第五章習(xí)題答案

1、第5章 集合與結(jié)構(gòu)5.1 選擇題1語(yǔ)句cout<<(1&2)<<", "<<(1&&2)<<endl; 的輸出結(jié)果是（ B ）。（A）0, 0（B）0, 1（C）1, 0（C）1, 12. 語(yǔ)句cout<<(1|2)<<", "<<(1|2)<<endl; 的輸出結(jié)果是（ D ）。（A）0, 0（B）1, 1（C）2, 0（D）3, 13. 有以下說(shuō)明語(yǔ)句：struct point int x; int y; p;則正確的賦值語(yǔ)句是（

2、C ）。（A）point.x = 1; point.y = 2;（B）point= 1, 2 ; （C）p.x = 1; p.y = 2;（D）p = 1, 2 ;4已知有職工情況結(jié)構(gòu)變量emp定義為：struct Date int year;int month; int day;strnct Employee char name20;long code;Date birth;Employee emp;下列對(duì)emp的birth正確賦值方法是（ D ）。（A）year=1980; month=5; day=1;（B）birth.year=1980; birth.month=5; birth.da

3、y=1;（C）emp.year=1980; emp.month=5; emp.day=1;（D）emp.birth.year=1980; emp.birth.month=5; emp.birth.day=1;5有以下說(shuō)明語(yǔ)句： struct Student int num;double score;Student stu3=1001,80, 1002,75, 1003,91, *p=stu;則下面引用形式錯(cuò)誤的是（ B ）。（A）p->num（B）(p+).num（C）(p+)->num（D）(*p).num6有以下說(shuō)明語(yǔ)句： struct Worker int no;char n

4、ame20;Worker w, *p &w;則下列錯(cuò)誤的引用是（ D ）。（A）w.no（B）p->no（C）(*p).no（D）*p.no7s1和s2是兩個(gè)結(jié)構(gòu)類型變量，若要使賦值s1=s2合法，則它們的說(shuō)明應(yīng)該是（ C ）。 （A）s1只能接收相同類型的數(shù)據(jù)成員 （B）結(jié)構(gòu)中的成員相同 （C）同一結(jié)構(gòu)類型的變量 （D）存儲(chǔ)字節(jié)長(zhǎng)度一樣的變量5.2 閱讀下列程序，寫出運(yùn)行結(jié)果。1 #include <iostream>using namespace std;struct Data int n; double score; ;int main() Data a3 =

5、1001,87,1002,72,1003,90 , *p = a; cout << (p+)->n << endl; cout << (p+)->n << endl; cout << p->n+ << endl; cout << (*p).n+ << endl;【解答】 10011002100310042 #include <iostream>using namespace std;struct Employee char name 20 ; char sex; ;vo

6、id fun( Employee *p )if( (*p).sex = 'm' ) cout << (*p).name << endl;int main() Employee emp5 = "Liming", 'm', "Wangxiaoping", 'f', "Luwei", 'm' ; int i; for( i=0; i<3; i+ ) fun( emp+i );【解答】 Liming Luwei3 #include <iostr

7、eam>using namespace std;struct Node char * s; Node * q; ;int main() Node a = "Mary", a+1 , "Jack", a+2 , "Jim", a ; Node *p = a; cout << p->s << endl; cout << p->q->s << endl; cout << p->q->q->s << endl; cout <

8、< p->q->q->q->s << endl;【解答】MaryJackJimMary5.3 思考題1判斷一個(gè)整數(shù)n的奇偶性，可以利用位運(yùn)算判斷嗎？請(qǐng)你試一試?！窘獯稹靠梢浴Ｒ粋€(gè)整數(shù)當(dāng)最低位為1時(shí)，它是奇數(shù)，否則為偶數(shù)。以下函數(shù)返回對(duì)參數(shù)k的奇偶判斷。bool odd( int k )return 1&k;2長(zhǎng)度為N的數(shù)組可以表示N個(gè)元素的集合，若有：Si=1，表示對(duì)應(yīng)元素在集合中如何實(shí)現(xiàn)集合的基本運(yùn)算？請(qǐng)你試一試。并從內(nèi)存和處理要求上與5.2.2節(jié)中集合的實(shí)現(xiàn)方法進(jìn)行比較。【解答】長(zhǎng)度為N的數(shù)組S可以表示有N個(gè)元素的集合。當(dāng)Si=1，表示元素

