C++ Given a positive integer, N, the ’3N+1’ sequence starting from N is defined as follows: If N is an even number, then divide N by two to get a new value for N If N is an odd number, then multiply N by 3 and add 1 to get a new value for N. Continue to generate numbers in this way until N becomes equal to 1 For example, starting from N = 3 the complete ’3N+1’ sequence would be: 3, 10, 5, 16, 8, 4, 2, 1 Write code to ask the user to enter a positive integer (N) in the main() function. Write a function sequence() that receives the integer value N and display the ‘3N+1’ sequence starting from the integer value that was received (entered by the user). The function must also count and return the numbers that the sequence consists of. The returned value must be displayed from the main() function.

C++ Given a positive integer, N, the ’3N+1’ sequence starting from N is defined as follows: If N is an even number, then divide N by two to get a new value for N If N is an odd number, then multiply N by 3 and add 1 to get a new value for N. Continue to generate numbers in this way until N becomes equal to 1 For example, starting from N = 3 the complete ’3N+1’ sequence would be: 3, 10, 5, 16, 8, 4, 2, 1 Write code to ask the user to enter a positive integer (N) in the main() function. Write a function sequence() that receives the integer value N and display the ‘3N+1’ sequence starting from the integer value that was received (entered by the user). The function must also count and return the numbers that the sequence consists of. The returned value must be displayed from the main() function.

