编写程序,返回某月某日是某一年中的第几天(1和366之间的整数)。示例如下:

思路如下:
月份用switch选择语句输出;
输出日期要注意后缀:后缀th在日期中仅用于4号~20号以及24号~30号;1和21号的后缀是st,2和22号的后缀是nd,3和23号的后缀是rd。
年份包括平年和闰年,平年的二月只有28天,而闰年的二月有29天。可以用数组存储平年和闰年每个月的天数,判断后逐个月份天数相加即可。要判断某一年是不是闰年,一般方法是用4或400去除这一年的年份数,如果除得的商是整数而没有余数,那么这一年是闰年;如果有余数,那么这一年是平年。
程序如下:
main() {
int month, day, year;
int i, sum = 0;
int pingnian[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int runnian[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
printf("Enter month:");
scanf("%d", &month);
printf("Enter day:");
scanf("%d", &day);
printf("Enter year:");
scanf("%d", &year);
switch(month){
case 1: printf("January"); break;
case 2: printf("February"); break;
case 3: printf("March"); break;
case 4: printf("April"); break;
case 5: printf("May"); break;
case 6: printf("June"); break;
case 7: printf("July"); break;
case 8: printf("August"); break;
case 9: printf("September"); break;
case 10: printf("October"); break;
case 11: printf("November"); break;
case 12: printf("December"); break;
}
if(day == 1 || day == 21) printf("%3dst", day);
else if(day == 2 || day == 22) printf("%3dnd", day);
else if(day == 3 || day == 23) printf("%3drd", day);
else printf("%3dth", day);
if(year % 4 == 0 || year % 400 == 0) {
for(i = 0; i < month - 1; i++)
sum += runnian[i];
sum += day;
}
else {
for(i = 0; i < month - 1; i++)
sum += pingnian[i];
sum += day;
}
printf(" is the %d day of the %d year.", sum, year);
return 0;
}