三天打渔两天晒网

古城微笑少年丶 2022-10-08 15:18 132阅读 0赞

[题目要求]
    某人三天打渔两天晒网，假设他从1990年1月1日开始打渔三天，然后晒网两天，请编程回答任意的一天他在打渔还是晒网。
    A boy works for 3 days while has a 2 days off. If he is working on 1st, Jan, 1990, then for a date entered from the keyboard, please write a program to determine what the boy is doing, working or resting?
    Examples of input and output:
    1)Input:
    1990-01-05
    Output:
    He is having a rest.
    2)Input:
    1990-01-07
    Output:
    He is working.
    3)Input:
    1990-01-33
    Output:
    Invalid input.
    ***输入数据格式***："%4d-%2d-%2d"
    ***输出数据格式***："Invalid input."或"He is having a rest." 或"He is working."

\[程序代码\]

#include<stdio.h>
    #define OK 1
    #define ERROR -1
    #define Yes 1
    #define No 0
    #define working 1
    #define resting 0
    typedef short Status;
    struct Date/* 日期信息 */
    {
        short year;
        short month;
        short day;
    };
    /* */
    Status IsLeapYear(short year);
    Status Input(struct Date *d1);
    short LeapYearCount(short year1, short year2);
    Status WorkingOrResting(struct Date *d1);
    /* */
    int main()
    {
        struct Date date;
        if(Input(&date) == OK)
        {
            /* 输入信息合法 */
            if(WorkingOrResting(&date) == working)
            {
                fprintf(stdout, "He is working.");
            }
            else
            {
                fprintf(stdout, "He is having a rest.");
            }
        }
        return 0;
    }
    /* 判断是打渔还是晒网 */
    Status WorkingOrResting(struct Date *d1)/* 要求给定信息合法 */
    {
        short leapyearInfo[12] = {0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335};
        short commonyearInfo[12] = {0 ,31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
        short year = d1 -> year;
        short month = d1 -> month;
        short day = d1 -> day;
        /* 用这个年份和1990年1月1日作比较 */
        short leapyearcount = LeapYearCount(1990, year);/* 求解1990年到给定年份间的闰年数 */
        short commonyearcount = year - 1990 - leapyearcount;/* 求解1990年到给定年份间的平年数 */
        long s;
        if(IsLeapYear(year) == Yes)
        {
            s = 365 * commonyearcount + 366 * leapyearcount + leapyearInfo[month - 1] + day - 1;
        }
        else
        {
            s = 365 * commonyearcount + 366 * leapyearcount + commonyearInfo[month - 1] + day - 1;
        }
        if(s % 5 <= 2)
        {
            return working;
        }
        else
        {
            return resting;
        }
    }
    /* 判断两个年份间有几个闰年年份 */
    short LeapYearCount(short year1, short year2)/* year <= year2 */
    {
        short y = year1;
        short leapyearcount = 0;
        while(y < year2)
        {
            if(IsLeapYear(y) == Yes)
            {
                leapyearcount ++;
            }
            y ++;
        }
        return leapyearcount;
    }
    /* 判断给定年份是否为闰年 */
    Status IsLeapYear(short year)
    {
        if(year % 4 == 0 && year % 100 != 0 || year % 400 == 0)
        {
            /* 是闰年 */
            return Yes;
        }
        else
        {
            /* 不是闰年 */
            return No;
        }
    }
    /* 输入日期信息 */
    Status Input(struct Date* d1)
    {
        fscanf(stdin, "%4d-%2d-%2d", &d1 -> year, &d1 -> month, &d1 -> day);
        if(d1 -> year < 1990)
        {
            fprintf(stdout, "Invalid input.");
            return ERROR;
        }
        if(d1 -> month < 1 || d1 -> month > 12)
        {
            fprintf(stdout, "Invalid input.");
            return ERROR;
        }
        if(IsLeapYear(d1 -> year) == Yes && d1 -> day > 31 || d1 -> day < 1)
        {
            fprintf(stdout, "Invalid input.");
            return ERROR;
        }
        if(IsLeapYear(d1 -> year) == No && d1 -> day > 30 || d1 -> day < 1)
        {
            fprintf(stdout, "Invalid input.");
            return ERROR;
        }
        /* 程序运行至此, 说明输入的信息已通过合法性检查 */
        return OK;
    }