Find Binary Number Of Any Given Decimal Number

/*Find Binary Number Of Any Given Decimal Number (In Reverse Order) */


#include <stdio.h>
int main(void)
{
    int num,j,k = 0;
    printf("Enter a number whose binary you want to know :");
    scanf("%d",&num);
    j = num ;

    while ( j > 0)
    {
        k = j % 2;
        printf("%d ",k );
        j = j / 2 ;
        continue;
    }

    printf("\n");
    return 0;

}

Euclid's Geometric Position

/*According to Euclid's Geometric Position find Integer Factorization/Prime Factorization of Given Number */


#include <stdio.h>

int main(void)
{
    int i,num,j;
    printf("Enter a number whose Prime factors you want to find :");
    scanf("%d",&num);
    j = num ;    //Assigning value of num to j

    for ( i = 2 ; i <= j-1 ;i++)
    {
         while( num % i == 0)
         {
             printf("\n%d ",i );
             num = num / i ;
             continue;
         }
       
    }
 
   printf("\n");
   return 0;

 }

According to the Gregorian calendar,it was Monday on the date 01/01/1900.If any year is input through the keyboard write a program to find out what is the day on 1st January of this year

Solution code:-
=========

Easy One (Irrespective of difference between year input from year 1901) : 

/*According to the Gregorian calender, it was Monday on the date 01/01/01. If any year is input
 * through the keyboard write a program to find out what is the day on 1st January of this year?*/



#include <stdio.h>
int main(void)
{
    int yr,diff,lpyrdays,normaldays,res;
    printf("\nEnter a year whose day of 1st Jan you want to know : ");
    scanf("%d",&yr);
   
    yr = (yr - 1) ; //removing 1 year as only 1 day we are calculating of current yr
    lpyrdays =  (yr/4)  + (yr / 400) - (yr / 100 ); //Calculating leap days in that particular year
    normaldays = (yr* 365 )+ 1 + lpyrdays ; //Calculating normal days in that year & adding 1st jan's 1 day in
    res = normaldays % 7;
   
   
    if(res==0)
    printf("\nSunday");
    if(res==1)
    printf("Monday");
    if(res==2)
    printf("Tuesday");
    if(res==3)
    printf("Wednesday");
    if(res==4)
    printf("Thursday");
    if(res==5)
    printf("Friday");
    if(res==6)
    printf("Saturday");
   
    return 0;
   
   
   
   
}

Tough One (Solution Code):


#include <stdio.h>

int main(void)          //main function.. starting of c code
{
    int year,differ,lp_year,day_type;
    long int days;
  
   
    printf("Please enter the year: ");
    scanf("%d",&year);
    year=year-1; //we will find days before given year so
    differ=year-1900;


   /*as leap year is not divisible by 100.so,create 2 condition
   one difference less than 100 and greater than 100*/

   if(differ<100)
   {
   lp_year=differ/4; //caln of total no. of leap year
   days=(366*lp_year)+((differ-lp_year)*365+365+1);//see Note1
   day_type=days%7; //caln of day type sun, mon......
   }
  
   if(differ>=100)
   {
   lp_year=(differ/4)-(differ/100)+1+((year-2000)/400);//see Note2           
   days=(366*lp_year)+((differ-lp_year)*365+365+1);//see Note3
   day_type=days%7;
   }

if(day_type==0)
printf("\nSunday");
if(day_type==1)
printf("Monday");
if(day_type==2)
printf("Tuesday");
if(day_type==3)
printf("Wednesday");
if(day_type==4)
printf("Thursday");
if(day_type==5)
printf("Friday");
if(day_type==6)
printf("Saturday");
    
    
  
   
    return 0;  //int main() is function so value must be return.
               //u will read in function chapter
}


/*Note1:
-leap year has 366 day so lp_year*366
-remaining year has 365 day so (differ-lp_year)*365
-add 365 because we reduce 1 year
-add 1 to make jan 1 on which we find day type
     
Note2:
-(leap year come in every 4 year so) (differ/4) for leap year
-(leap year isn't divisible by 100 so we subtract (differ/100)
  from counting as leap year
-(leap year will be if divisible by 400 so ((year-2000)/400)
  to count that year as leap year
- we calculate from 2000 so we add 1

Note3:
-leap year has 366 day so lp_year*366
-remaining year has 365 day so (differ-lp_year)*365
-add 365 because we reduce 1 year
-add 1 to make jan 1 on which we find day type */