9、i+1在集合中；當(dāng)Si=0，表示元素i+1不在集合中。集合運(yùn)算通過(guò)對(duì)數(shù)組元素操作完成。用數(shù)組實(shí)現(xiàn)集合運(yùn)算的空間和時(shí)間消耗高于用無(wú)符號(hào)整數(shù)和位運(yùn)算實(shí)現(xiàn)集合運(yùn)算。用數(shù)組實(shí)現(xiàn)集合運(yùn)算程序如下。#include<iostream>using namespace std;void setPut( unsigned *S ); /輸入集合S的元素void setDisplay( const unsigned *S ); /輸出集合S中的全部元素bool putX( unsigned *S, unsigned x );/元素x并入集合void Com( unsigned *C, const un

10、signed *A, const unsigned *B);/求并集C=ABvoid setInt( unsigned *C, const unsigned A, const unsigned B); /求交集C=ABvoid setDif( const unsigned A, const unsigned B ); /求差集C=A-Bbool Inc( const unsigned *A, const unsigned *B ); /判蘊(yùn)含bool In( const unsigned *S, const unsigned x ); /判屬于xSbool Null( const unsign

11、ed *S ); /判空集const int N=32;/輸入集合元素void setPut(unsigned *S) unsigned x;cin >> x;while( x>0&&x<=N) putX( S, x );/把輸入元素并入集合S cin >> x;/輸出集合S中的全部元素void setDisplay( const unsigned *S ) cout << " "if (Null(S) cout<<" n"elsefor( int i=0; i<N; i+

12、 )/輸出元素 if( Si )cout << i+1 << ", "cout << "bb n"/擦除最后的逗號(hào)return;/元素x并入集合Sbool putX( unsigned *S, unsigned x ) if(x>0&&x<=N) Sx-1 = 1; return true;return false;/求并集C=ABvoid Com( unsigned *C, const unsigned *A, const unsigned *B ) for( int i=0; i<

13、N; i+ )Ci= int( Ai | Bi ) ;/求交集C=ABvoid setInt( unsigned *C, const unsigned *A, const unsigned *B ) for( int i=0; i<N; i+ )Ci= int( Ai&&Bi ) ;/求差集C=A-Bvoid setDif( unsigned *C, const unsigned *A, const unsigned *B ) for( int i=0; i<N; i+ )Ci= int( Ai&&!(A1&&Bi) ) ;/判蘊(yùn)含，

14、A蘊(yùn)含于B時(shí)返回true bool Inc( const unsigned *A, const unsigned *B ) for(int i=0; i<N; i+) if(Ai&&!Bi) return false; return true;/判屬于，xS時(shí)返回truebool In( const unsigned *S, const unsigned x ) return Sx-1;/判空集，S為空集時(shí)返回truebool Null( const unsigned *S ) for( int i=0; i<N; i+) if( Si)return false;r

15、eturn true;int main() unsigned AN=0, BN=0, CN=0;unsigned x;cout << "Input the elements of set A, 1-"<<N<<", until input 0 :n"setPut( A );cout << "Input the elements of set B, 1-"<<N<<", until input 0 :n"setPut( B );cout<&

16、lt;"A = " setDisplay( A );cout<<"B = "setDisplay( B );cout << "Input x: "cin>>x;cout << "Put " << x << " in A = "putX( A, x) ;setDisplay(A);cout << "C = A+B = "Com( C, A, B );setDisplay( C );cout &l

17、t;< "C = A*B = "setInt( C, A, B );setDisplay( C );cout << "C = A-B = "setDif( C, A, B );setDisplay( C );if( Inc( A, B ) )cout << "A <= B is truen"else cout << "not A <= Bn"cout << "Input x: "cin >> x;if( In( A,

