C语言程序设计第五版谭浩强课后答案 第九章习题答案

C语言程序设计第五版谭浩强第九章答案

1.定义一个结构体变量(包括年、月、日)。计算该日在本年中是第几天,注意闰年问题。

解题思路及答案：

• 用一个日期数组保存每一月的天数，二月的天数记为28天，后面根据输入的时间确定是否是闰年的二月，如果是，天数在加1。

#include <stdio.h>

struct Date{
    int year;
    int month;
    int day;
};

int main(){
    struct Date date;
    printf("Please give date: ");
    scanf("%d%d%d", &date.year, &date.month, &date.day);
    int Days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    
    int i, days = 0;
    for (i = 1; i < date.month; i++)
        days += Days[i];
    days += date.day;
    //如果包含闰年的二月，天数加1
    if(date.month > 2)
    {
       if (date.year%400 == 0 || (date.year%4 == 0 && date.year%100 != 0)){
            ++days;
        } 
    }
    printf("It's day %d in the year.\n", days);
    return 0;
}

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2.写一个函数days,实现第1 题的计算。由主函数将年、月、日传递给days函数,计算后将日子数传回主函数输出。

#include <stdio.h>

struct Date{
  int year;
  int month;
  int day;
};

int Days(struct Date date)
{
  static int Days[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

  int i, days = 0;
  for (i = 1; i < date.month; i++)
    days += Days[i];
  days += date.day;
  //如果包含闰年的二月，天数加1
  if (date.month > 2)
  {
    if (date.year % 400 == 0 || (date.year % 4 == 0 && date.year % 100 != 0)){
      ++days;
    }
  }
  return days;
}

int main(){
  struct Date date;
  printf("Please give date: ");
  scanf("%d%d%d", &date.year, &date.month, &date.day);
  int days = Days(date);
  printf("It's day %d in the year.\n", days);
  return 0;
}

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3.编写一个函数print,打印一个学生的成绩数组，该数组中有5个学生的数据记录，每个记录包括num,name,score[3],用主函数输人这些记录,用print函数输出这些记录。

#include <stdio.h>

#define
//定义一个student结构体数组，包含5个元素
struct student_t{
  int num;
  char name[NAMLEN];
  int score[3];
} students[5];

void print(struct student_t *stu);

int main(){
  for (int i = 0; i < 5; i++){
    scanf("%d%s%d%d%d", &students[i].num, students[i].name, &students[i].score[0],
      &students[i].score[1], &students[i].score[2]);
  }
  print(students);
  return 0;
}

void print(struct student_t *stu){
  for (int i = 0; i < 5; i++){
    printf("%d %s %d %d %d\n", students[i].num, students[i].name, students[i].score[0],
      students[i].score[1], students[i].score[2]);
  }
}

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4.在第3题的基础上,编写一个函数input,用来输人5个学生的数据记录。

#include <stdio.h>

#define
//定义一个student结构体数组，包含5个元素
struct student_t{
  int num;
  char name[NAMLEN];
  int score[3];
} students[5];

void print(struct student_t *stu);
void input(struct student_t *stu);

int main(){
  input(students);
  print(students);
  return 0;
}

void input(struct student_t *stu)
{
  for (int i = 0; i < 5; i++){
    scanf("%d%s%d%d%d", &students[i].num, students[i].name, &students[i].score[0],
      &students[i].score[1], &students[i].score[2]);
  }
}

void print(struct student_t *stu){
  for (int i = 0; i < 5; i++){
    printf("%d %s %d %d %d\n", students[i].num, students[i].name, students[i].score[0],
      students[i].score[1], students[i].score[2]);
  }
}

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5.有10个学生,每个学生的数据包括学号、姓名、3门课程的成绩,从键盘输人10个学生数据,要求输出3门课程总平均成绩,以及最高分的学生的数据(包括学号、姓名、3门课程成绩、平均分数)。

