华南农业大学数据结构实验答案(包含STL版)

8576 顺序线性表的基本操作

时间限制:1000MS 内存限制:1000K

提交次数:9027 通过次数:2456

题型: 编程题语言: 无限制

Description

编写算法，创建初始化容量为LIST_INIT_SIZE的顺序表T，并实现插入、删除、遍历操作。本题目给出部分代码，请补全内容。

#include<>

#include<>

#define OK 1

#define ERROR 0

#define LIST_INIT_SIZE 100

#define LISTINCREMENT 10

#define ElemType int

typedef struct

{

int *elem;

int length;

int listsize;

}SqList;

int InitList_Sq(SqList &L)

{

n");

}

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d%d",&i,&x);

if(_________________________) printf("Insert Error!\n"); 1:Insert element 2:Delete element

3:Load all elements

0:Exit

Please choose:

The Element 2 is Successfully Inserted!

1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

The Element 3 is Successfully Inserted!

1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

The Element 3 is Successfully Deleted!

1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

The List is: 2

1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

作者

yqm

解法一：（正规解法）#include<>

#include<>

#define OK 1

#define ERROR 0

#define LIST_INIT_SIZE 100 #define LISTINCREMENT 10 #define ElemType int

typedef struct

{

int *elem;

int length;

int listsize;

}SqList;

int InitList_Sq(SqList &L)

{

n");

}

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d%d",&i,&x);

if(!ListInsert_Sq(T,i,x)) printf("Insert Error!\n"); n");

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1:

scanf("%d%d",&i,&x);

if(i<1||i>(int)()+1) printf("Insert Error!\n"); n");

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1:

scanf("%d%d",&i,&x);

if(i<1||i>k) printf("Insert Error!\n");

., ai,...an-1)，其逆顺序表定义为A'=( an-1,..., ai,...,a1, a0)。设计一个算法，将顺序表逆置，要求顺序表仍

占用原顺序表的空间。本题不提供代码，请同学们独立完成，所需子函数参考前面完成的内容。

输入格式

第一行：输入顺序表的元素个数

第二行：输入顺序表的各元素，用空格分开

输出格式

第一行：逆置前的顺序表元素列表

第二行：逆置后的顺序表元素列表

输入样例

10

1 2 3 4 5 6 7 8 9 10

输出样例

The List is:1 2 3 4 5 6 7 8 9 10

The turned List is:10 9 8 7 6 5 4 3 2 1

作者

yqm

解法一：（正规解法）

#include<>

#include<>

#define OK 1

#define ERROR 0

#define LIST_INIT_SIZE 100

#define LISTINCREMENT 10

#define ElemType int

typedef struct

{

int *elem;

int length;

int listsize;

}SqList;

int InitList_Sq(SqList &L,int n)

{

n");

LoadLink_L(T);

}

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d%d",&i,&x);

if(___________________________) printf("Insert Error!\n"); The LinkList is:3 6 9 1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

The LinkList is:3 6 9

1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

The Element 12 is Successfully Inserted! 1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

The Element 3 is Successfully Deleted! 1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

The LinkList is:6 9 12

1:Insert element

2:Delete element

3:Load all elements

0:Exit

Please choose:

作者

yqm

解法一：（正规解法）

#include<>

#include<>

#define ERROR 0

#define OK 1

#define ElemType int

typedef struct LNode

{

int data;

struct LNode *next;

}LNode,*LinkList;

int CreateLink_L(LinkList &L,int n){

n");

LoadLink_L(T);

}

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d%d",&i,&x);

if(!LinkInsert_L(T,i,x)) printf("Insert Error!\n"); n");

for(i=0;i

{

scanf("%d",&x);

(x);

}

load(T);

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1:

scanf("%d%d",&i,&x);

if(i<1||i>(int)()+1) printf("Insert Error!\n"); n");

if(k==1) printf("The List is empty!");

else

{

printf("The LinkList is:");

for(int j=1; j

}

printf("\n");

while(1)

{

printf("1:Insert element\n2:Delete element\n3:Load all elements\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1:

scanf("%d%d",&i,&x);

if(i<1||i>k) printf("Insert Error!\n");

., ai,...an-1)，其逆线性表定义为A'=( an-1,..., ai,...,a1, a0)，设计一个算法，将线性表逆置，要求线性表仍

占用原线性表的空间。本题不提供代码，请同学们独立完成，所需子函数参考之前完成的内容。

输入格式

第一行：输入n，表示单链表的元素个数

第二行：输入单链表的各元素，用空格分开

输出格式

第一行：输出单链表逆置前的元素列表

第二行：输出单链表逆置后的元素列表

输入样例

8

32 97 54 65 35 84 61 75

输出样例

The List is:32 97 54 65 35 84 61 75

The turned List is:75 61 84 35 65 54 97 32

作者

yqm

解法一：(正规解法)

#include<>

#include<>

#define ERROR 0

#define OK 1

#define ElemType int

typedef struct LNode

{

int data;

struct LNode *next;

}LNode,*LinkList;

int CreateLink_L(LinkList &L,int n){

n");

}

while(1)

{

printf("1:Push \n2:Pop \n3:Get the Top \n4:Return the Length of the Stack\n5:Load the Stack\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d", &x);

if(______________________) printf("Push Error!\n"); 1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack

5:Load the Stack

0:Exit

Please choose:

The Element 2 is Successfully Pushed!

