The following program has run time (number of operations) proportional to: n = int(input("Please enter n"))
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Please sttae clearly what should I choose from the multiple choice
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- int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;10Question 6 sm.Full explain this question and text typing work only We should answer our question within 2 hours takes more time then we will reduce Rating Dont ignore this line
- int n = 1; int k - 2; int r = n; if (k < n) { r - kCFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;The following function f uses recursion:def f(n):if n <= 1return nelsereturn f(n-1) + f(n-2)Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but withoutrecursion?a) def f(n):a <- 0b <- 1if n = 0return aelsif n = 1return belsefor i in 1..nc <- a + ba <- bb <- creturn bb) def f(n):a <- 0i <- nwhile i > 0a <- a + i + (i-1)return ac) def f(n):arr[0] <- 0arr[1] <- 1if n <= 1return arr[n]elsefor i in 2..narr[i] <- arr[i-1] + arr[i-2]return arr[n]d) def f(n):arr[0..n] <- [0, ..., n]if n <= 1return arr[n]elsea <- 0for i in 0..na <- a + arr[i]return a
- none#include double f(double x) { return 4 * exp(-x); } double trapezoidalRule(double a, double b, int N) { double h = (b - a) / N; double sum = 0.5 * (f(a) + f(b)); for (int i = 1; i < N; i++) { sum += f(a + i * h); } return h * sum; } int main() { double a = 0.0; // lower limit double b = 1.0; // upper limit int N = 5; // number of trapezoids double integral = trapezoidalRule(a, b, N); std::cout << "The estimate of the integral of f(x) = 4e^-x between " << a << " and " << b << " using " << N << " trapezoids is: " << integral << std::endl; return 0; } Please write down the explanation, step by step for this trapezoidal rule program, Emphasize on this part:for (int i = 1; i < N; i++) { sum += f(a + i * h); }In main() for now: do these - one at a time, each in it's own loop (we will make functions out of them later –Declare an array RandArray with 20 integers –Assign each of the 20 values with a random value from 0 to 99 Hint: Use rand()%100 For mimir: Do not call srand at the top of main. // normally: Call srand(time(0)) at the top of main –(you need #include<cstdlib>) –Write another loop that prints the array, with index values Important: If the output values do not match mimir, please add srand(17); // inside your main function - at the top code format: #include <iostream>using namespace std; #include <cstdlib> // required for rand() int main(){ srand(17); // define a constant ARRAYSIZE that is 20 // declare randArray // set the 20 elements in randArray to be a random number between 0 and 99 // hint: use rand()%100 // print the 20 values return 0;}
- please answerFun(n) { if (n ≤ 1){return;} else { Statement; Fun (√) return; } Assume "Statement" takes (1) time, Further assume that ʼn = 2m, then the time complexity of Fun() is:Language: C Pascal’s triangle is a triangular array, useful for calculating the binomial coefficients, n k , that are used in expanding binomials raised to powers, combinatorics and probability theory. 0 0 1 0 1 1 2 0 2 1 2 2 3 0 3 1 3 2 3 3 4 0 4 1 4 2 4 3 4 4 Evaluating the values of the binomial coefficients, you get the following pattern, 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 The number of the entries in each row is increased by one, as we move down. Each number in the triangle, is constructed by adding the number above it and to the left, with the number above it and to the right. The blank entries as treated as 0. Using the recursion, implement the function that computes the Pascal’s triangle. Pr