#include <stdio.h>

#define
#define

typedef struct student_t{
  int num;
  char name[NAMLEN];
  int score[3];
} student;

int main(){
  student students[STUCNT];
  int maxi = 0, maxsum = 0;
  double aver_0 = 0, aver_1 = 0, aver_2 = 0;
  for (int i = 0; i < STUCNT; i++){
    scanf("%d%s%d%d%d", &students[i].num, students[i].name, &students[i].score[0], &students[i].score[1], &students[i].score[2]);
    int sum = students[i].score[0] + students[i].score[1] + students[i].score[2];
    if (sum > maxsum){
      maxsum = sum;
      maxi = i;
    }
    aver_0 += students[i].score[0];
    aver_1 += students[i].score[1];
    aver_2 += students[i].score[2];
  }
  aver_0 /= STUCNT;
  aver_1 /= STUCNT;
  aver_2 /= STUCNT;
  printf("%lf %lf %lf\n", aver_0, aver_1, aver_2);
  printf("%d %s %d %d %d %lf\n", students[maxi].num, students[maxi].name, students[maxi].score[0], students[maxi].score[1], students[maxi].score[2],
    (students[maxi].score[0] + students[maxi].score[1] + students[maxi].score[2]) / 3.0);
  return 0;
}

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6.13个人围成一圈,从第1个人开始顺序报号1,2,3。凡报到3者退出圈子。找出最后留在圈子中的人原来的序号。要求用链表实现。

解题思路及答案：

创建一个环形链表，给链表中的每一个节点从1~13编号，然后开始淘汰过程，对于淘汰的节点，序号置为0，淘汰完成之后，找到序号不为0的即为最后留下的。

#include <stdio.h>
#define
typedef struct people
{
    int num;
    struct people *next;
} people;

int main()
{
    int count = NUM;
    people p[NUM];
    people *head;
    head = p; //head 指向p[0]
    //1~13编号
    for (int i = 0; i < NUM; i++)
    {
        head->num = i + 1;
        head->next = &p[i + 1];
        head = head->next;
    }
    //最后一个元素指向第一个元素 ， 形成环
    p[NUM - 1].next = p; 

    int i = 1;
    head = p;
    while (count > 1)
    {
        //跳过已经被淘汰的节点
        if (head->num == 0)
        {
            head = head->next;
            continue;
        }
        if (i == 3)
        {
            //被淘汰的节点，num置为0
            printf("第 %d 位置被淘汰\n", head->num);
            head->num = 0;
            count--;
        }
        head = head->next;
        i++;
        if (i > 3)
        {
            i = 1;
        }
    }
    printf("--------------\n");
    while (head->num == 0)
    {
        //非0节点即为最后留下的
        head = head->next;
        if (head->num != 0)
        {
            printf("留到最后的是 %d \n", head->num);
        }
    }

    return 0;
}

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7.在第9章例9.9和例9.10的基础上,写一个函数del,用来删除动态链表中指定的节点

解题思路及答案：

首先创建一个带头的单链表，然后让用户输入需要删除的节点，调用del函数，找到需要删除的节点，把待删除节点的前驱和后继重新链接。

#include <stdio.h>
#include <stdlib.h>

typedef struct LNode
{
  int num;
  struct LNode *next;
} LNode;

//创建含有n个值的节点
LNode* creat(int n)
{
  LNode *head, *p;
  head = (LNode *)malloc(sizeof(LNode));
  p = head; //头节点为0 加上头节点共n + 1个节点
  head->num = 0;
  head->next = NULL;
  for (int i = 1; i <= n; i++)
  {
    LNode *newNode = (LNode *)malloc(sizeof(LNode));
    newNode->num = i;
    newNode->next = NULL;
    p->next = newNode;
    p = p->next;
  }
  return head;
}

//删除值为n的节点
void del(int n, LNode *head)
{
  LNode *pre, *current;
  pre = head;
  current = head->next;
  //从第一个有效节点开始查找待删除节点
  printf("delete node %d\n", n);
  while (current != NULL)
  {
        //找到待删除节点，重新链接，释放待删除节点
    if (current->num == n)
    {
      pre->next = current->next;
            free(current);
      break;
    }
        //更新查找位置
    pre = current;
    current = current->next;
  }
}

int main()
{
  LNode *head, *p;
  int n;
  head = creat(10);
  printf("请输入需要删除的节点:1-10\n");
  scanf("%d", &n);
  del(n, head);
  int i = 1;
  p = head->next;
  while (p != NULL)
  {
    printf("p %d.num -> %d\n", i, p->num);
    p = p->next;
    i++;
  }
  return 0;
}

