C语言程序设计：现代设计方法习题笔记《chapter5》下篇

第七题

题目分析：求最大最小值转换为条件判断问题，最大值有四种可能，最小值相应有三种情况，给出下列代码。

示例代码：

#include <stdio.h>

int main() {
    int num1, num2, num3, num4; // 定义四个变量来存储输入的数字
    int max, min; // 定义变量来保存最大值和最小值

    printf("Enter four integere: ");
    scanf_s("%d %d %d %d", &num1, &num2, &num3, &num4);

    // 初始化最大值和最小值为第一个数字
    max = num1;
    min = num1;

    if (num2 > max || num3 > max || num4 > max) {
        max = num2 > num3 ? (num2 > num4 ? num2 : num4) : (num3 > num4 ? num3 : num4);
    }
    if (num2 < min || num3 < min || num4 < min) {
        min = num2 < num3 ? (num2 < num4 ? num2 : num4) : (num3 < num4 ? num3 : num4);
    }

    // 输出结果
    printf("Largest: %d\n", max);
    printf("Smallest: %d\n", min);

    return 0;
}

输出

第八题

题目分析：根据输入时间，匹配最接近的下一时刻，然后匹配对应的到达时间，得出代码。

示例代码

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

int main() {
    int hour, minu, input;
    char found[100], time[100];

    // 输入时间
    printf("Enter a 24-hour time: ");
    scanf_s("%d:%d", &hour, &minu);
    input = hour * 60 + minu;

    // 飞机起飞时间
    int fly_time1 = 8 * 60;
    int fly_time2 = 9 * 60 + 43;
    int fly_time3 = 11 * 60 + 19;
    int fly_time4 = 12 * 60 + 47;
    int fly_time5 = 14 * 60;
    int fly_time6 = 15 * 60 + 45;
    int fly_time7 = 19 * 60;
    int fly_time8 = 21 * 60 + 45;

    const char* time1 = "8:00 a.m.";
    const char* time2 = "9:43 a.m.";
    const char* time3 = "11:19 a.m.";
    const char* time4 = "12:47 p.m.";
    const char* time5 = "14:00 p.m.";
    const char* time6 = "15:45 p.m.";
    const char* time7 = "19:00 p.m.";
    const char* time8 = "21:45 p.m.";

    int closestGreater = -1; // 初始化为一个不可能的值

    // 检查每一个单独的值
    if (fly_time1 > input) {
        closestGreater = fly_time1;
        strcpy_s(time, time1);
        strcpy_s(found, "10:16 a.m.");
    }
    if (fly_time2 > input && (closestGreater == -1 || fly_time2 - input < closestGreater - input)) {
        closestGreater = fly_time2;
        strcpy_s(time, time2);
        strcpy_s(found, "11:52 a.m.");
    }
    if (fly_time3 > input && (closestGreater == -1 || fly_time3 - input < closestGreater - input)) {
        closestGreater = fly_time3;
        strcpy_s(time, time3);
        strcpy_s(found, "1:31 p.m.");
    }
    if (fly_time4 > input && (closestGreater == -1 || fly_time4 - input < closestGreater - input)) {
        closestGreater = fly_time4;
        strcpy_s(time, time4);
        strcpy_s(found, "3:00 p.m.");
    }
    if (fly_time5 > input && (closestGreater == -1 || fly_time5 - input < closestGreater - input)) {
        closestGreater = fly_time5;
        strcpy_s(time, time5);
        strcpy_s(found, "4:08 p.m.");
    }
    if (fly_time6 > input && (closestGreater == -1 || fly_time6 - input < closestGreater - input)) {
        closestGreater = fly_time6;
        strcpy_s(time, time6);
        strcpy_s(found, "5:05 p.m.");
    }
    if (fly_time7 > input && (closestGreater == -1 || fly_time7 - input < closestGreater - input)) {
        closestGreater = fly_time7;
        strcpy_s(time, time7);
        strcpy_s(found, "9:20 p.m.");
    }
    if (fly_time8 > input && (closestGreater == -1 || fly_time8 - input < closestGreater - input)) {
        closestGreater = fly_time8;
        strcpy_s(time, time8);
        strcpy_s(found, "11:58 p.m.");
    }

    if (closestGreater != -1) {
        printf("Closest departure time is %s, arriving at %s\n", time, found);
    }
    else {
        printf("没有找到大于输入时间的飞行时间。\n");
    }

    return 0;
}

输出

第九题

示例代码

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

// 函数声明
int compareDates(int m1, int d1, int y1, int m2, int d2, int y2);

int main() {
    int month1, day1, year1;
    int month2, day2, year2;

    // 提示用户输入第一个日期
    printf("Enter first date（mm/dd/yy）: ");
    if (scanf_s("%d/%d/%d", &month1, &day1, &year1) != 3) {
        printf("输入错误。\n");
        return 1;
    }

    // 提示用户输入第二个日期
    printf("Enter second date（mm/dd/yy）: ");
    if (scanf_s("%d/%d/%d", &month2, &day2, &year2) != 3) {
        printf("输入错误。\n");
        return 1;
    }