18、x ) )cout << x << " in An"elsecout << x << " not in An"3分析以下說(shuō)明結(jié)構(gòu)的語(yǔ)句：struct Node int data; Node error;/錯(cuò)誤 Node * ok;/正確;error和ok分別屬于什么數(shù)據(jù)類型？有什么存儲(chǔ)要求？error出錯(cuò)的原因是什么？【解答】error是Node結(jié)構(gòu)類型數(shù)據(jù)成員，錯(cuò)誤。原因是結(jié)構(gòu)定義的數(shù)據(jù)成員若為本身的結(jié)構(gòu)類型，是一種無(wú)窮遞歸。ok是指向Node類型的指針，定義正確，占4字節(jié)。4本章例5-8中用輔助數(shù)組對(duì)結(jié)

19、構(gòu)數(shù)組進(jìn)行關(guān)鍵字排序，有定義：person *index100;index數(shù)組存放結(jié)構(gòu)數(shù)組元素的地址。如果把index定義改為：int index100;用于存放結(jié)構(gòu)數(shù)組元素的下標(biāo)，可以實(shí)現(xiàn)對(duì)結(jié)構(gòu)數(shù)組的索引排序嗎？如何修改程序？請(qǐng)你試一試?！窘獯稹靠梢浴ｊP(guān)鍵是通過(guò)整型索引數(shù)組元素作為下標(biāo)訪問(wèn)結(jié)構(gòu)數(shù)組。表示為：allpii.idallpii.salary有關(guān)程序如下：#include<iostream>using namespace std;struct person/說(shuō)明結(jié)構(gòu)類型 char name10; unsigned int id; double sal

20、ary; ;void Input( person, const int );void Sort( person, int,const int );void Output( const person, int,const int );int main() person allone100 ;/說(shuō)明結(jié)構(gòu)數(shù)組 int index100;/說(shuō)明索引數(shù)組 int total ; for(int i=0; i<100; i+)/索引數(shù)組元素值初始化為結(jié)構(gòu)數(shù)組元素下標(biāo) indexi=i ; cout<<"輸入職工人數(shù)：" cin>>total; cout&l

21、t;<"輸入職工信息：n" Input(allone,total); cout<<"以工資做關(guān)鍵字排序n" Sort(allone,index, total); cout<<"輸出排序后信息：n" Output(allone,index,total);void Input( person all, const int n )int i ; for( i=0; i<n; i+ )/ 輸入數(shù)據(jù) cout<<i<<": 姓名: " cin>>alli

22、.name; cout<<"編號(hào): " cin >> alli.id; cout<<"工資: " cin >> alli.salary ; void Sort(person all, int pi, const int n) int i,j; int t; /交換用中間變量 for(i=1; i<n; i+)/以成員salary做關(guān)鍵字排序 for(j=0; j<=n-1-i; j+) if(allpij.salary>allpij+1.salary)/通過(guò)索引數(shù)組訪問(wèn)結(jié)構(gòu)數(shù)組元素 t=p

23、ij;/交換索引數(shù)組元素值pij=pij+1;pij+1= t; void Output(const person all, int pi, const int n) for( int i=0; i<n; i+ )/ 輸出排序后數(shù)據(jù) cout<<<<'t'<<allpii.id<<'t'<<allpii.salary<<endl;5有以下結(jié)構(gòu)說(shuō)明和遍歷單向鏈表的函數(shù)。函數(shù)內(nèi)有錯(cuò)誤嗎？是什么性質(zhì)的錯(cuò)誤？請(qǐng)上機(jī)驗(yàn)證你的分析。struct Node int data;

