The number of required [n x 1] multiplexer to implement one [m x 1] multiplexer.

 /******************************************************************************

This is a program to find the number of required (m x 1) multiplexer to implement one (n x 1) multiplexer. And m is always even, otherwise we are wasting the bit(s).
If your m is odd then increase it by one.

*******************************************************************************/

#include <stdio.h>


int main()

{

    int n, m, x, y = 0;

    printf("This is a program to find the number of required (m x 1) multiplexer to implement one (n x 1) multiplexer. And m is always even, otherwise, we are wasting the bit(s). If your m is odd then increase it by one.\n");

    printf("Enter the value of n:\t");

    scanf("%d", &n);

    printf("Enter the value of m:\t");

    scanf("%d", &m);

    if(n>m){

        x = n;

        while(x > m){

            //If x is odd then we need an another multiplexer.

            if( x%2 == 0 ){

            x = x/m;

            y = y + x;

            }

            else{

            x = x/m;

            y = y + x + 1;    

            }

        }

        printf("%d", y+1);

    }

    else

        printf("1");


    return 0;

}


Comments

Post a Comment

Popular posts from this blog

My very 1st contest! | Problem - 1 | Beginner level | C language | CodeChef

Link to the problem statement:   https://www.codechef.com/submit-v2/MY1STCONTEST Solution:  #include <stdio.h> int main(void) { int n, a, b; scanf("%d %d %d", &n, &a, &b); int c, d; c = n - a; //user made submissions and will get a rating. d = c - b; //user that were able to solve at least 1 problem and hence will get a            //rating greater than 1000. printf("%d %d", c, d); return 0; } You can watch the video solution of this problem: Hope this will help you, if you want anything else leave a comment. Thank you
Read more

Implementing linked list in C language

#include <stdio.h> #include <stdlib.h> // Define the structure for a node in the linked list struct node {     int data; // The data stored in the node     struct node *next; // Pointer to the next node in the list }; struct node* head = NULL; // Initialize the head of the list as NULL // Function to insert nodes at the end of the list void insert(struct node **head) {     int num = 0, i, data;     printf("\nEnter the number of elements to be inserted: ");     scanf("%d", &num); // Get the number of nodes to be inserted     // Loop to insert the nodes     for(i = 0; i<num; i++) {         // Allocate memory for a new node and get the data for it         struct node* newNode = malloc(sizeof(struct node));         struct node *last = *head;         printf("\nEnter the element %d: ", i+1);         scanf("%d", &data); // Get the data for the node         // Initialize the new node with the given data and set its next pointer to
Read more

Score High | Problem - 4 | Beginner level | C language | CodeChef

Link to the problem statement:  https://www.codechef.com/submit-v2/HIGHSCORE Solution:  1st Method: #include <stdio.h> int main(void) { int t, i, n, a, b, c, d, s; scanf("%d", &t); for(i=0;i<t;i++) {     if(t>=1 && t<=1000)     {         scanf("%d",&n);         scanf("%d %d %d %d", &a, &b, &c, &d);         s = a+b+c+d;         if( s == n && n>=a && n>=b && n>=c && n>=d )         {             if(a>=b && a>=c && a>=d)             {                 printf("%d\n", a);             }             else if(b>=a && b>=c && b>=d)             {                 printf("%d\n", b);             }             else if(c>=a && c>=b && c>=d)             {                 printf("%d\n", c);             }             else             {
Read more