我已经编写了代码,通过输入整数月和整数年来打印一个月的日历。但是我想用字符输入月份,月份的前三个字母如1月和2月2月。如图所示,月份由字符输入。 Here is the Image。所以请更改代码,这样我就可以在字符中输入月份。谢谢
#include <stdio.h>
int isLeapYear(int y); /* True if leap year */
int leapYears(int y); /* 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 365Q * 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");
}
答案 0 :(得分:1)
有很多方法可以做到这一点。它们通常涉及获取输入字符串并查找某些表格。
一种有效的方法是根据字符串输入计算哈希,而不是最多12个字符串比较。使用哈希查看月份名称并查看它是否匹配。下面的哈希需要月份名称的ASCII编码。 char
可能已签名或未签名。
当然如果月份名称改变(另一种语言?),下面的具体值和哈希方法需要调整。
// Assume month is any 3 ASCII characters (either case)
int month2int_chux(const char *month) {
if (month[0] && month[1] && month[2]) {
unsigned m0 = month[0] | 0x20;
unsigned m1 = month[1] | 0x20;
unsigned m2 = month[2] | 0x20;
unsigned m = (14 * m2) ^ (47 * m1); // magic computation does not use m0.
m %= 13;
const unsigned char hash[] = { 9, 11, 5, 12, 0, 7, 2, 1, 3, 4, 8, 10, 6 };
m = hash[m % 13u];
if (m && (NameOfMonth[m][0] | 0x20) == m0 &&
NameOfMonth[m][1] == m1 && NameOfMonth[m][2] == m2) {
return m;
}
}
return 0;
}