write about the iterative function factorial to a recursive function. function shall calculate the faculty for a number, e.g.
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C++
write about the iterative function factorial to a recursive function. function shall calculate the faculty for a number, e.g.
factorial (3) -> 6
Input: 0 output: 1
input: 3 output: 6
The code :
int factorial (int n) {
int f = 1;
for (int i = n; i> 1; i -) {f * = i; }
return f;
}
int main () {
// get input from cin
// call your function and send results to cout
return 0;
}
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- Java C++ Write a recursive function that accepts two arguments into the parameters x and y. The function should return the value of x times y. Remember, multiplication can be performed as repeated addition: 7∗4=4+4+4+4+4+4+47 * 4=4+4+4+4+4+4+47∗4=4+4+4+4+4+4+4in C++ please 1. Write a recursive function called sumover that has one argument n, which is an unsigned integer. The function returns a double value, which is the sum of the reciprocals of the first n positive integers. (The reciprocal of x is the fraction 1/ x .) For example, sumover(1) returns 1.0 (which is 1/1); sumover(2) returns 1.5 (which is 1/1 + 1/2); sumover(3) returns approximately 1.833 (which is 1/1 + 1/2 + 1/3). Define sumover(0) to be zero. Do not use any local vari- ables in your function. Submit output for when n=290 2. isPal funcitonImplement the recursive function isPal(), which determines whether a string str isa simple palindrome. A simple palindrome is a string consisting entirely of thecharacters a-z that reads the same forward and backward. For instance, theupcoming are palindromes: dad, level, mom, madamimadam,gohangasalamiimalasagnahog.Use the following declaration of isPal():bool isPal (const string& str, int startIndex, intendIndex);It returns true…Write a C++ recursive function PrintPattern3 to print following pattern using recursion. No loops allowed whatsoever, and you can write maximum two functions apart from main function. For example, calling your function with these argument PrintPattern1(1,5) should print following pattern. Your function prototype must be as follows recursive function.void PrintPattern3(int start, int end);
- In C++ Write a recursive function called PrintNumPattern() to output the following number pattern. Given a positive integer as input (Ex: 12), subtract another positive integer (Ex: 3) continually until 0 or a negative value is reached, and then continually add the second integer until the first integer is again reached. For this lab, do not end output with a newline. Ex. If the input is: 123 the output is: 12 9 6 3 0 3 6 9 12 #include <iostream> using namespace std; // TODO: Write recursive PrintNumPattern() function int main(int argc, char* argv[]) { int num1; int num2; cin >> num1; cin >> num2; PrintNumPattern(num1, num2); return 0;}(C Language) Write a recursive function called DrawTriangle() that outputs lines of '*' to form a right side up isosceles triangle. Function DrawTriangle() has one parameter, an integer representing the base length of the triangle. Assume the base length is always odd and less than 20. Output 9 spaces before the first '*' on the first line for correct formatting. Hint: The number of '*' increases by 2 for every line drawn.In C programing Write a recursive function that returns the product of the digits of its integer input parameter, n. You may assume that n is non-negative. For example, productDigits(243) should return 24, since 2 x 4 x 3 = 24.int productDigits (int n) {
- Write a recursive function power( base, exponent ) that when invoked returns baseexponentFor example, power( 3, 4 ) = 3 * 3 * 3 * 3. Assume that exponent is an integer greater than orequal to 1. Hint: The recursion step would use the relationship baseexponent = base * baseexponent1and the terminating condition occurs when exponent is equal to1 becausebase1 = base c language(C Language) Write a recursive function called PrintNumPattern() to output the following number pattern. Given a positive integer as input (Ex: 12), subtract another positive integer (Ex: 3) continually until a negative value is reached, and then continually add the second integer until the first integer is again reached. For this lab, do not end output with a newline.Question: First write a function mulByDigit :: Int -> BigInt -> BigInt which takes an integer digit and a big integer, and returns the big integer list which is the result of multiplying the big integer with the digit. You should get the following behavior: ghci> mulByDigit 9 [9,9,9,9] [8,9,9,9,1] Your implementation should not be recursive. Now, using mulByDigit, fill in the implementation of bigMul :: BigInt -> BigInt -> BigInt Again, you have to fill in implementations for f , base , args only. Once you are done, you should get the following behavior at the prompt: ghci> bigMul [9,9,9,9] [9,9,9,9] [9,9,9,8,0,0,0,1] ghci> bigMul [9,9,9,9,9] [9,9,9,9,9] [9,9,9,9,8,0,0,0,0,1] ghci> bigMul [4,3,7,2] [1,6,3,2,9] [7,1,3,9,0,3,8,8] ghci> bigMul [9,9,9,9] [0] [] Your implementation should not be recursive. Code: import Prelude hiding (replicate, sum)import Data.List (foldl') foldLeft :: (a -> b -> a) -> a -> [b] -> afoldLeft =…
- IN C Programing Let us define the weighted sum of an integer array a[0], a[1], a[2], …, a[n-1] be a[0]*1 + a[1]*2 + a[2]*3 + …+a[n-1]*n. For example, the weighted sum of the array [5,2,6] would be 5*1+2*2+6*3 = 27. Write a recursive function that takes in an array numbers and its length n, and returns its weighter sum. You can assume n is non-negative integer. int weightedSum(int numbers[], int n) {Write a program in C that includes the following functions: A function elementsSummation() that takes an integer array and the array size as arguments and prints the summation of the array’s elements. A function smallestElement() that takes the array and its size as arguments and returns the smallest element of the array. A recursive function integerReverse() that takes the array, start index, and the array size as arguments and prints the array back to front.Write a C++ function using recursion that returns the Greatest Common Divisor of two integers. The greatest common divisor (gcd) of two integers, which are not zero, is the largest positive inteteger that divides each of the integers. The code should follow the exact function declaration: int GCD(int number1, int number2);