实现了打印日历给定月份和年份的函数。首先，提示用户：

 输入月份和年份：
 

一旦用户输入一个有效的输入（两个整数用空格隔开），打印出的格式日历是类似的UNIX CAL 命令的输出。例如，如果用户输入 03 2014年，输出应该是：

我需要能够询问用户特定的输入，这个问题正在寻求帮助。我还具有与创建code表示将能够打印基于所输入的不同月份，如每个月开始在不同的一天的麻烦。我不能使用任何过于复杂，因为我是在一个初级课程的规划。

在code我到目前为止只打印出月：

 的#include＆LT; stdio.h中＆GT;

诠释的main（）
{
    INT K，RMD;

    的printf（2014年3月的\ n）;
    的printf（苏周一周二周三周四周五周六\ N）;

    为（K = 1; K＆所述; 32; ++ k）的{
         如果（K == 1）{
             的printf（％2D \ N，K）;
         }
         否则如果（K％7 == 1）{
             的printf（％2D \ N，K）;
         }
         其他 {
             的printf（％2d中，k）的;
         }
    }
    返回0;
}
 

解决方案

 的#include＆LT; stdio.h中＆GT;

INT isLeapYear（INT年）; / *如果真闰年* /
INT leapYears（INT年）; / *闰年数* /
INT todayOf（INT Y，INT男，INT D）; / *天自今年年初数* /
长天（INT Y，INT男，INT D）; / *总天数* /
无效日历（INT Y，INT M）; / *显示日历在M Y * /

诠释的主要（无效）{
    INT年，月;

    的printf（请输入月份和年份：）;
    scanf函数（％D，和放大器;当月，和放大器;一年）;

    日历（年，月）;

    返回0;
}

INT isLeapYear（INT Y）/ *如果真闰年* /
{
    收益率（Y％400 == 0）|| （（Y％4 == 0）及和放大器;！（Y％100 = 0））;
}

INT leapYears（INT Y）/ *闰年数* /
{
    返回Y / 4  -  Y / 100 + Y / 400;
}

INT todayOf（INT Y，INT男，INT D）/ *天自今年年初数* /
{
    静态INT DAYOFMONTH [] =
        {1 / *虚* /，0,31,59,90,120,151,181,212,243,273,304,334};

    返回DAYOFMONTH [米] + D +（（米→2＆安培;＆安培; isLeapYear（y））为1：0）;
}

长天（INT Y，INT男，INT D）/ *总天数* /
{
    INT lastYear;

    lastYear = Y  -  1;

    返回365L * lastYear + leapYears（lastYear）+ todayOf（Y，M，D）;
}

无效日历（INT Y，INT M）/ *显示日历我的* /
{
    为const char * NameOfMonth [] = {NULL / * dummp * /，
        月，月，月，月，五一，六一，
        七五，八五，九五，十月，月，月
    };
    焦炭周[] =苏周一周二周三周四周五周六;
    INT DAYOFMONTH [] =
        {1 / *虚* /，31,28,31,30,31,30,31,31,30,31,30,31};
    INT weekOfTopDay;
    INT I，日;

    weekOfTopDay =天（Y，M，1）％7;
    如果（isLeapYear（y））为
        DAYOFMONTH [2] = 29;
    的printf（\ñ％S％D \ñ％S \ N，NameOfMonth [M]，Y，周）;

    对于（i = 0; I＆LT; weekOfTopDay;我++）
        的printf（）;
    对于（I = weekOfTopDay，天= 1;天＆LT; = DAYOFMONTH [M];我++，一天++）{
        的printf（％2D，日）;
        如果（I％7 == 6）
            的printf（\ N）;
    }
    的printf（\ N）;
}
 

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

Enter the month and year:

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:

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.

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

#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;
}

#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");
}