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Write a C program to print all permutations of a given string

A permutation, also called an “arrangement number” or “order,” is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation.



Below are the permutations of string ABC.
ABC, ACB, BAC, BCA, CAB, CBA
Here is a solution using backtracking.

// C program to print all permutations with duplicates allowed #include <stdio.h> #include <string.h>   /* Function to swap values at two pointers */ void swap(char *x, char *y) {     char temp;     temp = *x;     *x = *y;     *y = temp; }   /* Function to print permutations of string    This function takes three parameters:    1. String    2. Starting index of the string    3. Ending index of the string. */ void permute(char *a, int l, int r) {    int i;    if (l == r)      printf("%s\n", a);    else    {        for (i = l; i <= r; i++)        {           swap((a+l), (a+i));           permute(a, l+1, r);           swap((a+l), (a+i)); //backtrack        }    } }   /* Driver program to test above functions */ int main() {     char str[] = "ABC";     int n = strlen(str);     permute(str, 0, n-1);     return 0; }

Output:

ABC
ACB
BAC
BCA
CBA
CAB

Algorithm Paradigm: Backtracking
Time Complexity: O(n*n!)

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programe to find date and time of system

 #include <time.h>
 #include <stdio.h> 

          int main(void)
 {
  time_t mytime;
  mytime = time(NULL);
  printf(ctime(&mytime));

  return 0;
 }

The string that is returned will have the following format:

 Www Mmm dd hh:mm:ss yyyy

 Www = which day of the week.
 Mmm = month in letters.
 dd = the day of the month.
 hh:mm:ss = the time in hour, minutes, seconds.
 yyyy = the year.

 Output example:
  Tue Feb 26 09:01:47 2009

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c programe to count no of set bit in an integer

#anki 



#include<stdio.h>

void main()
{
 int i,count=0;
 printf("enter the value");
 scanf("%d",&i);
 while(i)
 {
   if((i & 1)==1) count++;
  i=i>>1;
   }
   printf("count =%d",count);
   getch();   
}

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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.

?

#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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