C Program to display current date

#include <dos.h>
#include <stdio.h>
int main(void)
{
struct date d;
getdate(&d);
printf(“The current year is: %d\n”, d.da_year);
printf(“The current day is: %d\n”, d.da_day);
printf(“The current month is: %d\n”, d.da_mon);
return 0;
}

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C Program to Set / Change current system date

#include <stdio.h>
#include <dos.h>
int main(void)
{
struct time t;
gettime(&t);
printf(“The current hour is: %d\n”, t.ti_hour);
printf(“The current min is: %d\n”, t.ti_min);
printf(“The current second is: %d\n”, t.ti_sec);
/* Add one to the hour,minute & sec struct element and then call settime */
t.ti_hour++;
t.ti_min++;
t.ti_sec++;
settime(&t);
printf(“The current hour is: %d\n”, t.ti_hour);
printf(“The current min is: %d\n”, t.ti_min);
printf(“The current second is: %d\n”, t.ti_sec);
return 0;
}

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Enter and print details of n employees using structures and dynamic memory allocation

 Enter and print details of n employees using structures and dynamic memory allocation

This program is used to show the most basic use of structures. The structure is made of a character array name[], integer age and float salary. We’ll make an array of the structure for the employees. We’ll also use dynamic memory allocation using malloc. You can also use a linked list to the same which is a better option. Let’s check the code.

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#include #include typedef struct{ //structure of emp  char name[30];  int age;  float salary; }emp; int main(){  int n,i;  emp *employee;  printf("Enter no of employees: ");  scanf("%d",&n);  employee=(emp*)malloc(n*sizeof(emp)); //dynamic memory allocation using malloc()  for(i=0;i<n;i++){  printf("\n\nEnter details of employee %d\n",i+1);  printf("Enter name: ");  scanf("%s",employee[i].name);  printf("Enter age: ");  scanf("%d",&employee[i].age);  printf("Enter salary: ");  scanf("%f",&employee[i].salary);  }  printf("\nPrinting details of all the employees:\n");  for(i=0;i<n;i++){  printf("\n\nDetails of employee %d\n",i+1);  printf("\nName: %s",employee[i].name);  printf("\nAge: %d",employee[i].age);  printf("\nSalary: %.2f",employee[i].salary);  }  getch();  return 0; }

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Patterns (one of their favorites)

 Patterns (one of their favorites)

If the interviewer asks you a pattern and you don’t know how to do that you are screwed big time. Often you might make a very simple mistake which the interviewer was actually looking for. Here we’ll find out how to print this pattern

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1 2 3 4 5

A B C D E D C B A   A B C D C B A     A B C B A       A B A         A

And here is the code.

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01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21

//Printing pattern #include #include int main(){  int n,i,j;  printf("Enter no of lines: ");  scanf("%d",&n);  for(i=0;i<n;i++){  for(j=0;j<i;j++){ //for printing spaces  printf(" ");  }  for(j=0;j<n-i;j++){ //for printing the left side   printf("%c ",'A'+j); //the value of j is added to 'A'(ascii value=65)   }   for(j=n-i-2;j>=0;j--){ //for printing the right side     printf("%c ",'A'+j);   }  printf("\n");  }  getch(); }

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easiest way to sort

#include <stdio.h>
#include <stdlib.h>

int values[] = { 88, 56, 100, 2, 25 };

int cmpfunc (const void * a, const void * b)
{
   return ( *(int*)a - *(int*)b );
}

int main()
{
   int n;

   printf("Before sorting the list is: \n");
   for( n = 0 ; n < 5; n++ ) 
   {
      printf("%d ", values[n]);
   }

   qsort(values, 5, sizeof(int), cmpfunc);

   printf("\nAfter sorting the list is: \n");
   for( n = 0 ; n < 5; n++ ) 
   {   
      printf("%d ", values[n]);
   }
  
   return(0);
}

output

Before sorting the list is: 
88 56 100 2 25 
After sorting the list is: 
2 25 56 88 100 

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number of ways a particular no can formed..

#include<stdio.h>


// Returns the count of ways we can sum S[0...m-1] coins to get sum n

int count( int S[], int m, int n )
{
    // If n is 0 then there is 1 solution (do not include any coin)
    if (n == 0)
        return 1;
   
    // If n is less than 0 then no solution exists
    if (n < 0)
        return 0;

    // If there are no coins and n is greater than 0, then no solution exist
    if (m <=0 && n >= 1)
        return 0;

    // count is sum of solutions (i) including S[m-1] (ii) excluding S[m-1]
    return count( S, m - 1, n ) + count( S, m, n-S[m-1] );
}

// Driver program to test above function
int main()
{
    int i, j;
    int arr[] = {1, 3, 5};
    int m = sizeof(arr)/sizeof(arr[0]);
    printf("%d ", count(arr, m, 5));
    getchar();
    return 0;
}



It should be noted that the above function computes the same subproblems again and again. See the following recursion tree for S = {1, 2, 3} and n = 5.
The function C({1}, 3) is called two times. If we draw the complete tree, then we can see that there are many subproblems being called more than once.

C() --> count()
                              C({1,2,3}, 5)                     
                           /                \
                         /                   \              
             C({1,2,3}, 2)                 C({1,2}, 5)
            /     \                        /         \
           /        \                     /           \
C({1,2,3}, -1)  C({1,2}, 2)        C({1,2}, 3)    C({1}, 5)
               /     \            /    \            /     \
             /        \          /      \          /       \
    C({1,2},0)  C({1},2)   C({1,2},1) C({1},3)    C({1}, 4)  C({}, 5)
                   / \      / \       / \        /     \    
                  /   \    /   \     /   \      /       \ 
                .      .  .     .   .     .   C({1}, 3) C({}, 4)
                                               /  \
                                              /    \  

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