【C语言】学习笔记第13章字符串编程题

ゝ一世哀愁。 2024-03-22 06:39 11阅读 0赞

### 【C语言】学习笔记 ###

#### 文章目录 ####

*   *  【C语言】学习笔记
     *   *  第13章 字符串
         *   *  编程题

#### 第13章 字符串 ####

![在这里插入图片描述][3f299ec085fd460abac11fdba109628c.png_pic_center]

字符串总是“板着一副面孔”，但它是我们唯一能指望的交流纽带。

##### 编程题 #####

【1】编写程序找出一组单词中“最小”单词和“最大”单词。用户输入单词后，程序根据字典顺序决定 排在最前面和最后面的单词。当用户输入 4 个字母的单词时，程序停止读入。假设所有单词都不超 过 20 个字母。程序会话如下：

Enter word: dog
    Enter word: zebra
    Enter word: rabbit
    Enter word: catfish
    Enter word: walrus
    Enter word: cat
    Enter word: fish
    
    Smallest word: cat 
    Largest word: zebra

提示：使用两个名为 smallest\_word 和 largest\_word 的字符串来分别记录所有输入中的“最小” 单词和“最大”单词。用户每输入一个新单词，都要用 strcmp 函数把它与 smallest\_word 进行比 较。如果新的单词比 smallest\_word“小”，就用 strcpy 函数把新单词保存到 smallest\_word 中。用类似的方式与 larges\_word 进行比较。用 strlen 函数来判断用户是否输入了 4 个字母的 单词。

#include <stdio.h>
    #include <ctype.h>
    #include <string.h>
    #include <stdbool.h>
    
    int main(void) {
        
    
        char smallest_word[20], largest_word[20], temp[20];
    
        printf("Enter word: ");
        scanf("%s", temp);
    
        strcpy(smallest_word, temp);
    
        while (strlen(temp) != 4) {
        
            if (strcmp(smallest_word, temp) > 0) strcpy(smallest_word, temp);
            if (strcmp(largest_word, temp) < 0) strcpy(largest_word, temp);
    
            printf("Enter word: ");
            scanf("%s", temp);
        }
    
        printf("Smallest word: %s\n", smallest_word);
        printf("Largest word: %s\n", largest_word);
    
        return 0;
    }

![在这里插入图片描述][08159649d0664891946874224e9dcb8c.png_pic_center]

【2】按如下方式改进 13.5 节的 remind.c 程序。

(a) 如果对应的日期为负数或大于 31，程序显示出错消息，并忽略提醒。提示：使用 continue 语句。

(b) 允许用户输入日期、24 小时格式的时间和提醒。显示的提醒列表必须先按日期排序，然后再按 时间排序。（原始的 remind.c 程序允许用户输入时间，但是它把时间作为提醒的一部分来处理。）

#include <stdio.h> 
    #include <string.h> 
    
    #define MAX_REMIND 50 /* maximum number of reminders */ 
    #define MSG_LEN 60 /* max length of reminder message */ 
    
    int read_line(char str[], int n);
    
    int main(void)
    {
        
    
        char reminders[MAX_REMIND][MSG_LEN + 3];
        int msg_str[MSG_LEN + 1];
    
        int month, day, i, j, num_remind = 0;
    
        int hour, minute;
        char time_str[12];
    
        for (;;) {
        
    
            if (num_remind == MAX_REMIND) {
        
                printf("-- No space left --\n");
                break;
            }
    
            printf("Enter day and reminder: ");
            scanf("%2d/%2d", &month, &day);
            scanf("%2d:%2d", &hour, &minute);
    
            if (day == 0)
                break;
    
            if (day < 0 || day > 3) {
        
                printf("\nIgnore.\n");
                while (getchar() != '\n');
                continue;
            }
    
