Computer Science, asked by aishwaryad333, 4 months ago

3.5 marks
- Ubiem statement
A Farm is represented by the following structure:
struct Farm
{
int hens:
int cows
Is
You are given a function,
int MaxLegs (struct Farm farms[], int
m);
5398
The function accepts an array 'farms' of type 'Farm' consisting of 'm' elements as its argument. A 'Farm' consists of hens which have two legs
and cows which have 4 legs. Implement the function to count the total number of legs for each array element of 'farms' and return the
maximum leg count.
Note:
Computed value lies within integer range.
Return -1 if 'farms' is empty.
Example:
Input:
hens
27
Output:
398​

Answers

Answered by Agastya0606
1

The C program for the given problem statement is

#include <stdio.h>

#include<stdlib.h>

#include <math.h>

int main()

{

int L,E,x,y;

printf("total number of legs L = \n");

scanf("%d",&L);

printf("total number of eyes E =\n");

scanf("%d",&E);

if(L%2 == 0 & E%2 == 0){

y = (L - E)/2;

x = (L-(4*y))/2;

printf("%d ",y);

printf("%d",x);

}

else {

if(L%2 != 0){

printf("Number of legs should be even number");

}

else {

printf("Number of eyes should be even number");

}

}

return 0;

}

  • Different libraries are included at the beginning of the code.
  • The variables are declared and the conditions according to the problem statement are implemented.
  • If else condition statements are used and at last the result id printed on the screen.

#SPJ2

Answered by hemakumar0116
0

Answer:

#include <stdio.h>

#include<stdlib.h>

#include <math.h>

int main()

{

int L,E,x,y;

printf("total number of legs L = \n");

scanf("%d",&L);

printf("total number of eyes E =\n");

scanf("%d",&E);

if(L%2 == 0 & E%2 == 0){

y = (L - E)/2;

Explanation:

#include <stdio.h>

#include<stdlib.h>

#include <math.h>

int main()

{

int L,E,x,y;

printf("total number of legs L = \n");

scanf("%d",&L);

printf("total number of eyes E =\n");

scanf("%d",&E);

if(L%2 == 0 & E%2 == 0){

y = (L - E)/2;

x = (L-(4*y))/2;

printf("%d ",y);

printf("%d",x);

}

else {

if(L%2 != 0){

printf("Number of legs should be even number");

}

else {

printf("Number of eyes should be even number");

}

}

return 0;

}

The code starts off by including a number of libraries.

The problem statement's conditions are put into practise, and the variables are stated.

Finally, the result is printed on the screen using if-else condition statements.

#SPJ2

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