打印一个日历月 [英] Printing a calendar month

问题描述

实现一个打印给定月份和年份的日历的函数.首先，提示用户:

Implement a function that prints the calendar for a given month and year. First, prompt the user:

Enter the month and year:

一旦用户输入了一个有效的输入(两个用空格分隔的整数)，以类似于 UNIX cal 命令输出的格式打印出日历.例如，如果用户输入 03 2014，输出应该是:

Once the user enters a valid input (two integers separated by a space), print out the calendar in a format be similar to the output of the UNIX cal command. For example, if the user enters 03 2014, the output should be:

我需要帮助才能向用户询问此问题所要求的特定输入.我在创建能够根据输入打印不同月份的代码时也遇到了麻烦，因为每个月都从不同的一天开始.因为我正在学习编程初学者课程，所以我不能使用任何太复杂的东西.

I need help with being able to ask the user for the specific input that this question is asking for. I am also having trouble with creating code that will be able to print different months based on the input, as each month starts on a different day. I cannot use anything too complex as I'm taking a beginner course in programming.

到目前为止我只打印出 March 的代码:

The code I have so far for only printing out March:

#include <stdio.h>

int main()
{
    int k, rmd;

    printf("     March 2014
");
    printf(" Su Mo Tu We Th Fr Sa
");

    for(k = 1; k < 32; ++k) {
         if(k == 1){
             printf("                   %2d
", k); 
         }
         else if(k % 7 == 1) {
             printf(" %2d
", k);
         }
         else {
             printf(" %2d", k);
         }
    }
    return 0;
}

推荐答案

#include <stdio.h>

int isLeapYear( int year );        /* True if leap year */
int leapYears( int year );         /* The number of leap year */
int todayOf( int y, int m, int d); /* The number of days since the beginning of the year */
long days( int y, int m, int d);   /* Total number of days */
void calendar(int y, int m);       /* display calendar at m y */

int main(void){
    int year,month;

    printf("Enter the month and year: ");
    scanf("%d %d", &month, &year);

    calendar(year, month);

    return 0;
}

int isLeapYear( int y ) /* True if leap year */
{
    return(y % 400 == 0) || ((y % 4 == 0) && (y % 100 != 0));
}

int leapYears( int y ) /* The number of leap year */
{
    return y/4 - y/100 + y/400;
}

int todayOf( int y, int m, int d) /* The number of days since the beginning of the year */
{
    static int DayOfMonth[] = 
        { -1/*dummy*/,0,31,59,90,120,151,181,212,243,273,304,334};

    return DayOfMonth[m] + d + ((m>2 && isLeapYear(y))? 1 : 0);
}

long days( int y, int m, int d) /* Total number of days */
{
    int lastYear;

    lastYear = y - 1;

    return 365L * lastYear + leapYears(lastYear) + todayOf(y,m,d);
}

void calendar(int y, int m) /* display calendar at m y */
{
    const char *NameOfMonth[] = { NULL/*dummp*/,
        "January", "February", "March", "April", "May", "June",
        "July", "August", "September", "October", "November", "December"
    };
    char Week[] = "Su Mo Tu We Th Fr Sa";
    int DayOfMonth[] =
        { -1/*dummy*/,31,28,31,30,31,30,31,31,30,31,30,31 };
    int weekOfTopDay;
    int i,day;

    weekOfTopDay = days(y, m, 1) % 7;
    if(isLeapYear(y))
        DayOfMonth[2] = 29;
    printf("
     %s %d
%s
", NameOfMonth[m], y, Week);

    for(i=0;i<weekOfTopDay;i++)
        printf("   ");
    for(i=weekOfTopDay,day=1;day <= DayOfMonth[m];i++,day++){
        printf("%2d ",day);
        if(i % 7 == 6)
            printf("
");
    }   
    printf("
");
}

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