1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack

5:Load the Stack

0:Exit

Please choose:

The Element 4 is Successfully Pushed! 1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

The Element 6 is Successfully Pushed! 1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

The Stack is: 6 4 2

1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

The Top Element is 6!

1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

The Length of the Stack is 3!

1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

The Element 6 is Successfully Poped! 1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

The Stack is: 4 2

1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack

5:Load the Stack

0:Exit

Please choose:

The Element 4 is Successfully Poped! 1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

The Element 2 is Successfully Poped! 1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

Pop Error!

1:Push

2:Pop

3:Get the Top

4:Return the Length of the Stack 5:Load the Stack

0:Exit

Please choose:

作者

yqm

解法一：（正规解法）

#include<>

#include<>

#include<>

#define OK 1

#define ERROR 0

#define STACK_INIT_SIZE 100 n");

}

while(1)

{

printf("1:Push \n2:Pop \n3:Get the Top \n4:Return the Length of the Stack\n5:Load the Stack\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d", &x);

if(!Push(S,x)) printf("Push Error!\n"); n");

while(1)

{

printf("1:Push \n2:Pop \n3:Get the Top \n4:Return the Length of the Stack\n5:Load the Stack\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1:

scanf("%d", &x);

(x);

printf("The Element %d is Successfully Pushed!\n", x);

break;

case 2:

if()) printf("Pop Error!\n"); int i;

i=;

if(______________________)printf("The Queue is Empty!"); n");

}

while(1)

{

printf("1:Enter \n2:Delete \n3:Get the Front \n4:Return the Length of the Queue\n5:Load the Queue\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d", &x);

if(______________________) printf("Enter Error!\n"); 1:Enter

2:Delete

3:Get the Front

4:Return the Length of the Queue

5:Load the Queue

0:Exit

Please choose:

The Element 1 is Successfully Entered!

1:Enter

2:Delete

3:Get the Front

4:Return the Length of the Queue

5:Load the Queue

0:Exit

Please choose:

The Element 2 is Successfully Entered! 1:Enter

2:Delete

3:Get the Front

4:Return the Length of the Queue 5:Load the Queue

0:Exit

Please choose:

The Element 3 is Successfully Entered! 1:Enter

2:Delete

3:Get the Front

4:Return the Length of the Queue 5:Load the Queue

0:Exit

Please choose:

The Queue is: 1 2 3

1:Enter

2:Delete

3:Get the Front

4:Return the Length of the Queue 5:Load the Queue

0:Exit

Please choose:

The Element 1 is Successfully Deleted! 1:Enter

3:Get the Front

4:Return the Length of the Queue 5:Load the Queue

0:Exit

Please choose:

The Head of the Queue is 2!

1:Enter

2:Delete

3:Get the Front

4:Return the Length of the Queue 5:Load the Queue

0:Exit

Please choose:

The Queue is: 2 3

1:Enter

2:Delete

3:Get the Front

4:Return the Length of the Queue 5:Load the Queue

0:Exit

Please choose:

作者

yqm

解法一：（正规解法）

#include<>

#include<>

#define OK 1

#define ERROR 0

typedef int Status; int i;

i=;

if==printf("The Queue is Empty!"); n");

}

while(1)

{

printf("1:Enter \n2:Delete \n3:Get the Front \n4:Return the Length of the Queue\n5:Load the Queue\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1: scanf("%d", &x);

if(!EnQueue(S,x)) printf("Enter Error!\n"); n");

while(1)

{

printf("1:Enter \n2:Delete \n3:Get the Front \n4:Return the Length of the Queue\n5:Load the Queue\n0:Exit\nPlease choose:\n");

scanf("%d",&a);

switch(a)

{

case 1:

scanf("%d", &x);

(x);

printf("The Element %d is Successfully Entered!\n", x);

break;

case 2:

if()) printf("Delete Error!\n");

lf",d*n);

return 0;

}

8591 计算next值

时间限制:1000MS 内存限制:1000K

提交次数:3227 通过次数:1617

题型: 编程题语言: 无限制

Description

编写算法，录入多个字符串计算并验证NEXT值，输入0结束。本题目给出部分代码，请补全内容。] #include ""

#include ""

#include ""

#define MAXSTRLEN 255 // 用户可在255以内定义最大串长

typedef unsigned char SString[MAXSTRLEN+1]; // 0号单元存放串的长度

void get_next(SString T,int next[]){

// 算法

// 求模式串T的next函数值并存入数组next

// 请补全代码