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8.写一个函数insert,用来向一个动态链表插入结点

#include <stdio.h>
#include <stdlib.h>

typedef struct LNode
{
  int num;
  struct LNode *next;
} LNode;

void insert(int n, LNode *node)
{
  //创建新节点
  LNode *newNode = (LNode *)malloc(sizeof(LNode));
  newNode->num = n;

  LNode* next = node->next;

  // node ---> newNode  ---> next
  newNode->next = next;
  node->next = newNode;
}

LNode* creat(int n)
{
  LNode *head, *p;
  head = (LNode *)malloc(sizeof(LNode));
  p = head; //头节点为0 加上头节点共11个节点
  head->num = 0;
  head->next = NULL;
  for (int i = 1; i <= n; i++)
  {
    LNode *newNode = (LNode *)malloc(sizeof(LNode));
    newNode->num = i;
    newNode->next = NULL;
    p->next = newNode;
    p = p->next;
  }
  return head;
}

void printNode(LNode* head)
{
  LNode* p = head->next;
  while (p != NULL)
  {
    printf("num -> %d\n", p->num);
    p = p->next;
  }
}

int main()
{
  LNode *head;
  int n;
  head = creat(10);
  printNode(head);
  printf("请输入需要插入的节点:\n");
  scanf("%d", &n);
  insert(n, head);
  printf("链表的新内容：\n");
  printNode(head);
  return 0;
}

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9.综合本章例9.9(建立链表的函数creat)、例9.10(输出链表的函数print)和本章习题第7题(删除链表中结点的函数del)、第8题(插入结点的函数insert),再编写一个主函数，先后调用这些函数。用以上5个函数组成一个程序,实现链表的建立、输出、删除和插入,在主函数中指定需要删除和插人的结点的数据。

#include <stdio.h>
#include <stdlib.h>
#define
typedef struct LNode
{
    int num;
    struct LNode *next;
} LNode;

LNode* create(int n)
{
    LNode *head, *p;
    head = (LNode *)malloc(sizeof(LNode));
    p = head; //头节点为0 加上头节点共11个节点
    head->num = 0;
    head->next = NULL;
    for (int i = 1; i <= n; i++)
    {
        LNode *newNode = (LNode *)malloc(sizeof(LNode));
        newNode->num = i;
        newNode->next = NULL;
        p->next = newNode;
        p = p->next;
    }
    return head;
}
//在指定位置插入数据
void insert(int n, int positon, LNode *root)
{
    //首先找到指定位置
    while (positon--)
    {
        root = root->next;
    }
    //插入新的数据，重新链接插入点的前后节点关系
    LNode *newNode = (LNode *)malloc(sizeof(LNode));
    newNode->num = n;
    newNode->next = root->next;
    root->next = newNode;
}
void del(int n, LNode *root)
{
    LNode *pre;
    while (root->num != n)
    {
        pre = root;
        root = root->next;
    }
    pre->next = root->next;
}
void printList(LNode *root)
{
    printf("----\n");
    int i = 0;
    while (root != NULL)
    {
        printf("node %d -> %d\n", i, root->num);
        root = root->next;
        i++;
    }
}

int main()
{
    int n, position;
    printf("请输入插入/删除的数，及插入的位置，位置最大为:%d\n", COUNT);
    scanf("%d %d", &n, &position);
    LNode *head = create(COUNT);
    printList(head->next);
    insert(n, position, head->next);
    printList(head->next);
    del(n, head->next);
    printList(head->next);
    return 0;
}

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10.已有a,b两个链表,每个链表中的结点包括学号、成绩。要求把两个链表合并, 按学号升序排列。

解题思路及答案：

首先合并两个链表，然后采用选择排序，给合并之后的链表进行排序。

#include <stdio.h>
typedef struct student
{
    int num;
    double grade;
    struct student *next;
} student;

student *merge(student *a, student *b)
{
    //先合并，后排序
    student *head = a;
    while (a->next != NULL)
    {
        a = a->next;
    }
    a->next = b;
    //选择排序，每次选最小的，放在未排序的链表头部
    student *pre;
    pre = head;
    while (pre->next != NULL)
    {
        a = pre->next;
        while (a != NULL)
        {
            if (pre->num > a->num)
            {
                int num = pre->num;
                double grade = pre->grade;
                pre->num = a->num;
                pre->grade = a->grade;
                a->num = num;
                a->grade = grade;
            }
            a = a->next;
        }
        pre = pre->next;
    }
    return head;
}
int main()
{
    student a[3] = {{1, 79}, {4, 36}, {5, 79}};
    for (int i = 0; i < 2; i++)
    {
        a[i].next = &a[i + 1];
    }