    
    
            sprintf(time_str, "%2d/%2d %2d:%2d", month, day, hour, minute);
    
            read_line(msg_str, MSG_LEN);
    
            for (i = 0; i < num_remind; i++)
                if (strcmp(time_str, reminders[i]) < 0)
                    break;
    
            for (j = num_remind; j > i; j--)
                strcpy(reminders[j], reminders[j - 1]);
    
            strcpy(reminders[i], time_str);
            strcat(reminders[i], msg_str);
            num_remind++;
        }
    
        printf("\nDay Reminder\n");
    
        for (i = 0; i < num_remind; i++)
            printf(" %s\n", reminders[i]);
    
        return 0;
    }
    
    int read_line(char str[], int n)
    {
        
        int ch, i = 0;
    
        while ((ch = getchar()) != '\n')
            if (i < n)
                str[i++] = ch;
        str[i] = '\0';
    
        return i;
    }

【3】修改 8.2 节的 deal.c 程序，使它显示出牌的全名：

Enter number of cards in hand: 5
    Your hand: 
    Seven of clubs 
    Two of spades 
    Five of diamonds 
    Ace of spades 
    Two of hearts

提示：用指向字符串的指针的数组来替换数组 rank\_code 和数组 suit\_code。

#include <stdbool.h> 
    #include <stdio.h> 
    #include <stdlib.h> 
    #include <time.h> 
    
    #define NUM_SUITS 4 
    #define NUM_RANKS 13 
    
    int main(void)
    {
        
    
        bool in_hand[NUM_SUITS][NUM_RANKS] = {
         false };
        int num_cards, rank, suit;
    
        const char rank_code[][8] = {
         "Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten","Jack","Queen","King","Ace" }; // 点数
        const char suit_code[][8] = {
         "clubs","diamond","hearts","spades" }; // 花色
    
        srand((unsigned)time(NULL)); // time函数返回当前的时间, 用一个数表示; srand 函数初始化C语言的随机数生成器【避免每次运行程序都发到一样的牌】
    
        printf("Enter number of cards in hand: ");
        scanf("%d", &num_cards);
    
        printf("Your hand:");
    
        while (num_cards > 0) {
        
    
            suit = rand() % NUM_SUITS; /* picks a random suit */
            rank = rand() % NUM_RANKS; /* picks a random rank */
    
            if (!in_hand[suit][rank]) {
        
                in_hand[suit][rank] = true;
                num_cards--;
                printf(" %s of %s\n", rank_code[rank], suit_code[suit]);
            }
        }
    
        printf("\n");
        return 0;
    }

![在这里插入图片描述][ef4409892c1d4c0b9e9af672ac201b1e.png_pic_center]

【4】编写名为 reverse.c 的程序，用来逆序输出命令行参数。如果输入

reverse void and null

运行程序，产生的输出应为

null and void

#include <stdio.h>
    
    int main(int argc, char* argv[])
    {
        
    
        for (int i = argc - 1; i > 0; i--) {
        
            printf(" %s ", argv[i]);
        }
    
        printf("\n");
    
        return 0;
    }

![在这里插入图片描述][6e0dd68e316a4c69a72efc44a2ba7ef0.png_pic_center]

【5】编写名为 sum.c 的程序，用来对命令行参数（假设都是整数）求和。如果输入

sum 8 24 62

运行程序，产生的输出应为

Total: 94

提示：用 atoi 函数把每个命令行参数从字符串格式转换为整数格式。

#include <stdio.h>
    
    int main(int argc, char const* argv[])
    {
        
    
        int result = 0;
    
        for (int i = 0; i < argc; i++) {
        
            result += atoi(argv[i]);
        }
    
        printf("Result: %d\n", result);
    
        return 0;
    }

![在这里插入图片描述][bd34721627b94c19a9fe07327854f15b.png_pic_center]

【6】改进 13.7 节的程序 planet.c，使它在对命令行参数和 planets 数组中的字符串进行比较时忽略大 小写。

#include <stdio.h> 
    #include <string.h> 
    #include <ctype.h>
    
    #define NUM_PLANETS 9 
    
    int main(int argc, char* argv[])
    {
        
    
        char* planets[] = {
         "Mercury", "Venus", "Earth","Mars", "Jupiter", "Saturn", "Uranus", "Neptune", "Pluto" };
        int i, j;
    
        for (i = 1; i < argc; i++) {
        
            argv[i][0] = toupper(argv[i][0]);
    
            for (j = 0; j < strlen(argv[i]); j++)
                argv[i][j] = tolower(argv[i][j]);
        }
    
        for (i = 1; i < argc; i++) {
        
            for (j = 0; j < NUM_PLANETS; j++)
                if (strcmp(argv[i], planets[j]) == 0) {
        
                    printf("%s is planet %d\n", argv[i], j + 1);
                    break;
                }
            if (j == NUM_PLANETS)
                printf("%s is not a planet\n", argv[i]);
        }
    
        return 0;
    }

![在这里插入图片描述][e4e438f8060343bd9300822e74f875e1.png_pic_center]

