2.给出距离1900年1月1日的天数，求日期

Posted 2020-08-22 不忘初心 方得始终

 1 #include <iostream>
 2 #include <assert.h>
 3 
 4 
 5 //判断是否闰年
 6 bool IsLeapYear(unsigned int year)
 7 {
 8   if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
 9   {
10     return true;
11   }
12   else
13   {
14     return false;
15   }
16 }
17 
18 //根据一年中的第几天(0起始)，返回月份(1起始)，并且置天数为其在本月的索引(0起始)
19 unsigned int GetMonthByDay(unsigned int &day, bool leap)
20 {
21   assert(day < (leap ? 366 : 365));
22   static unsigned int monthtable[] = { 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
23   monthtable[1] = leap ? 29 : 28;
24 
25   unsigned int  month = 0;
26   unsigned int count = 0;
27   unsigned int n = day;
28 
29   do
30   {
31     n = day - count;
32     count += monthtable[month++];
33   } while (day >= count);
34   day = n;
35   return month;
36 }
37 
38 //距离1900年1月1日的天数(0起始)，返回年，并且置天数为其在本年的索引(0起始)
39 unsigned int GetYearByDay(unsigned int &day)
40 {
41   unsigned int Y = day / 365;
42   int D = day % 365;
43   int  YY = Y - 1;
44   D -= YY / 4;
45   D += YY / 100;
46   D -= (YY + 300) / 400;
47   while (D < 0)
48   {
49     Y--;
50     D += IsLeapYear(Y + 1900) ? 366 : 365;
51   }
52   day = D;
53   return Y + 1900;
54 }
55 
56 //距离1900年1月1日的天数(0起始)，计算这一天的日期(年，月，日)(1起始)。
57 void foo(unsigned int &day, unsigned int &year, unsigned int &month)
58 {
59   year = GetYearByDay(day);
60   month = GetMonthByDay(day, IsLeapYear(year));
61   day += 1;
62 }
63 
64 int main()
65 {
66   unsigned int day = 42696;
67   unsigned int year = 0;
68   unsigned int month = 0;
69   foo(day, year, month);
70   std::cout << year << "/" << month << "/" << day << std::endl;
71   getchar();
72   return 0;
73 }