C language to determine whether a year is a leap years of various implementation program code

The introduction of this article:C language to determine whether a year is a leap years of various implementation program code 1, the Gregorian Leap year calculation principle (according to a regression years 365 days 5 hours 48 minutes 45.5 seconds) 1) ordinary year can divide 4 and cannot divide 100 into leap years. (such as 2004 years is a leap year, 1900 is not a leap year) 2) century can divide 400 is leap years. (such as 2000 is a leap year ...)

C language to determine whether a year is a leap years of various implementation program code

1. The Gregorian leap year calculation principle (according to a regression years 365 days 5 hours 48 minutes 45.5 seconds)

1) The average year can be divisible by 4 and not divisible by 100 for leap years. (For example, 2004 is a leap year, 1900 is not a leap)

2) century can be divisible by a leap year. (as 2000 is a leap year, 1900 is not a leap year)

3) for a large number of years, if the year can be divisible by 3200, and can be divisible by 172800 is a leap. If 172,800 is a leap year, 86,400 years is not a leap year (because it can be divisible by 3200, but not divisible by 172800) (this is calculated by a regression of 365 days 5h48 ' 45.5 ").

2, the Gregorian leap year program judgment statement

if ((0 = = year%4) && (0! = year%100)) | | (0 = = year%400))

{

The yeat that satisfy this condition are leap years.

}

3, the Gregorian Leap Year program code (by www. 169it. COM collected from the Internet )

Gregorian leap Year implementation code one:

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#include <stdio.h>void main(){ int year,leap; scanf("%d",&year); if(year%4==0) {  if(year%100!=0)   leap=1;  else  {   if(year%400==0)    leap=1;   else    leap=0;  } } if(leap==1)  printf("%d是闰年n",year); else  printf("%d不是闰年n",year);}

Gregorian leap Year implementation code two:

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#include <stdio.h>void main(){ int year,leap; scanf("%d",&year); if(year%400==0||year%4==0&&year%100!=0)  printf("%d是闰年n",year); else  printf("%d不是闰年n",year);}

Gregorian leap Year implementation code three:

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#include<stdio.h>int main(){int year;year=1900;while(year<=2000){if(year%400==0||year%4==0&&year%100!=0){printf("%d是闰年n",year);year++;}else year++;}return 0;}

The above code is for reference only.

C language to determine whether a year is a leap years of various implementation program code