【7】修改第 5章的编程题 11，用字符串指针数组取代 switch 语句。例如，现在不再用 switch 语句来显示 第一位数字对应的单词，而把该数字用作下标，从包含"twenty"、"thirty"等字符串的数组中搜索。

#include<stdio.h>
    
    
    int main(void) {
        
    
        int number;
    
        const char digit_one[][10] = {
         "","One","Two","Three","Four","Five","Six","Seven","Eight","Nine" };
        const char digit_two[][10] = {
         "","","Twenty-","Thirthy-","Forty-","Fifty-","Sixty-","Seventy-","Eighty-","Ninety" };
        const char ten[][10] = {
         "Ten","Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen" };
    
        int one_digit, two_digit;
    
        printf("Enter a two-digit number: ");
        scanf("%d", &number);
    
        printf("You entered the number ");
    
        if (number >= 20 || number < 10) {
        
            two_digit = number / 10;
            one_digit = number % 10;
            printf("%s%s", digit_two[two_digit], digit_one[one_digit]);
        }
        else {
        
            printf("%s", ten[number - 10]);
        }
    
        return 0;
    }

![在这里插入图片描述][2ab79d26b560405fb683f21ab866c487.png_pic_center]

【8】修改第 7 章的编程题 5，使其包含如下函数：

int compute_scrabble_value(const char *word);

函数返回 word 所指向的字符串的拼字值。

#include <stdio.h>
    
    int compute_scrabble_value(const char* word);
    
    int main(void) {
        
    
        char words[100];
    
        printf("Enter a word: ");
        scanf("%s", words);
    
        printf("Scrabble value: %d", compute_scrabble_value(words));
    
        return 0;
    }
    
    int compute_scrabble_value(const char* word) {
        
    
        int i = 0;
        char c = *(word + i++);
        int sum = 0;
    
        while (c != '\0') {
        
            c = toupper(c);
            if (c == 'D' || c == 'G') sum += 2;
            else if (c == 'J' || c == 'X') sum += 8;
            else if (c == 'Q' || c == 'Z') sum += 10;
            else if (c == 'K') sum += 5;
            else if (c == 'B' || c == 'C' || c == 'M' || c == 'P') sum += 3;
            else if (c == 'F' || c == 'H' || c == 'V' || c == 'W' || c == 'Y') sum += 4;
            else sum += 1;
    
            c = *(word + i++);
        }
    
        return sum;
    
    }

![在这里插入图片描述][4fe9fcf30cdd49f59d233171b9221f66.png_pic_center]

【9】修改第 7 章的编程题 10，使其包含如下函数：

int compute_vowel_count(const char *sentence);

函数返回 sentence 所指向的字符串中元音字母的个数。

#include <stdio.h>
    #include <ctype.h>
    
    int read_line(char str[], int n);
    int compute_vowel_count(const char* sentence);
    
    int main(void) {
        
    
        char sentence[100];
    
        printf("Enter a sentence: ");
    
        read_line(sentence, 100);
        puts(sentence);
    
        printf("Your sentence contains %d vowels.", compute_vowel_count(sentence));
    
        return 0;
    }
    
    int read_line(char str[], int n) {
        
    
        int ch, i = 0;
    
        while ((ch = getchar()) != '\n' && ch != EOF)
            if (i < n)
                str[i++] = ch;
    
        str[i] = '\0';
        return i;
    }
    
    int compute_vowel_count(const char* sentence) {
        
    
        int counter = 0;
    
        int i = 0;
    
        char c = *(sentence + i++);
    
        while (c != '\0') {
        
            c = toupper(c);
    
            if (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U') counter++;
    
            c = *(sentence + i++);
        }
        return counter;
    }

![在这里插入图片描述][664c025c44c541b68b24624cabea0684.png_pic_center]

