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Aleks [24]
3 years ago
15

Create a function (prob3_6) that will do the following: Input a positive scalar integer x. If x is odd, multiply it by 3 and add

1. If the given x is even, divide it by 2. Repeat this rule on the new value until you get 1, if ever. Your program will output how many operations it had to perform to get to 1 and the largest number along the way.
Computers and Technology
1 answer:
aniked [119]3 years ago
5 0

Answer:

Following are the program in python language

def prob3_6(k): #function definition

  c = 0 #variable declaration

  while k != 1: #iterating the while loop

      print(k) #print k

      if k % 2 == 0:#check if condition  

          k= k // 2 #divisible by 2

      else: #else condition

          k = k * 3 + 1  

      c = c + 1

  print(k) #print k

  print c #print count

prob3_6(3)#function call

Output:

3

10

5

16

8

4

2

1

7

Explanation:

Following are the description of program

  • Create a function "prob3_6" in this function we passing an integer parameter of type "int" named "k".
  • Inside that function we declared a variable "c" of type "int" that is used for counting purpose .
  • After that we iterated the while for print the value of when it is not equal to 1 .
  • We check if condition when k gives 0 on modulus then k is divisible by 2 otherwise else block will be executed in the else part we multiply by 3 to k and add 1 to the k variable .
  • Finally print "k" and "c"

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What did czarnowski and triantafyllou learn from observing boat propulsion systems?
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3 years ago
Write a C++ program that reads from the standard input and counts the number of times each word is seen. A word is a number of n
Rufina [12.5K]

Answer:

#include <stdio.h>

#include <string.h>

#include <ctype.h>

 

struct detail

{

   char word[20];

   int freq;

};

 

int update(struct detail [], const char [], int);

 

int main()

{

   struct detail s[10];

   char string[100], unit[20], c;

   int i = 0, freq = 0, j = 0, count = 0, num = 0;

 

   for (i = 0; i < 10; i++)

   {

      s[i].freq = 0;

   }

   printf("Enter string: ");

   i = 0;

   do

   {

       fflush(stdin);

       c = getchar();

       string[i++] = c;

 

   } while (c != '\n');

   string[i - 1] = '\0';

   printf("The string entered is: %s\n", string);

   for (i = 0; i < strlen(string); i++)

   {

       while (i < strlen(string) && string[i] != ' ' && isalnum(string[i]))

       {

           unit[j++] = string[i++];

       }

       if (j != 0)

       {

           unit[j] = '\0';

           count = update(s, unit, count);

           j = 0;

       }

   }

 

   printf("*****************\nWord\tFrequency\n*****************\n");

   for (i = 0; i < count; i++)

   {

       printf("%s\t   %d\n", s[i].word, s[i].freq);

       if (s[i].freq > 1)

       {

           num++;

       }

   }

   printf("The number of repeated words are %d.\n", num);

 

   return 0;

}

 

int update(struct detail s[], const char unit[], int count)

{

   int i;

 

   for (i = 0; i < count; i++)

   {

       if (strcmp(s[i].word, unit) == 0)

       {

           s[i].freq++;

 

           return count;

       }

   }

   /*If control reaches here, it means no match found in struct*/

   strcpy(s[count].word, unit);

   s[count].freq++;

 

   /*count represents the number of fields updated in array s*/

   return (count + 1);

}

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Answer:

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6 0
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