24、Node * next; ;void ShowList( Node *head ) while( head ) cout << head->date << 'n' head +; 【解答】head+錯(cuò)誤，原因是動(dòng)態(tài)鏈表的結(jié)點(diǎn)存放不是連續(xù)順序的內(nèi)存空間，它們是逐個(gè)結(jié)點(diǎn)通過(guò)new建立的，所以不能用+做地址偏移運(yùn)算。應(yīng)該用：head=head->next;5.4 編程題1編寫程序，將一個(gè)整型變量右移4位，并以二進(jìn)制形式輸出該整數(shù)在移位前和移位后的數(shù)值。觀察系統(tǒng)填補(bǔ)空缺的數(shù)位情況。#include<iostream>using names

25、pace std;void bitDisplay(unsigned value);int main() unsigned x;cout << "Enter an unsigned integer: "cin >> x;bitDisplay(x);x>>=4;cout<<"Right 4-bitn"bitDisplay(x);void bitDisplay(unsigned value) unsigned c;unsigned bitmask = 1<<31;cout << value

26、 << " = t"for( c=1; c<=32; c+ ) cout << ( value&bitmask ? '1' : '0' );value <<= 1;if( c%8 = 0 ) cout << ' 'cout << endl;2整數(shù)左移一位相當(dāng)于將該數(shù)乘以2。編寫一個(gè)函數(shù)unsigned power2( unsigned number, unsigned pow );使用移位運(yùn)算計(jì)算number*2pow，并以整數(shù)形式輸出計(jì)算結(jié)果。注意考慮

27、數(shù)據(jù)的溢出。【解答】unsigned power2( unsigned number, unsigned pow )unsigned c=1;unsigned bitmask = 1<<31;while(c<31)/溢出判斷if( number&bitmask ) break;/查找最高位的1 即判斷c為何值時(shí)最高位為1，判斷可左移的最大次數(shù)c+;bitmask>>=1;if(pow<c)為此處做準(zhǔn)備return number<<pow;else cout<<"overflow!n"return 0;3使用

28、按位異或（）運(yùn)算，可以不使用中間變量，快速交換兩個(gè)變量的值。設(shè)計(jì)一個(gè)函數(shù)，實(shí)現(xiàn)快速交換兩個(gè)整型變量的值?！窘獯稹縱oid Swap( int &A, int &B) A=AB;B=AB;A=AB;4. 集合的元素通常是字符。設(shè)計(jì)程序，用無(wú)符號(hào)整數(shù)表示ASCII碼字符集合，用位運(yùn)算實(shí)現(xiàn)各種基本集合運(yùn)算?！窘獯稹緼SCII碼是0127的整數(shù)，可以用長(zhǎng)度為4的無(wú)符號(hào)整型數(shù)組表示集合，如教材例5-4所示。區(qū)別是，在輸入集合元素時(shí)，需要把字符轉(zhuǎn)換成整型數(shù)據(jù)，在輸出操作中，把整型集合元素轉(zhuǎn)換成字符型數(shù)據(jù)。程序略。5使用結(jié)構(gòu)類型表示復(fù)數(shù)。設(shè)計(jì)程序輸入兩個(gè)復(fù)數(shù)，可以選擇進(jìn)行復(fù)數(shù)的、×

29、或÷運(yùn)算，并輸出結(jié)果。【解答】#include <iostream>#include <iomanip>using namespace std;struct complex double re, im; ;int main() complex a,b,c; char oper; cout << "輸入復(fù)數(shù)a的實(shí)部和虛部: " cin >> a.re >> a.im; cout << "輸入復(fù)數(shù)b的實(shí)部和虛部:" cin >> b.re >> b.im

30、; cout << "輸入運(yùn)算符: " cin >> oper; switch ( oper ) case '+': c.re=a.re+b.re; c.im=a.im+b.im; break; case '-': c.re=a.re-b.re; c.im=a.im-b.im; break; case '*': c.re=a.re*b.re-a.im*b.im; c.im=a.im*b.re+a.re*b.im; break; case '/': c.re=(a.re*b.re+a.im

31、*b.im)/(b.re*b.re+b.im*b.im); c.im=(a.im*b.re-a.re*b.im)/(b.re*b.re+b.im*b.im);break; default: cout << "input error!" << endl; return 0;cout << "c=" << c.re;cout << setiosflags( ios:showpos );cout << c.im << "i" << endl;re