【10】修改第 7 章的编程题 11，使其包含如下函数：

void reverse_name(char *name);

在参数 name 指向的字符串中，名在前、姓在后。在修改后的字符串中，姓在前，其后跟一个逗号和 一个空格，然后是名的首字母，最后加一个点。原始的字符串中，名的前面、名和姓之间、姓的后 面都可以有额外的空格。

#include <stdio.h>
    #include <string.h>
    
    void reverse_name(char* name);
    int read_line(char str[], int n);
    
    int main(void) {
        
    
        char name[100];
    
        printf("Enter a first and last name: ");
    
        read_line(name, 100);
        reverse_name(name);
    
        printf("%s\n", name);
    
        return 0;
    }
    
    void reverse_name(char* name) {
        
        char fname[2];
        char lname[strlen(name)];
    
        char* p = name;
    
        while (*p++ == ' ')
            fname[0] = *p;
    
        while (*p++ != ' ');
    
        strcpy(lname, p);
        strcpy(name, lname);
        strcat(name, ".");
        strcat(name, fname);
    }
    
    int read_line(char str[], int n) {
        
    
        int ch, i = 0;
    
        while ((ch = getchar()) != '\n' && ch != EOF)
            if (i < n)
                str[i++] = ch;
    
        str[i] = '\0';
        return i;
    }

【11】修改第 7 章的编程题 13，使其包含如下函数：

double compute_average_word_length(const char *sentence);

函数返回 sentence 所指向的字符串中单词的平均长度。

#include <stdio.h>
    
    double compute_average_word_length(const char* sentence);
    int read_line(char str[], int n);
    
    int main(void) {
        
    
        char str[100];
    
        printf("Enter a sentence: ");
        read_line(str, 100);
    
        printf("Average word length: %.1f\n", compute_average_word_length(str));
    
        return 0;
    }
    
    double compute_average_word_length(const char* sentence) {
        
    
        int words = 1;
        int allchars = 0;
        int i = 0;
    
        char c = *sentence;
    
        while (c != '\0') {
        
            if (c == ' ') words++;
            else allchars++;
    
            i++;
            c = *(sentence + i);
        }
    
        return 1.0 * allchars / words;
    }
    
    int read_line(char str[], int n) {
        
    
        int ch, i = 0;
    
        while ((ch = getchar()) != '\n' && ch != EOF)
            if (i < n)
                str[i++] = ch;
    
        str[i] = '\0';
        return i;
    }

![在这里插入图片描述][da9d72be6c3248f8bf0b1e0e3c6ab0d8.png_pic_center]

【12】修改第 8 章的编程题 14，读取句子时把单词存储在一个二维的 char 类型数组中，每行存储一个单 词。假定句子中的单词数不超过30，且每个单词的长度都不超过20个字符。注意，要在每个单词的 后面存储一个空字符，使其可以作为字符串处理。

#include <stdio.h>
    
    int main(void) {
        
    
        char words[30][20];
        char c;
    
        int i = 0, pos = 0;
    
    
        printf("Enter a sentence: ");
        c = getchar();
    
        while (c != '.' && c != '!' && c != '?') {
        
            if (c != ' ') {
        
                words[pos][i++] = c;
            }
            else {
        
                words[pos][i] = '\0';
                pos++;
                i = 0;
            }
            c = getchar();
        }
    
        printf("Reversal of sentence: ");
    
        while (pos >= 0) {
        
            printf("%s ", words[pos--]);
        }
    
        printf("%c", c);
    
        return 0;
    }

![在这里插入图片描述][6fa91f309bb4447aa10c0dddb165ae0b.png_pic_center]

【13】修改第 8 章的编程题 15，使其包含如下函数：

void encrypt(char *message, int shift);

