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关于C语言题602

发布时间 2023-06-06 20:09:49作者: 风中凌乱的猪头

1、编写函数void count(char a[],char w[][10],int n,int b[])。功能是：统计w指向的数组中的n个单词在a指向的字符串中各自出现的次数（将非字母字符看作单词分割符），拧将统计结果依次保存在b指向的数组中。

 

#include<stdio.h>
#include<string.h>

#define N 10
int main()
{
    void count(char a[],char w[][N],int  n,int b[]);
    char a[N*4];
    char w[N][N];
    int n;
    int i = 0;
    int b[N];
    
    printf("please input the num of strings:");
    scanf("%d",&n);
    printf("please input some strings:");
    for(;i<n;i++)
    {
        scanf("%s",w[i]);
    }
    getchar();
    printf("please input a string:");
    scanf("%[^\n]",a);

    count(a,w,n,b);

    for(i = 0;i < n;i++)
    {
        printf("%4d",b[i]);
    }
    printf("\n");
    return 0;
    
}

void count(char a[],char w[][N],int n,int b[])
{
    char temp[N][N];
    int i = 0;
    int j = 0;
    int k = 0;
    int z = 0;
    int count;
    while(a[i] != '\0')
    {
        if((a[i] <= 'z'&&a[i] >= 'a')||(a[i] <= 'Z'&&a[i] >= 'A'))
        {
            temp[j][k] = a[i];
            k++;
        }
        else
        {
            j++;
            k = 0;
        }

        i++;
    }
    for(i = 0;i < n;i++)
    {
        count = 0;
        for(k = 0;k <= j;k++)
        {
            if(strcmp(w[i],temp[k])==0)
            {
                count++;
            }
        }
        b[z] = count;
        z++;
    }
}

2、 编写函数int stat(int a[],int n,int c[][2])。a指向的数组中保存了由n个1位整数组成的数列（n为偶数）。函数从前至后依次将a数组中每两个相邻元素拼成一个不超过2位的整数，从而生成有n/2个元素组成的整数数列；统计该数列中不同整数各自出现的次数，并将统计结果保存到c指向的二维数组中。函数返回不同整数的个数。

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

#define N 30

int main()
{
    int stat(int a[],int n,int c[][2]);
    srand((unsigned)time(NULL));
    int a[N];
    int n;
    int i;
    int c[N/3][2] = {0};
    int b;
    printf("please input a number of integer(even number):");
    scanf("%d",&n);
    
    for(i = 0;i < n;i++)
    {
        a[i] = rand()%10;
    }

    for(i = 0;i < n;i++)
    {
        printf("%4d",a[i]);
    }
    printf("\n");

    b = stat(a,n,c);
    printf("there are %d numbers\n",b);

    for(i = 0;i < b;i++)
    {
        printf("%d : %d\n",c[i][0],c[i][1]);
    }
    
    return 0;

}

int stat(int a[],int n,int c[][2])
{
    int i = 0;
    int j = 0;
    int k = 0;
    for(;i<n;)
    {
        a[j] = a[i]*10+a[i+1];//直接放入a数组中
        j++;
        i = i+2;

    }
    for(i = 0;i < n/2;i++)
    {
        j = 0;
        for(;j < k;j++)
        {
            if(a[i]==c[j][0])
            {
                c[j][1]++;
                break;
            }
        }
        if(j == k)//放在一个for循环里就很好使
        {
            c[k][0] = a[i];
            c[k][1]++;
            k++;
        }
        

    }
    return k;
}

3、编写函数fun(int *a, int n, int *odd, int *even)，功能是：求出数组a[]中所有奇数之和以及所有偶数之和。并利用指针odd返回奇数之和，利用指针even返回偶数之和。 例如：a[]的值依次为：1，9，2，3，11，6；则利用指针odd返回奇数之和24；利用指针even 返回偶数之和 8。

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

#define N 20

int main()
{
    void fun(int *a,int n,int *odd,int *even);
    srand((unsigned)time(NULL));
    int n;
    int a[N];
    int i;
    int odd = 0,even = 0;
    printf("please input number:");
    scanf("%d",&n);

    for(i = 0;i < n;i++)
    {
        a[i] = rand()%20;
    }
    for(i = 0;i < n;i++)
    {
        printf("%4d",a[i]);
    }
    printf("\n");

    fun(a,n,&odd,&even);

    printf("odd sum is %d , even sum is %d\n",odd,even);
}
void fun(int *a,int n,int *odd,int *even)
{
    int i;
    for(i = 0;i < n;i++)
    {
        if(a[i]%2)
        {
            *odd+=a[i];
        }
        else
        {
            *even+=a[i];
        }
    }
}

