I am only given one hour to code this. [NOTE : NO arrays, pointers, recursion an

Question

I am only given one hour to code this.
[NOTE : NO arrays, pointers, recursion and strings should be used in this code.]
1) How do I go about doing this is under an hour? Please include some thought processes.
2) What's the best way to savage the maximum amount of marks if I don't complete this in under an hour?

Exercise 1: Collatz's Problem[50 marks] Problem Statement Following is the control flow graph describing the computation of Collatz's sequence. False True True False n is even printn Given an integer value n, a program with the above control-flow graph prints out the current value ot at every iteration, until control reaches the exit point. Both the division and multiplication operate on integer value Write a program collatz.c that reads in two positive integers, and count, where n is the input to the Collatr's seqence, and count is the maximum number of iterations (and thus the maximum number of times the value of n is printed) required far execution The prograshold call a function displayCollotzinrernal which does the fallowing: .Compute Collatz's sequence by printing out the values of at each iteration, for up to cvunt number of iterations Print out the number of times the division statement "n=n÷2" and the multiplication statement "n = n × 3 + 1" have been executed. You are to decide the appropriate parameterís), return type and precondition f anyl for this function. You may define additional functions as needed You may assume that the inputs are valid .e., positive integers)

Explanation / Answer

#include <stdio.h>

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

* Name : displayCollatzInternal(n, count)

* Description : Generates the Collatz Sequence, limiting to the count.

* Also, prints the sequence, count of the division excecutions

* & multiplication executions.

* Parameters :

* n - number of which Collatz Sequence is to be generated.

* count - maximum iterations to be performed.

* Output :

* None

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

void displayCollatzInternal(int n, int count) {

int divCount = 0; // division execution counter

int mulCount = 0; // multiplication execution counter

int itrCount = 0; // iteration counter

  

// print the first line outside all conditions and loop, so it gets printed

// in the edge case (input as 1)

printf("The collatz sequence is: ");

  

// exit condition for while loop either n is 1 or

// iteration counter equals the maximum count of iterations passed as input

while ((n != 1) && (itrCount < count)) {

// increment iteration counter

itrCount++;

// mod operator checks for remainder

// so condition to check even numbers

if (n % 2 == 0) {

// even number

// reassign n & increment division execution counter

n = n / 2;

divCount++;

} else {

// odd number

// reassign n & increment multiplication execution counter

n = (3 * n) + 1;

mulCount++;

}

// print value of n for this iteration

// note no new line is added here

// also this should be inside the while loop and at the end, to have modified value

// of n printed

printf("%d ", n);

}

  

// print division count & multiplication count outside loop, so as to be in the end

// note for new line to be inserted before actually printing the count

printf(" The division statement has been executed %d times.", divCount);

printf(" The multiplication statement has been executed %d times.", mulCount);

}

int main() {

int n=0, count=0;

// get user input

printf("Enter positive numbers for n and count: ");

printf(" ");

scanf("%d %d", &n, &count);

// invoke displayCollatzInternal()

displayCollatzInternal(n, count);

return 0;

}

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