32、turn 0;6把一個(gè)班的學(xué)生姓名和成績(jī)存放到一個(gè)結(jié)構(gòu)數(shù)組中，尋找并輸出最高分者。【解答】#include <iostream>using namespace std;int main() struct data char name12; double score;a = "李小平",90,"何文章",66,"劉大安",87,"汪立新",93,"羅建國(guó)",78, "陸豐收",81,"楊勇",85,"吳一兵",55,"

33、;伍曉笑",68,"張虹虹",93; double max = a0.score; int i,n = sizeof(a) / sizeof(data); for( i=1; i<n; i+ ) if( ai.score > max ) max = ai.score; for( i=0; i<n; i+ ) if( ai.score = max ) cout << << endl;同樣高分！7使用結(jié)構(gòu)表示XY平面直角坐標(biāo)系上的點(diǎn)，編寫程序順序讀入一個(gè)四邊形的4個(gè)頂點(diǎn)坐標(biāo)，判別由這個(gè)頂點(diǎn)的連線構(gòu)成的圖形是否為

34、正方形、矩形或其他四邊形。要求：定義求兩個(gè)點(diǎn)距離的函數(shù)使用結(jié)構(gòu)參數(shù)?！窘獯稹?include <iostream>#include <cmath>using namespace std;struct point double x; double y; ;double d( point p1, point p2 ) return sqrt( pow( p1.x-p2.x,2 )+pow( p1.y-p2.y,2 ) ); int main() int i; point p5; for( i=1; i<=4; i+ ) cout << "輸入第&

35、quot; << i << "個(gè)頂點(diǎn)的橫坐標(biāo)和縱坐標(biāo): " cin >> pi.x >> pi.y; if( fabs( d( p1,p2 ) - d( p3,p4 )<=1e-8 && fabs( d( p1,p4 ) - d( p2,p3 )<=1e-8 && fabs( d( p1,p3 ) - d( p2,p4 )<=1e-8) if( fabs( d( p1,p2 ) - d( p2,p3 )<1e-8 ) cout << "四個(gè)頂點(diǎn)構(gòu)成的

36、圖形為正方形！" << endl; else cout << "四個(gè)頂點(diǎn)構(gòu)成的圖形為矩形！" << endl; else cout << "四個(gè)頂點(diǎn)構(gòu)成的圖形為其它四邊形！" << endl;8建立一個(gè)結(jié)點(diǎn)包括職工的編號(hào)、年齡和性別的單向鏈表，分別定義函數(shù)完成以下功能：（1）遍歷該鏈表輸出全部職工信息；（2）分別統(tǒng)計(jì)男、女職工的人數(shù)；（3）在鏈表尾部插入新職工結(jié)點(diǎn)；（4）刪除指定編號(hào)的職工結(jié)點(diǎn)；（5）刪除年齡在60歲以上的男性職工或55歲以上的女性職工結(jié)點(diǎn)，并保存在另一個(gè)鏈表中。要求：

37、用主函數(shù)建立簡(jiǎn)單菜單選擇，并測(cè)試程序。【解答】#include <iostream>using namespace std;struct employeeint num; int age; char sex; employee *next; ;employee *head, *head1;/建立單向鏈表employee *create() employee *head, *p, *pend; char ch; head = NULL; cout << "t輸入數(shù)據(jù)?(y/n)" cin >> ch; if( ch = 'y'

38、; ) p = new employee; cout << "t編號(hào):" cin >> p->num; cout << "t年齡:" cin >> p->age; cout << "t性別:" cin >> p->sex; else goto L0; while( ch = 'y' ) if( head = NULL ) head = p; else pend->next = p; pend = p; cout <&l

39、t; "t輸入數(shù)據(jù)?(y/n)" cin>>ch; if( ch = 'y' ) p = new employee; cout << "t編號(hào):" cin >> p->num; cout << "t年齡:" cin >> p->age; cout << "t性別:" cin >> p->sex; pend->next = NULL; L0: return head;/顯示單向鏈表中全部職工信息