    student b[2] = {{2, 38}, {6, 98}};
    for (int i = 0; i < 1; i++)
    {
        b[i].next = &b[i + 1];
    }
    student *combine = merge(a, b);
    while (combine != NULL)
    {
        printf("%d -> %.2lf\n", combine->num, combine->grade);
        combine = combine->next;
    }

    return 0;
}

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11.有两个链表a和b,设结点中包含学号、姓名。从a链表中删去与b链表中有相同学号的那些结点。

解题思路及答案：

对于b链表中的每一个节点，都从a链表的表头开始查找，如果可以找到，直接删除，如果找不到，继续从a链表表头找下一个b的节点。

#include <stdio.h>
typedef struct student
{
  int num;
  double grade;
  struct student *next;
} student;
student *del(student *a, student *b)
{
  student *pre, *current, *head;
  head = a;

  while (b != NULL)
  {
    //重置指针指向a链表的头部
    pre = head;
    current = head->next;
    //a 链表的头等于b
    if (pre->num == b->num)
    {
      pre->next = NULL;
      pre = current;
      current = current->next;
      //更新表头
      head = pre;
    }
    else
    {
      while (pre->next != NULL)
      {
        if (current->num == b->num)
        {
          //找到就删除
          pre->next = current->next;
          break;
        }
        else
        {
          //否则继续遍历
          pre = pre->next;
          current = current->next;
        }
      }
    }
    b = b->next;
  }
  return head;
}

void printList(student *root)
{
  printf("----\n");
  int i = 0;
  while (root != NULL)
  {
    printf("student %d -> %d -> %.2lf\n", i, root->num, root->grade);
    root = root->next;
    i++;
  }
}

int main()
{
  student a[3] = { { 1, 79 }, { 4, 36 }, { 5, 79 } };
  for (int i = 0; i < 2; i++)
  {
    a[i].next = &a[i + 1];
  }
  a[2].next = NULL;
  printf("链表a:\n");
  printList(&a[0]);

  student b[2] = { { 5, 38 }, { 4, 98 } };
  for (int i = 0; i < 1; i++)
  {
    b[i].next = &b[i + 1];
  }
  b[1].next = NULL;
  printf("链表b:\n");
  printList(&b[0]);
  student *combine = del(a, b);
  printf("删除之后：\n");
  while (combine != NULL)
  {
    printf("%d -> %.2lf\n", combine->num, combine->grade);
    combine = combine->next;
  }

  return 0;
}

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12.建立一个链表,每个结点包括:学号、姓名、性别、年龄。输入一个年龄，如果链表中的结点所包含的年龄等于此年龄,则将此结点删去。

#include <stdio.h>
#include <stdio.h>
typedef struct student
{
  int num;
  char sex[10];
  char name[100];
  int age;
  struct student *next;
} student;

void printList(student *root)
{
  printf("----\n");
  while (root != NULL)
  {
    printf("num:%d, sex: %s, name: %s, age: %d\n", root->num, root->sex, root->name, root->age);
    root = root->next;
  }
}

int main()
{
  student a[] = { { 1, "woman", "apple", 12 }, { 4, "woman", "banbana", 36 }, { 5, "man", "candy", 79 }, { 5, "man", "danny", 36 }, { 4, "man", "enjoy", 98 } };
  for (int i = 0; i < 4; i++)
  {
    a[i].next = &a[i + 1];
  }
  a[4].next = NULL;

  printList(&a[0]);

  int n;
  printf("请输入要删除的年龄：\n");
  scanf("%d", &n);
  student *pre = a, *current = a->next, *head;

  head = a;
  while (current != NULL)
  {
    //如果头结点需要删除，则更新头结点
    if (head->age == n)
    {
      pre->next = NULL;
      pre = current;
      current = current->next;
      head = pre;
    }
    else
    {
      //删除节点，重新链接
      if (current->age == n)
      {
        pre->next = current->next;
      }
      pre = current;
      current = current->next;
    }
  }
  printList(head);

  return 0;
}

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