    // 比较两个日期
    if (compareDates(month1, day1, year1, month2, day2, year2)) {
        printf("%02d/%02d/%02d is earlier than %02d/%02d/%02d\n", month1, day1, year1, month2, day2, year2);
    }
    else {
        printf("%02d/%02d/%02d is earlier than %02d/%02d/%02d\n", month2, day2, year2, month1, day1, year1);
    }

    return 0;
}

// 比较两个日期，返回 1 表示第一个日期更早，0 表示第二个日期更早
int compareDates(int m1, int d1, int y1, int m2, int d2, int y2) {
    if (y1 < y2 || (y1 == y2 && m1 < m2) || (y1 == y2 && m1 == m2 && d1 < d2)) {
        return 1;
    }
    else if (y1 > y2 || (y1 == y2 && m1 > m2) || (y1 == y2 && m1 == m2 && d1 > d2)) {
        return 0;
    }
    else {
        return 0; // 如果两个日期相同
    }
}

输出

第十题

示例代码

#include <stdio.h>

int main() {
    int score; // 存储成绩分数

    // 提示用户输入成绩分数
    printf("Enter a numerical grade: ");
    scanf_s("%d", &score);
    // 检查输入的成绩分数是否在合理范围内
    if (score < 0 || score > 100) {
        printf("错误：成绩分数应在 0 到 100 之间。\n");
        return 1; // 返回错误代码
    }
    // 使用 switch 语句判断成绩等级
    switch (score / 10) {
    case 10:
        printf("Letter grade: A\n");
        break;
    case 9:       
        printf("Letter grade: A\n");
        break;
    case 8:
        printf("Letter grade: B\n");
        break;
    case 7:
        printf("Letter grade: C\n");
        break;
    case 6:
        printf("Letter grade: D\n");
        break;
    default:
        printf("Letter grade: F\n");
        break;
    }

    return 0;
}

输出

第十一题

示例代码

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

int main() {
    int number; // 存储输入的两位数
    char result[50] = ""; // 存储英文单词的结果

    // 提示用户输入一个两位数
    printf("Enter a two-digit number: ");
    if (scanf_s("%d", &number) != 1) {
        printf("输入错误。\n");
        return 1;
    }

    // 检查输入的数字是否在合理范围内
    if (number < 10 || number > 99) {
        printf("错误：输入的数字应为两位数（10-99）。\n");
        return 1; // 返回错误代码
    }

    // 获取十位和个位上的数字
    int tens = number / 10;
    int ones = number % 10;

    // 特殊处理 11 到 19 的情况
    if (tens == 1 && ones >= 1) {
        switch (number) {
        case 11:
            strcpy_s(result, sizeof(result), "eleven");
            break;
        case 12:
            strcpy_s(result, sizeof(result), "twelve");
            break;
        case 13:
            strcpy_s(result, sizeof(result), "thirteen");
            break;
        case 14:
            strcpy_s(result, sizeof(result), "fourteen");
            break;
        case 15:
            strcpy_s(result, sizeof(result), "fifteen");
            break;
        case 16:
            strcpy_s(result, sizeof(result), "sixteen");
            break;
        case 17:
            strcpy_s(result, sizeof(result), "seventeen");
            break;
        case 18:
            strcpy_s(result, sizeof(result), "eighteen");
            break;
        case 19:
            strcpy_s(result, sizeof(result), "nineteen");
            break;
        }
    }
    else {
        // 处理其他情况
        switch (tens) {
        case 2:
            strcpy_s(result, sizeof(result), "twenty");
            break;
        case 3:
            strcpy_s(result, sizeof(result), "thirty");
            break;
        case 4:
            strcpy_s(result, sizeof(result), "forty");
            break;
        case 5:
            strcpy_s(result, sizeof(result), "fifty");
            break;
        case 6:
            strcpy_s(result, sizeof(result), "sixty");
            break;
        case 7:
            strcpy_s(result, sizeof(result), "seventy");
            break;
        case 8:
            strcpy_s(result, sizeof(result), "eighty");
            break;
        case 9:
            strcpy_s(result, sizeof(result), "ninety");
            break;
        }

        // 如果个位数不为零，追加连字符和个位数的英文单词
        if (ones != 0) {
            char temp[10];
            strcpy_s(temp, sizeof(temp), "-");
            strcat_s(result, sizeof(result), temp);

            switch (ones) {
            case 1:
                strcat_s(result, sizeof(result), "one");
                break;
            case 2:
                strcat_s(result, sizeof(result), "two");
                break;
            case 3:
                strcat_s(result, sizeof(result), "three");
                break;
            case 4:
                strcat_s(result, sizeof(result), "four");
                break;
            case 5:
                strcat_s(result, sizeof(result), "five");
                break;
            case 6:
                strcat_s(result, sizeof(result), "six");
                break;
            case 7:
                strcat_s(result, sizeof(result), "seven");
                break;
            case 8:
                strcat_s(result, sizeof(result), "eight");
                break;
            case 9:
                strcat_s(result, sizeof(result), "nine");
                break;
            }
        }
    }

    // 对于 10 的情况
    if (number == 10) {
        strcpy_s(result, sizeof(result), "ten");
    }

    // 输出结果
    printf("You entered the number：%s\n", result);

    return 0;
}

输出