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter8: Arrays And Strings

Section: Chapter Questions

Problem 21PE

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le.com/forms/d/e/1FAlpQLSc6PlhZGOLJ4LOHo5cCGEf9HDChfQ-tT1bES-BKgkKu44eEnw/formResponse The following iterative sequence is defined for the set of positive integers: Sn/2 3n +1 ifn is odd if n is even Un = Using the rule above and starting with 13, we generate the following sequence: 13 u13 = 40 u40 =20 u20 = 10→ u10 =5 u5 = 16 u16 = 8 ug = 4 → Us =2 u2 =1. It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. The below function takes as input an integer n and returns the number of terms generated by the sequence starting at n. function i-Seq (n) u=n; i=%3; while u =1 if statement 1 u=u/2; else statement 2 end i=i+1; end statement 1 and statement 2 should be replaced by: None of the choices statement 1 is "mod(u,2)=D%3D0" and statement 2 is "u = 3*u+1;" statement 1 is "u%2" and statement 2 is "u = 3*u+1;" O statement 1 is "mod(n,2)=30" and statement 2 is "u = 3*n+1;"
c++ Given two sequences with n integers between 0 and 9, interpreted as two n-digit integers, calculate a sequence of numbers that represents the sum of the two integers. Example: n = 8, 1st sequence 8 2 4 3 4 2 5 1 2nd sequence + 3 3 7 5 2 3 3 7 1 1 6 1 8 6 5 8 8
JAVA CODE PLEASE Functions With No Parameters and Return Values Quiz by CodeChum Admin Create a function named banner that prints “CodyChum” with a newline. In the main function, write a program that accepts an integer input which would serve as an inclusive stopping point of a loop iteration that starts at the value of 1. If the number that loops through the range is divisible by either 2 or 3, then call the banner function, otherwise just print the number. Input 1. One line containing an integer Output First line is the integer user input. Succeeding multiple lines containing an integer or “CodyChum” 10 1 CodyChum CodyChum CodyChum 5 CodyChum 7 CodyChum CodyChum CodyChum
1.Given a positive integer, N, the ’3N+1’ sequence starting from N is defined as follows:If N is an even number, then divide N by two to get a new value for NIf N is an odd number, then multiply N by 3 and add 1 to get a new value for N.Continue to generate numbers in this way until N becomes equal to 1For example, starting from N = 3 the complete ’3N+1’ sequence would be:3, 10, 5, 16, 8, 4, 2, 1Write c++ code to ask the user to enter a positive integer (N) in the main() function. Write a function sequence()that receives the integer value N and display the ‘3N+1’ sequence starting from the integer value that wasreceived (entered by the user). The function must also count and return the numbers that the sequenceconsists of. The returned value must be displayed from the main() function.
#Solve it with C programing Mr. X is a student of Computer Science. He is facing a problem and needs your help to solve it.The problem is, you will be given N integer numbers( 4<n<=100) which are arranged side by side (x1,x2,x3,x4,.....).Now you have to sort those items in ascending order. After doing this , your job is to print the items in a sorted way and find the median.[NB: when N is odd , Median= (N+1)/2; When N is even Median= (N/2)+1 ] Sample Input: 74 7 11 2 9 3 5Sample Output:2 3 4 5 7 9 115
hexadecimal We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different “digits” {0, 1, . . . , 9}. Sometimes though it is useful to write numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0,1,...,9,A,B,C,D,E,F}. So for example, a 3 digit hexadecimal number might be 2B8. (d) How many 3-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) (or both)? Explain (please answer it the way i can make myself understand)
C++ program help Write a program to transform n mixed numbers into improper fractions. A mixed number is a number consisting of a whole number and a proper fraction. Given the mixed number: 3 3/5 The improper fraction representing this mixed number is: 18/5 Convert 3 3/5 into an improper fraction, we: Multiply the whole number with the denominator of the proper fraction. Add the numerator of the proper fraction to the product from step 1 to obtain the numerator of the improper fraction. The denominator of the improper fraction will be the same as the denominator of the proper fraction. The input for your program will be: An integer value that represents the number mixed numbers to transform. Three integer values per mixed number representing the whole number, the numerator, and the denominator of the mixed number. Given the following input: 33 3 54 1 23 1 3 The expected output is: 3 3/5 = 18/5 4 1/2 = 9/2 3 1/3 = 10/3
In C++ Write a program that prints the count of all prime numbers between A and B (inclusive), where A and B are defined as follows: A = 12345 B = A + 5000 Note: A prime number is any number, greater or equal to 2, that is divisible ONLY by 1 and itself. Here are the first 10 prime numbers: 2, 5, 7, 11, 13, 17, 19, 23, and 29. Rules: You should first create a boolean function called isPrime and use that function in your program. This function should take in any int and return true if the number is prime, otherwise, return a false. In the main body of your program, you should create a loop from A to B (inclusive) and use isPrime function to determine if the loop number should be counted or not. Your program SHOULD NOT PRINT the individual prime numbers. You can print them for your own testing purpose, but in the final submission, comment out such print statements. Your program SHOULD ONLY PRINT the answer -- which is a number. Do not print extra characters in your answer (E.g.…
Sevens rule, zeros drooldef seven_zero(n):Seven is considered a lucky number in Western cultures, whereas zero is what nobody wants to be. We now bring these two opposites briefly together with positive integers that consist of some solid sequence of sevens, followed by some (possibly empty) solid sequence of zeros. Examples of such integers are 7, 77777, 7700000, 77777700, or 700000000000000. A surprising theorem proves that for any positive integer n, there exist infinitely many integers of such seven-zero form that are divisible by n. This function should return the smallest such seven-zero integer. This exercise is about efficiently generating all numbers of the constrained form of sevens and zeros in strictly ascending order to guarantee finding the smallest working such number. This logic might be best written as a generator to yield such numbers. The body of this generator consists of two nested loops. The outer loop iterates through the number of digits d in the current number.…
7. We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different "digits" {0,1,.9). Sometimes though it is useful to write numbers in hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0,1,..,9,A,B,C,D,E,F}. So, for example, a 3 digit hexadecimal number might be 3B8. (a) How many 2-digit hexadecimals are there in which the first digit is E or F? Explain your answer in terms of the additive principle (using either events or sets). (b) Explain why your answer to the previous part is correct in terms of the multiplicative principle (using either events or sets). Why do both the additive and multiplicative principles give you the same answer? (c) How many 3-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)? Explain. (d) How many 3-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) (or both)? Explain.
//Can you please debug this program. Thank you #include<stdio.h>#include<stdlib.h> //This function takes n and k and computes n using Pascal’s Rule.long choose(int n, int k); //This functions below create a memoization table containing//long values of dimension (n+1)×(k+1) long chooseWithMemoization(int n, int k); //This is a new recursive function that also takes the table as a parameter.//When the function needs to compute (n , k )it checks the table first, and//if the value has already been comput(is not -1) then it returns that value. long chooseWithMemoizationRecursive(int n, int k, long **tableau); int main(int argc, char **argv) {//variable declaration int n; int k; scanf("%d",&n); scanf("%d",&k ); long choice = -1;//printing out the out put choice = chooseWithMemoization(n,k); printf("choose(%d,%d) = %ld\n", n, k, choice); return 0;} /** * This function takes n and k and computes n using Pascal’s Rule.*/long choose(int…
Write a function that takes in an integer n and computes n!. Do this without recursion. In [ ]: deffactorial_iter(n):"""Takes in an integer n>0 and returns the product of all integers from 1 to n."""# YOUR CODE HEREraiseNotImplementedError() In [ ]: In [ ]: assert factorial_iter(6) == 720 assert factorial_iter(7) == 5040 assert factorial_iter(10) == 3628800