如何将整数值转换为年，月和日？ [英] How Do I Convert An Integer Value To Year, Month And Days?

问题描述

  #include   <   stdio.h  >  
   #include   <   conio.h  >  
   #define DAYSINYEAR 365 
  < span class =code-preprocessor> #define DAYSINMONTH 30 
   #define DAYSINWEEK 7 
   void  main（）
 {
  int  year =  0 ，month =  0 ，week =  0 ，day =  0 ; 
 clrscr（）; 
 printf（ 输入天数：）; 
 scanf（ ％d，& day）; 
  while （day> = DAYSINYEAR）
 {
 day = day-DAYSINYEAR; 
年++; 
} 
  while （day> = DAYSINMONTH）
 {
 day = day-DAYSINMONTH; 
 month ++; 
} 
 
 而（day> = DAYSINWEEK）
 {
 day = day-DAYSINWEEK; 
周++; 
} 
 printf（  \ n％d年，％d月，％d周和％d天，年，月，周，日）; 
  //   return 0;  
 getch（）; 
} 

我怎样才能将简单的整数值转换为年，月和日照顾闰年，以及31个月的月份。

就像我已经输入364然后回答将是11个月和30天..

或如果输入366那么它应该是1年，0个月和1天..

请帮帮我。我已经有了简单的代码，但它并没有给我正确的答案。

如果我在12个月和4天内输入364就会给出答案，所以它错了，但似乎有很大的价值。所以我需要确切的代码..

代码如上所示。



Thanx

解决方案

没有直接的解决方案：如果不知道开始日期（包括年份），就不能计算月数 - 因为2月的长度每四年变化一次。



但是......如果你想接受不准确的答案，假设你的代码中有30个月的月份，那么它比你想象的更容易 -

 years = days /  365 ; 
天=天％ 365 ; 
个月=天/  30 ; 
天=天％ 30 ; 

由于你在整个过程中使用整数，因此除去任何分数，然后模数就会丢失你计算的那一点！



但请注意：它会在现实世界中产生垃圾结果：例如1年12个月，如果输入729则为4天！

#include<stdio.h>
#include<conio.h>
#define DAYSINYEAR 365
#define DAYSINMONTH 30
#define DAYSINWEEK 7
void main()
{
   int year=0, month=0, week=0, day=0;
   clrscr();
   printf("Enter the number of days:");
   scanf("%d",&day);
   while(day>=DAYSINYEAR)
   {
     day = day-DAYSINYEAR;
     year++;
   }
   while(day>=DAYSINMONTH)
   {
     day = day-DAYSINMONTH;
     month++;
   }

 while(day>=DAYSINWEEK)
 {
   day = day-DAYSINWEEK;
   week++;
 }
 printf("\n%d year, %d Month, %d Weeks and %d Days", year,month,week,day);
// return 0;
 getch();
}

how can i convert simple integer value to years, months and days with taking care of leap year, and also 31 day of months.
like i have entered 364 then answer will be 11 months and 30 days..
or if entered 366 then it should be 1 year, 0 month and 1 day..
please help me. i have already the simple code but it is not giving me correct answer.
it is giving the answer if i entered 364 the 12 months and 4 days so its wrong but it seems right for big amount of value. so i need the exact code..
the code is given above.

Thanx

解决方案

There is no "direct" solution: you can't have a "months" count without knowing the start date, including the year - as the length of February changes every four years.

But... if you want to accept inaccurate answers which assume a 30 day month as in your code, then it's easier than your think - sort of.

years = days / 365;
days = days % 365;
months = days / 30;
days = days % 30;

Since you are using integers throughout, the divide throws away any fractions, then the modulus throws away the bit you just calculated!

But do be aware: it produces rubbish results in the real world: such as 1 year, 12 months, 4 days if you enter 729!

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