参数 message 指向一个包含待加密消息的字符串，shift 表示消息中每个字母需要移动的位数。

#include <stdio.h>
    #include <string.h>
    
    void encrypt(char* message, int shift);
    
    int main(void) {
        
    
        char message[80];
        char ch;
        int n;
        int length = 0;
    
        printf("Enter message to be encrypted: ");
        while ((ch = getchar()) != '\n') message[length++] = ch;
    
        printf("Enter shift amount (1-25): ");
        scanf("%d", &n);
    
        encrypt(message, n);
    
        printf("Encrypted message: %s", message);
    
        return 0;
    }
    
    void encrypt(char* message, int shift) {
        
    
        int length = strlen(message);
        for (int i = 0; i < length; i++) {
        
            if (message[i] >= 'a' && message[i] <= 'z') {
        
                message[i] = ((message[i] - 'a') + shift) % 26 + 'a';
            }
            else if (message[i] >= 'A' && message[i] <= 'Z') {
        
                message[i] = ((message[i] - 'A') + shift) % 26 + 'A';
            }
        }
    }

![在这里插入图片描述][a6766279e94f49c7bda67cf9244b1208.png_pic_center]

【14】修改第 8 章的编程题 16，使其包含如下函数：

bool are_anagrams(const char *word1, const char *word2);

如果 word1 和 word2 指向的字符串是变位词，函数返回 true。

#include <stdio.h>
    #include <ctype.h>
    #include <stdbool.h>
    #include <string.h>
    
    bool are_anagrams(const char* word1, const char* word2);
    
    int main(void) {
        
    
        char word1[20], word2[20];
    
        printf("Enter first word: ");
        scanf("%s", word1);
    
        printf("Enter second word: ");
        scanf("%s", word2);
    
        if (are_anagrams(word1, word2)) {
        
            printf("The words are anagrams.");
        }
        else {
        
            printf("The words are NOT anagrams.");
        }
    
        return 0;
    }
    
    bool are_anagrams(const char* word1, const char* word2) {
        
    
        int words[26] = {
         0 };
    
        for (int i = 0; i < strlen(word1); i++) {
        
            if (isalpha(*(word1 + i))) words[toupper(*(word1 + i)) - 'A']++;
        }
    
        for (int i = 0; i < strlen(word2); i++) {
        
            if (isalpha(*(word2 + i))) words[toupper(*(word2 + i)) - 'A']--;
        }
    
        for (int i = 0; i < 26;i++) {
        
            if (words[i] != 0) return false;
        }
    
        return true;
    }

![在这里插入图片描述][aada88df09ff4a1fa9fd1bda63bf1c47.png_pic_center]

【15】修改第 10 章的编程题 6，使其包含如下函数：

int evaluate_RPN_expression(const char *expression);

函数返回 expression 指向的 RPN 表达式的值。

#include <stdbool.h>
    #include <stdio.h>
    #include <stdlib.h>
    
    #define STACK_SIZE 100
    
    int contents[STACK_SIZE];
    int top = 0;
    
    void make_empty(void);
    bool is_empty(void);
    bool is_full(void);
    void push(int i);
    int pop(void);
    int read_line(char str[], int n);
    int evaluate_RPN_expression(const char* expression);
    
    int main(void) {
        
    
        char exp[50];
    
        printf("Enter an RPN expression: ");
    
        read_line(exp, 50);
        printf("Result: %d", evaluate_RPN_expression(exp));
    
        return 0;
    }
    
    int evaluate_RPN_expression(const char* expression) {
        
    
        char ch = *expression;
    
        int i = 0;
        int result = 0;
    
        int input, first, second;
    
        while (ch != '\n') {
        
            if ((ch - '0') > 0 && (ch - '9' <= 0)) {
        
                input = ch - '0';
                i++;
                ch = *(expression + i);
                while ((ch - '0') >= 0 && (ch - '9' <= 0)) {
        
                    input = input * 10 + ch - '0';
                    i++;
                    ch = *(expression + i);
                }
                push(input);
                continue;
            }
            else if (ch == '+') push(pop() + pop());
            else if (ch == '*') push(pop() * pop());
            else if (ch == '-') {
        
                second = pop();
                first = pop();
                push(first - second);
            }
            else if (ch == '/') {
        
                second = pop();
                first = pop();
                push(first / second);
            }
            else if (ch == '=') {
        