40、void show( employee *head ) employee *p = head;if( !head ) cout << "t空鏈表！" << endl; goto L1; cout << "t鏈表中的數(shù)據(jù)是: n" while( p )NULL cout << 't' << p->num << "," << p->age << "," << p->sex <

41、;< endl; p = p->next; L1: /統(tǒng)計(jì)男女職工人數(shù)void count( employee *head ) employee *p = head; int m, f; m = 0; f = 0; while( p ) if( p->sex = 'm' ) m+; else f+; p = p->next; cout << "t男職工人數(shù)：" << m << endl; cout << "t女職工人數(shù)：" << f << en

42、dl; /在鏈表尾部插入新結(jié)點(diǎn) employee *insert(employee *&head) employee *pend = head, *p; /在空鏈表尾部插入新結(jié)點(diǎn) if( !head ) p = new employee; cout << "t編號(hào):" cin >> p->num; cout << "t年齡:" cin >> p->age; cout << "t性別:" cin >> p->sex; head = p; p

43、->next = NULL; return head; /在鏈表尾部插入新結(jié)點(diǎn) while( pend->next != NULL ) pend = pend->next; p = new employee; cout << "t編號(hào):" cin >> p->num; cout << "t年齡:" cin >> p->age; cout << "t性別:" cin >> p->sex; pend->next = p; pe

44、nd = p; pend->next = NULL; return head; /刪除指定編號(hào)的結(jié)點(diǎn)employee *del(employee *&head， int bh ) employee *p, *q; if ( !head ) cout << "t空鏈表！" << endl; goto L2; /刪除鏈?zhǔn)捉Y(jié)點(diǎn) if( head->num = bh ) p = head; head = head->next; delete p; cout << "t結(jié)點(diǎn)已被刪除！" <<

45、 endl; goto L2; /刪除非鏈?zhǔn)捉Y(jié)點(diǎn) q = head; while( q->next != NULL ) if ( q->next->num = bh ) p = q->next; /待刪除結(jié)點(diǎn) q->next = p->next; delete p; cout << "t結(jié)點(diǎn)已被刪除！" << endl; goto L2; q = q->next; cout << "t找不到需刪除結(jié)點(diǎn)！" << endl;L2: return head ;/刪除指定

46、年齡段的結(jié)點(diǎn)，并把被刪除結(jié)點(diǎn)保存在另一鏈表中employee *delcreate(employee *&head) employee *p, *pd, *p1, *q; int flag; /建立新鏈表 if ( head = NULL ) cout << "t空鏈表！" << endl; goto L3; head1 = NULL; pd = new employee; p = head; flag = 0; while ( p != NULL ) if( p->age >= 55 && p->age &

47、lt;=60 ) pd->num = p->num; pd->age = p->age; pd->sex = p->sex; if( head1 = NULL ) head1 = pd; else p1->next = pd; p1 = pd; pd = new employee; flag = 1; p = p->next; if ( flag = 0 ) cout << "t沒(méi)有需刪除的結(jié)點(diǎn)！" << endl; goto L3; p1->next = NULL;/顯示新鏈表cout <

48、<"t新鏈表中的數(shù)據(jù)是: n"p = head1;while( p ) cout << 't' << p->num << "," << p->age << "," << p->sex << endl; p = p->next; /刪除指定年齡的結(jié)點(diǎn)p = head; q = p;while ( p != NULL ) if( p->age >= 55 && p->age &

49、lt;= 60) if( head->age = p->age ) pd = head; /待刪除結(jié)點(diǎn) head = head->next;delete pd;p = head;continue; elseif( p->next = NULL ) pd = p; /待刪除結(jié)點(diǎn) q->next = NULL; delete pd; goto L3; elsepd = p; /待刪除結(jié)點(diǎn)q->next = p->next;delete pd;p = q->next;continue;q = p;p = p->next; L3: return ( head );int main()int choice, bh ; L:cout << "ntt請(qǐng)鍵入操作選擇n" << endl;cout << "t 1 - 建立單向鏈表"