程序设计上机实验课后答案(7)

33. date=2005/6/15

34. 101 Zhang M 19 95.00 64.00 0.00

102 Wang F 18 92.00 97.00 0.00 103 Zha0 M 19 85.00 78.00 0.00 104 Li M 20 96.00 88.00 0.00 105 Gou M 19 91.00 96.00 0.00 106 Lin M 18 93.00 78.00 0.00 107 Ma F 18 98.00 97.00 0.00 108 Zhen M 21 89.00 93.00 0.00 109 Xu M 19 88.00 90.00 0.00 110 Mao F 18 94.00 90.00 0.00 ―――――――――――――――――――

101 Zhang M 19 95.00 64.00 79.50 102 Wang F 18 92.00 97.00 94.50 103 Zha0 M 19 85.00 78.00 81.50 104 Li M 20 96.00 88.00 92.00 105 Gou M 19 91.00 96.00 93.50 106 Lin M 18 93.00 78.00 85.50 107 Ma F 18 98.00 97.00 97.50 108 Zhen M 21 89.00 93.00 91.00 109 Xu M 19 88.00 90.00 89.00 110 Mao F 18 94.00 90.00 92.00 35. st= 101 Zhang M 19 89.00

t= 101 Zhang M 19 89.00 t= 101 Zhang M 19 95.00 st= 101 Zhang M 19 95.00 36. 0 7 15

0 1 15

37. No. Name Sex Age Score

103 Zhao M 19 85.70 109 Xu M 19 89.80 108 Zhen M 21 90.10 105 Gou M 19 90.20 106 Lin M 18 91.50 102 Wang F 18 92.40 110 Mao F 18 94.90 101 Zhang M 19 95.60 104 Li M 20 96.30 107 Ma F 17 98.70 38. sequence that persons leave the circle:

3 6 9 12 2 7 11 4 10 5 1 8 The last one is 13 四、程序设计题 1. 参考程序：

#include

#define N 5/*学生的人数*/

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struct student { char num[6]; char name[8]; int score[4]; }stu[N]; void main()

{void print(struct student stu[6]); int i,j;

for (i=0;i

{printf(\ printf(\

scanf(\ printf(\

scanf(\ for (j=0;j<3;j++)

{printf(\

scanf(\ }

printf(\}

print(stu); }

void print(struct student stu[6]) {int i,j;

printf(\ for (i=0;i

{printf(\ for (j=0;j<3;j++)

printf(\ printf(\ } }

2. 编程思路:

定义结构体变量a、b、c和指针变量p。

当p指向a时，将a.score表示为(*p).score。 #include void main()

{ typedef struct students { char name[10]; char sex; int score; }STDTS;

STDTS s1={\’f’,78},

s2={\’m’,88},s3={\’f’,83}; STDTS *p=&s1;

printf(\printf(\

(*p).name,(*p).sex,(*p).score); printf(\

s2.name,s2.sex,s2.score);

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printf(\

s3.name,s3.sex,s3.score); if(p->score

if(p->score

p->name,p->sex,p->score); }

3. 编程思路

#include

#define COUNTS struct counts void main( )

{ int i, ch;

COUNTS { char vowel; int num; };

static COUNTS figure[5]={‘a’,0,’e’,0,’i’,0,’o’,0,’u’,0}; printf(\请输入一行文字(以’!’结束):\\n\ ch=getchar( );

while ( ch!=‘!’ ) { switch ( ch ) { case ‘a’:

case ‘A’: figure[0].num++; break; case ‘e’:

case ‘E’: figure[1].num++; break; case ‘i’:

case ‘I’: figure[2].num++; break; case ‘o’:

case ‘O’: figure[3].num++; break; case ‘u’:

case ‘U’: figure[4].num++; }

ch=getchar( ); }

printf(\输入的文字中各元音的个数分别为:\ for(i=0; i<5; i++)

printf(\ } 4. 参考答案

#include void main( ) { int n;

enum weekdays{sun,mon,tue,wed,thu,fri,sat} today; printf(\请输入一整数(0～6):\ scanf(\

today=(enum weekdays)n; switch(today) { case mon: case tue: case wed:

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case thu:

case fri: printf(\工作日\ case sat:

case sun: printf(\休息日\ default: printf(\输入错\ } }

5. 解题思路。显然，需要设一个数组，有3个元素，每个元素中的信息应包括候选人的姓名（字符型）和得票

数（整型），只有结构体类型才能胜任。据此编写程序如下： #include #include

struct person { char name[20]; int count; }

leader[3]={\void main( )

{ int i,j;char leader_name[20]; for(i=0;i<=10;i++)

{ scanf(\ for(j=0;j<3;j++) if(strcmp(leader_name,leader[j].name)==0) leader[j].count++; } printf(\ for(i=0;i<3;i++)

printf(\

6. 分析：链表由头指针代表，要释放链表可以定义一个工作指针得到头指针，然后头指针和工作指针均不断下

移，同时释放结点。

void fun(struct node *head) {struct node *p; while(head!=NULL) {p=head;

head=head->next; free(p); } }

7. 参考答案

#include #include #include #include typedef struct stu { char name[20]; double score; struct stu *next; }NODE;

NODE *create(int n);

void insertlist(NODE *h, char *s); void deletelist(NODE *h, char *s); void printlist(NODE *h); void main()

{ char chose,sname[20];

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int n;

NODE *head=NULL; for(;;)

{ system(\

printf(\创建单链表.\\n\ printf(\在单链表中插入结点.\\n\ printf(\在单链表中删除结点.\\n\ printf(\输出单链表.\\n\ printf(\退出.\\n\

printf(\请选择命令号:\ chose=getchar(); getchar(); switch(chose)

{ case ‘1’: printf(\输入结点个数: \ scanf(\ getchar();

head=create(n); break;

case ‘2’: printf(\输入用于插入的姓名: \ gets(sname);

insertlist(head,sname); break;

case ‘3’: printf(\输入用于删除的姓名:\ gets(sname);

deletelist(head,sname); break;

case ‘4’: printlist(head);

printf(\按任意键返回主菜单.\\n\ getch(); break; case ‘0’: exit(1); default: break; } } }

NODE *create(int n) { NODE *p,*h; int i;

char inbuf[10];

h=(NODE *)malloc(sizeof(NODE)); h->next=NULL; for(i=n;i>0;i--)

{ p=(NODE *)malloc(sizeof(NODE)); printf(\输入新结点值:\\n\ gets(p->name); gets(inbuf);

p->score=atof(inbuf); p->next=h->next; h->next=p; }

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