                result = pop();
                if (is_empty()) {
        
                    return result;
                }
                else {
        
                    printf("Error!");
                    return 0;
                }
            }
    
            i++;
            ch = *(expression + i);
        }
    }
    
    void make_empty(void)
    {
        
        top = 0;
    }
    
    bool is_empty(void)
    {
        
        return top == 0;
    }
    
    bool is_full(void)
    {
        
        return top == STACK_SIZE;
    }
    
    
    void push(int i)
    {
        
        if (is_full()) {
        
            printf("Expression is too complex\n");
            exit(0);
        }
        else
            contents[top++] = i;
    }
    
    int pop(void)
    {
        
        if (is_empty()) {
        
            printf("Not enough operands in expression\n");
            exit(0);
        }
        else
            return contents[--top];
    }
    
    int read_line(char str[], int n) {
        
    
        int ch, i = 0;
    
        while ((ch = getchar()) != '\n' && ch != EOF)
            if (i < n)
                str[i++] = ch;
    
        str[i] = '\0';
        return i;
    }

![在这里插入图片描述][def7fe54e9374213a4458648681c6a18.png_pic_center]

【16】修改第 12 章的编程题 1，使其包含如下函数：

void reverse(char *message);

函数的作用是反转 message 指向的字符串。提示：使用两个指针，初始时一个指向字符串的第一个 字符，另一个指向最后一个字符；交换这两个字符，然后让两个指针相向移动；重复这一过程直到 两个指针相遇。

#include <stdio.h>
    #include <string.h>
    
    #define MAX 100
    
    void reverse(char* message);
    
    int main() {
        
    
        char str[MAX];
        char ch;
    
        int i = 0;
        printf("Enter a message: ");
    
        while ((ch = getchar()) != '\n' && i < MAX)
            str[i++] = ch;
    
        reverse(str);
    
        printf("Reversal is: %s", str);
        return 0;
    }
    
    void reverse(char* message) {
        
    
        char* phead = message;
        char* ptail = message + strlen(message) - 1;
    
        char temp;
    
        while (phead != ptail) {
        
            temp = *phead;
            *phead = *ptail;
            *ptail = temp;
            phead++;
            ptail--;
        }
    }

【17】修改第 12 章的编程题 2，使其包含如下函数：

bool is_palindrome(const char *message);

如果 message 指向的字符串是回文，函数返回 true。

#include <stdio.h>
    #include <stdbool.h>
    #include <ctype.h>
    #include <string.h>
    
    #define MAX 100
    
    bool is_palindrome(const char* message);
    
    int main(void) {
        
    
        char str[MAX];
        char ch;
        int i = 0;
    
        printf("Enter a message: ");
    
        while ((ch = getchar()) != '\n' && i < MAX) {
        
            if (isalpha(ch)) str[i++] = ch;
        }
    
        if (is_palindrome(str))
            printf("Palindrome.\n");
        else printf("Not a palindrome.\n");
    
        return 0;
    }
    
    bool is_palindrome(const char* message) {
        
    
        int i = 0;
        int n = strlen(message);
    
        while (i < n / 2) {
        
            if (toupper(*(message + i)) != toupper(*(message + n - 1 - i)))
                return false;
            i++;
        }
    
        return true;
    }

![在这里插入图片描述][644498f212f74430884ec59c17e3f619.png_pic_center]

【18】编写程序，按“月/日/年”的格式接受用户输入的日期，然后按“月 日，年”的格式显示，其中 “月”用英文全名：

Enter a date (mm/dd/yyyy): 2/17/2011
    You entered the date February 17, 2011

用字符串指针数组存储月份的名字。

#include <stdio.h>
    #include <stdbool.h>
    #include <ctype.h>
    #include <string.h>
    
    #define MAX 100
    
    char Month[][12] = {
         "January","February","March","April","May","June","July","August","September","October","November","December" };
    
    int main(void) {
        
    
        int month, day, year;
    
        printf("Enter a date (mm/dd/yyyy): ");
        scanf("%d/%d/%d", &month, &day, &year);
    
        printf("You entered the date %s %d, %d", Month[month - 1], day, year);
    
        return 0;
    
    }

![在这里插入图片描述][d54e82aa95a74622acfa594cadf01e23.png_pic_center]

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