4、 程序功能：建立一个带有头结点的单向链表，并将存储在数组中的字符依次转储到链表的各个结点中。

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

#define N 20

typedef struct Lnode
{
    char data;
    struct Lnode *next;
}LNode;

int main()
{
    char a[N];
    int i = 0;
    LNode *L = (LNode *)malloc(sizeof(LNode));
    L->next = NULL;
    LNode *p;
    p = L;
    printf("please input a string:");
    scanf("%s",a);

    while(a[i] != '\0')
    {
        LNode *q = (LNode *)malloc(sizeof(LNode));// 尾插法；
        if(!q)
        {
            exit(0);
        }
        q->data = a[i];
        p->next = q;
        p = p->next;
        i++;
    }
    p->next = NULL;

    p = L->next;
    while(p)
    {
        printf("%c ",p->data);
        p = p->next;
    }
    printf("\n");

    return 0;
}

5、编写程序STUDENT *Create(STUDENT studs[],int n)。STUDENT是一个结构类型，包含姓名、成绩和指针域。studs数组中存储了n个STUDENT记录。create函数的功能是根据studs数组建立一个链表，链表中结点按成绩降序排列，函数返回链表头指针。

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

#define N 20

typedef struct Lnode
{
    char name[20];
    int grade;
    struct Lnode* next;
}STUDENT;


int main()
{
    STUDENT *Create(STUDENT studs[],int n);
    STUDENT studs[N];
    STUDENT *p;
    int n;
    int i;
    printf("please input the number of students:");
    scanf("%d",&n);
    printf("please input the imformation:");
    for(i = 0;i < n;i++)
    {
        scanf("%s %d",studs[i].name,&studs[i].grade);
        getchar();
    }


   
    p = Create(studs,n);
   
    while(p)
    {
        printf("%20s%5d\n",p->name,p->grade);
        p = p->next;
    }
    printf("\n");


    return 0;

}

void INsert(STUDENT *L,STUDENT *s)
{
    STUDENT *p;
    STUDENT *q = NULL;
    p = L;
    s->next = NULL;
    while(p&&s->grade < p->grade)
    {
        q = p;
        p = p->next;
    }
    if(!p)
    {
        q->next = s;
    }
    else
    {
        q->next = s;
        s->next = p;
    }
}

STUDENT *Create(STUDENT studs[],int n)
{
    int i = 0;
    int max;
    int k;
    max = studs[0].grade;
    for(;i < n;i++)
    {
        if(max <= studs[i].grade)
        {
            max = studs[i].grade;
            k = i;
        }
    }
    studs[k].next = NULL;
    for(i = 0;i < n;i++)
    {
        if(i == k)
        {
            continue;
        }
        INsert(&studs[k],&studs[i]);
    }
    return &studs[k];
}

 

6、 题目:有n个人围成一圈,顺序排号,从第一个开始报数(从1到3报数),凡报到3的人退出圈子,问最后最后留下的是原来第几号的那位.

 

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

#define OK 1
#define ERROR 0

typedef struct  Lnode
{
    int data;
    struct Lnode *next;
    struct Lnode *prior;
}LNode;

int Init_L(LNode* L,int n)
{
    int i;
    LNode *q;
    q = L;
    for(i = 1; i <= n;i++)//尾插法
    {
        LNode *p = (LNode*)malloc(sizeof(LNode));
        if(!p)
        {
            return ERROR;
        }
        p->data = i;
        p->next = NULL;
        q->next = p;
        p->prior = q;
        q = q->next; 
    }
    q->next = L;//循环
    L->prior = q;

    return OK;  

}
void Output(LNode *L)
{
    LNode *p;
    p = L->next;
    while(p != L)
    {
        printf("%4d",p->data);
        p = p->next;
    }
    printf("\n");

}

int main()
{
    LNode *L = (LNode*)malloc(sizeof(LNode));
    L->next = L;
    L->prior = L;
    LNode *p = NULL;
    LNode *q = NULL;
    int n;
    int count = 0;
    printf("please input the number of people:");
    scanf("%d",&n);

    Init_L(L,n);
    p = L->next;
    Output(L);

    while(L->next->next != L)
    {
        count++;
        if(p == L)
        {
            count--;
        }
        if(count == 3)
        {
            count = 0;
            p->prior->next = p->next;
            p->next->prior = p->prior;
            q = p;
            p = p->next;
            free(q);
        }
        else
        {
            p = p->next;
        }
        
