Program to find day on a particular date

Program to find day on a particular date

CODE

👇

1. #include<stdio.h>
2. 
   
3. int isLeapYear(int year){
4.     if(year%4==0){
5.         if(year%100==0 && year%400!=0){
6.             return 0;
7.         }
8.         else{
9.             return 1;
10.         }
11.     }
12.     else{
13.         return 0;
14.     }
15. }
16. 
    
17. int main() {
18.     int year, i;
19.     int refYear=1600, leap=0;
20.     int diff, totalDays,day,month, oddDays;
21.     int lYear[]={3,1,3,2,3,2,3,3,2,3,2,3};
22.         int nYear[]={3,0,3,2,3,2,3,3,2,3,2,3};
23.     char  week[7][10]={"sunday",
24.                    "monday",
25.                    "tuesday",
26.                    "wednesday"
27.                    ,"thursday"
28.                    ,"friday",
29.                    "saturday"};
30.      printf("Enter a date between  1600 to 3000\n");
31.       scanf("%d%d%d",&day,&month,&year);
32.     diff = year - refYear;
33.     while(refYear < year){
34.         if(isLeapYear(refYear))
35.             leap++;
36.         refYear++;
37.     }
38.     totalDays = leap*366 + (diff-leap)*365;
39.     oddDays = totalDays%7;
40.      
41.     if(isLeapYear(year)){
42.         for( i=0;i<month-1;i++){
43.             oddDays+=lYear[i];
44.         }
45.         oddDays+=day%7;
46.     }
47.     else{
48.         for( i=0;i<month-1;i++){
49.             oddDays+=nYear[i];
50.         }
51.         oddDays+=day%7;
52.     }
53.     printf("Day on %d/%d/%d ",day,month,year);
54.     printf("%s",week[(5+oddDays)%7]);
55.     return 0;
56. }