打印日历月-Java 学习之路

-1

实现打印给定月份和年份的日历的功能 . 首先，提示用户：

Enter the month and year:

一旦用户输入有效输入（由空格分隔的两个整数），就以与UNIX cal 命令的输出类似的格式打印日历 . 例如，如果用户输入 03 2014 ，则输出应为：

我需要帮助才能向用户询问此问题所要求的具体输入 . 我也无法创建能够根据输入打印不同月份的代码，因为每个月都会在不同的一天开始 . 我不能使用任何太复杂的东西，因为我正在编程的初学者课程 .

我到目前为止只编写了3月份的代码：

#include <stdio.h>

int main()
{
    int k, rmd;

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

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

1 回答

#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("\n     %s %d\n%s\n", 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("\n");
    }   
    printf("\n");
}

回复于 2024-05-05T16:57:07+08:00

打印日历月

1 回答

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