C程式設計百例之第四例

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題目：輸入某年某月某日，判斷這一天是這一年的第幾天？

1.程式分析：以3月5日為例，應該先把前兩個月的加起來，然後再加上5天即本年的第幾天，特殊情況，閏年且輸入月份大於3時需考慮多加一天。

2.程式原始碼：

#include <stdio.h>

int main(void)
{ 
 int day, month, year, countDay, leap;
 printf("\nPlease input year, month and day\n");
 scanf("%d %d %d", &year, &month, &day);
 if((year % 4 == 0 && year % 100 != 0) || (year % 400 ==0))
  leap = 1;
 else
  leap = 0;
 switch(month)
 {
  case 1:
   countDay = 0;
   if(day <= 0 || day > 31)
    return 1;
   break;
  case 2:
   countDay = 31;
   if(leap != 1)
   {
    if(day <= 0 || day > 28)
     return 1;
   }
   else
   {
    if(day <= 0 || day > 29)
     return 1;
   }
   break;
  case 3:
   countDay = 59;
   if(day <= 0 || day > 31)
    return 1;
   break;
  case 4:
   countDay = 90;
   if(day <= 0 || day > 30)
    return 1;
   break;
  case 5:
   countDay = 120;
   if(day <= 0 || day > 31)
    return 1;
   break;
  case 6:
   countDay = 151;
   if(day <= 0 || day > 30)
    return 1;
   break;
  case 7:
   countDay = 181;
   if(day <= 0 || day > 31)
    return 1;
   break;
  case 8:
   countDay = 212;
   if(day <= 0 || day > 31)
    return 1;
   break;
  case 9:
   countDay = 243;
   if(day <= 0 || day > 30)
    return 1;
   break;
  case 10:
   countDay = 273;
   if(day <= 0 || day > 31)
    return 1;
   break;
  case 11:
   countDay = 304;
   if(day <= 0 || day > 30)
    return 1;
   break;
  case 12:
   countDay = 334;
   if(day <= 0 || day > 31)
    return 1;
   break;
  default:
   printf("Input month error!\n");
   return 1;
   break;
 }
 countDay = countDay + day + 1;
 printf("This is the %dth of this years!\n", countDay);
 return 0;
}