断，函数结束，返回实际通话时长
通话正常完成，函数结束，返回m */
};
phone::phone(){ strcpy(number, "08600000000"); strcpy(city, "***"); fee = 0; }
void phone::insertcard(const char *num, const char *ct){
strcpy(number, num);
strcpy(city, ct);
}
void phone::recharge(double m){ fee += m; }
double phone::getfee(){ return fee; }
char* phone::getno(){ return number; }
char* phone::getcity(){ return city; }
int phone::call(phone &other, int m){
double cost1, cost2 = 0;
int minute = 0;
if (!strcmp(other.city, city)){
cost1 = 0.7;
} else {
cost1 = 1.2;
cost2 = 0.7;
}
while (minute < m){
if (fee > cost1 && other.fee > cost2){
minute++;
fee -= cost1;
other.fee -= cost2;
} else
break;
}
return minute;
}
void display( phone t)
{
cout << "手机号码：" << t.getno()
<< "；归属地：" << t.getcity()
<< "；话费余额：" << t.getfee()
<< endl;
}
int main()
{
phone p1;
display(p1);
p1.insertcard("13458901211", "027");
p1.recharge(40);
phone p2, p3;
p2.insertcard("13652901219", "021");
p2.recharge(30);
p3.insertcard("15651004523", "027");
p3.recharge(50);
char line[] = "*-------------------------------------------------*";
int talk, m;
m = 35;
talk = p1.call(p2, m);
cout << line << endl;
cout << "p1主叫p2......." << endl;
if (talk == m)
cout << "通话时长" << talk << "分钟。通话结束,祝您愉快" << endl;
else
cout << "通话时长" << talk << "分钟。余额不足,请尽快预存话费" << endl;
display(p1); display(p2);
cout << line << endl;
p1.recharge(30);
m = 40;
talk = p3.call(p1, 40);
cout << line << endl;
cout << "p3主叫p1......." << endl;
if (talk == m)
cout << "通话时长" << talk << "分钟。通话结束,祝您愉快........" << endl;
else
cout << "通话时长" << talk << "分钟。电话余额不足.......请尽快预存话费." << endl;
display(p3); display(p1);
cout << line << endl;
return 0;
}
这道题唯一的难点也就在call函数了。不过call函数硬用if-else语句也能凑出来，就是会麻烦点。
由于C++11标准不允许把字符串常量传递给char*，我把insertcard函数的形参改成了const char*，当然一般来说不改也能通过编译，就是一些新的编译器会警告⚠️，看起来不爽。
有时间会写写提高题和第九章
学期也快结束了，提高题也懒得写了
(第九章已更新）
Created by goolwind on 2021/1/6
Edited on 2021/1/11
扫一扫
3274
到【灌水乐园】发言