        
    }
    printf("the last one is %d\n",L->next->data);

    return 0;

}

7、 题目:创建两个学生链表,含有姓名、年龄的信息，一个链表存放男生，一个链表存放女生，将上面两个链表合并，按学生的年龄进行排序，合成新的链表，将上题中建立的链表进行反转，实现按年龄的逆序排列

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

#define ERROR 0
#define OK 1

typedef struct LNode
{
    char name[20];
    int age;
    struct LNode *next;
}LNode;

int main()
{
    int Init_L(LNode *L,char *name,int age);
    void Output(LNode *L);
    void Merge_L(LNode *L1,LNode *L2,LNode *L);
    void Verse(LNode *L);

    LNode *ML = (LNode*)malloc(sizeof(LNode));
    LNode *WL = (LNode*)malloc(sizeof(LNode));
    LNode *L = (LNode*)malloc(sizeof(LNode));
    ML->next = NULL;
    WL->next = NULL;
    L->next = NULL;

    int age;
    char a[20];
    char ch;
    while(1)
    {
        printf("input input the choice(m is man ,w is woman,q is quit):");
        scanf("%c",&ch);
        getchar();
        if(ch == 'q')
        {
            break;
        }
        printf("please input the imformation:");
        scanf("%s %d",a,&age);
        getchar();
        if(ch == 'm')
        {
            Init_L(ML,a,age);
        }
        else if(ch == 'w')
        {
            Init_L(WL,a,age);
        }
        
    }

    Output(ML);
    Output(WL);

    Merge_L(ML,WL,L);//合并按照年龄顺序从小到大把
    Output(L);
    //头插法反转，比较简单 ，这里写的是就地翻转
    Verse(L);
    Output(L);

    free(ML);
    free(WL);
    free(L);
    return 0;

}

//构造链表
int Init_L(LNode *L,char *name,int age)//头插法
{
    LNode *p = (LNode *)malloc(sizeof(LNode));
    if(!p)
    {
        return ERROR;
    }
    p->age = age;
    strcpy(p->name,name);

    p->next = L->next;
    L->next = p;
    return OK;

}

//输出
 void Output(LNode *L)
 {
    LNode *p;
    p = L->next;
    while(p)
    {
        printf("%15s  %5d\n",p->name,p->age);
        p = p->next;
    }
    printf("\n");
 }

//判断插入
int Insert_L(LNode *L,LNode *s)
 {
    LNode *p;
    if(!L->next)
    {
        L->next = s;
        s->next = NULL;
        return OK;
    }
    p = L;
    while(p->next&&s->age > p->next->age)
    {
        p = p->next;
    }
    if(!p->next)
    {
        p->next = s;
        s->next = NULL;
    }
    else
    {
        s->next = p->next;
        p->next = s;
    }

    return OK;
    
 }

//有序合并
 void Merge_L(LNode *L1,LNode *L2,LNode *L)
 {  
    int min;
    LNode *p1;
    LNode *p2;
    LNode *p;
    p1 = L1->next;
    p2 = L2->next;

    p = p1;
    while(p1)
    {
        p = p1;
        p1 = p1->next;
        Insert_L(L,p);  
    }
    p = p2;
    while(p2)
    {
        p = p2;
        p2 = p2->next;
        Insert_L(L,p);   
    }
   
 }

//就地反转
void Verse(LNode *L)
{
    LNode *p;
    LNode *q;
    p = L->next;
    while(p->next)
    {
        q = p->next;
        p->next = q->next;
        q->next = L->next;
        L->next = q;
    }
}

8、在上面的实现的新链表中，给定一个学生的年龄，迅速查找和该学生年龄最接近的学生姓名（和当前年龄差最小）使用双向链表

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

#define ERROR 0
#define OK 1

typedef struct LNode
{
    char name[20];
    int age;
    struct LNode *next;
    struct LNode *prior;
}LNode;

int main()
{
    int Init_L(LNode *L,char *name,int age);
    void Output(LNode *L);
    void Merge_L(LNode *L1,LNode *L2,LNode *L);
    void Verse(LNode *L);

    LNode *ML = (LNode*)malloc(sizeof(LNode));
    LNode *WL = (LNode*)malloc(sizeof(LNode));
    LNode *L = (LNode*)malloc(sizeof(LNode));
    ML->next = NULL;
    ML->prior = NULL;
    WL->next = NULL;
    WL->prior = NULL;
    L->next = NULL;
    L->next = NULL;

    int age;
    char a[20];
    char ch;
    int x;
    LNode *p;
    while(1)
    {
        printf("input input the choice(m is man ,w is woman,q is quit):");
        scanf("%c",&ch);
        getchar();
        if(ch == 'q')
        {
            break;
        }
        printf("please input the imformation:");
        scanf("%s %d",a,&age);
        getchar();
        if(ch == 'm')
        {
            Init_L(ML,a,age);
        }
        else if(ch == 'w')
        {
            Init_L(WL,a,age);
        }
        
    }

    Output(ML);
    Output(WL);

    Merge_L(ML,WL,L);//合并按照年龄顺序从小到大把
    Output(L);
    //头插法反转，比较简单 ，这里写的是就地翻转
    Verse(L);
    Output(L);

    printf("please input the age you want find:");
    scanf("%d",&x);
    p = L->next;

    while(p)
    {
        if(x == p->age)
        {
            break;
        }
        p = p->next;
    }
    if(p->next != NULL&&p->prior != NULL)
    {
        if((p->next->age-x)>(x-p->prior->age))
        {
            printf("%s\n",p->next->name);
        }
        else
        {
            printf("%s\n",p->prior->name);
        }
    }
    else if(p->next==NULL)
    {
        printf("%s\n",p->prior->name);
    }
    else
    {
        printf("%s\n",p->next->name);
    }
    
    free(ML);
    free(WL);
    free(L);
    return 0;

}

//构造链表
int Init_L(LNode *L,char *name,int age)//头插法
{
    LNode *p = (LNode *)malloc(sizeof(LNode));
    if(!p)
    {
        return ERROR;
    }
    p->age = age;
    strcpy(p->name,name);

    p->next = L->next;
    if(L->next != NULL)
    {
        L->next->prior = p;
    }
    L->next = p;
    p->prior = L;
    return OK;

}

//输出
 void Output(LNode *L)
 {
    LNode *p;
    p = L->next;
    while(p)
    {
        printf("%15s  %5d\n",p->name,p->age);
        p = p->next;
    }
    printf("\n");
 }

//判断插入
int Insert_L(LNode *L,LNode *s)
 {
    LNode *p;
    if(!L->next)
    {
        L->next = s;
        s->prior = L;
        s->next = NULL;
        return OK;
    }
    p = L;
    while(p->next&&s->age > p->next->age)
    {
        p = p->next;
    }
    if(!p->next)
    {
        p->next = s;
        s->next = NULL;
        s->prior = p;
    }
    else
    {
        s->next = p->next;
        p->next->prior = s;
        p->next = s;
        s->prior = p;
    }

    return OK;
    
 }

//有序合并
 void Merge_L(LNode *L1,LNode *L2,LNode *L)
 {  
    int min;
    LNode *p1;
    LNode *p2;
    LNode *p;
    p1 = L1->next;
    p2 = L2->next;

    p = p1;
    while(p1)
    {
        p = p1;
        p1 = p1->next;
        Insert_L(L,p);  
    }
    p = p2;
    while(p2)
    {
        p = p2;
        p2 = p2->next;
        Insert_L(L,p);   
    }
   
 }

//就地反转
void Verse(LNode *L)
{
    LNode *p;
    LNode *q;
    p = L->next;
    while(p->next)
    {
        q = p->next;
        p->next = q->next;
        if(q->next != NULL)
        {
            q->next->prior = p;
        }
        q->next = L->next;
        L->next->prior = q;
        L->next = q;
        q->prior = L;
    }
}

9、利用链表实现一个先入后出的栈结构,并提供栈操作的push和pop的接口（头插法最后进最先出）

#include<stdio.h>//头插法先进先出
#include<stdlib.h>//晚上改个标准的

#define Elemtype int

typedef struct SNode
{
    Elemtype data;
    struct SNode *next;
}SNode;

void push(SNode *L,Elemtype e)
{
    SNode *p = (SNode *)malloc(sizeof(SNode));
    p->data = e;
    p->next = L->next;
    L->next = p;
}

void pop(SNode *L,Elemtype *e)
{
    SNode *q;
    if(!L->next)
    {
        exit(0);
    }
    q = L->next;
    *e = q->data;
    L->next = q->next;
    free(q);
}

int main()
{
    SNode *L = (SNode *)malloc(sizeof(SNode));
    SNode *p = NULL;
    L->next = NULL;
    int i;
    for(i = 0;i < 5;i++)
    {
        push(L,i+1);
    }
    p = L->next;
    while(p)
    {
        printf("%4d",p->data);
        p = p->next;
    }
    printf("\n");
    
    while(L->next)
    {
        pop(L,&i);
        printf("%4d",i);
    }
    printf("\n");

    return 0;
}