int arr[4]; for (int i = 0; i<4; i++) if (i%2 == 0) arr[i]=2*i; else arr[i]=i; What value is assigned to arr[0]? What value is assigned to arr[1]? What value is assigned to arr[2]? What value is assigned to arr[3]?
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Refer to the following code segment. show your solution.
int arr[4];
for (int i = 0; i<4; i++)
if (i%2 == 0)
arr[i]=2*i;
else
arr[i]=i;
- What value is assigned to arr[0]?
- What value is assigned to arr[1]?
- What value is assigned to arr[2]?
- What value is assigned to arr[3]?
Step by step
Solved in 2 steps
- Describe what problem occurs in the following code. What modifications should be made to it to eliminate the problem? int[] sums = {6, 12, 3, 32, 12, 10, 9, 6}; for (int index = 1; index <= sums.length; index += 1) { System.out.println(sums[index]); }def find_root4(x, epsilon): ''' IN PYTHON Assume: x, epsilon are floating point numbers and epsilon > 0 Use bisection search to find the following root of x such that If x >=0, return y such that x - epsilon <= y ** 2 <= x + epsilon Else, return y such that x - epsilon <= y ** 7 <= x + epsilon Note: You must use bisection search to implement the function. ''' passmplement a Java program that applies the Newton-Raphson's method xn+1 = xn – f(xn) / f '(xn) to search the roots for this polynomial function ax6 – bx5 + cx4 – dx3+ ex2 – fx + g = 0. Fill out a, b, c, d, e, f, and g using the first 7 digits of your ID, respectively. For example, if ID is 4759284, the polynomial function would be 4x6 – 7x5 + 5x4 – 9x3+ 2x2 – 8x + 4 = 0. The program terminates when the difference between the new solution and the previous one is smaller than 0.00001 within 2000 iterations. Otherwise, it shows Not Found as the final solution.
- Trace through (show all the steps as we did in class) for the following java code for the factorial function. Show all the steps (illustrate, do not code) that lead to the final answer when myFactorial(6) is called: int myFactorial( int integer) { if( integer == 1) return 1; else { return(integer*(myFactorial(integer-1); } }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. PrUsing C language, trace this: void trace(int x, int *y, int z) {x = 1; *y=2;z=4;printf("%2d %2d %2d\n", x, *y, z);}main() {int x=1, y=3,z=4;clrscr();printf("%2d %2d %2d\n",x,y,z);trace(y,&x,z);printf("%2d %2d %2d\n",x,y,z);trace(x,&z,y);printf("%2d %2d %2d\n",x,y,z);trace(z,&y,x);printf("%2d %2d %2d\n",x,y,z);getch();return 0; }(c)#include<s
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- please help solve this code using c++ arrays and pointers, and incorporate the following code: for (int i = 0 ; i < course ; i++) { // tell user which course is at cout << "Course "<< i+1 << endl ; // make pointer point to row arr[i] = new float [i+1] ; and then record assignment marks adding to the total for (int j = 0 ; j < entry ; j++) { // prompt for each course's entry cout << "Enter entry " <<j+1<< ": " ; cin >> score ; // store the score to each row arr[i][j] = score ; courseTol += score; }Debug the following program and answer the following questions. #include <stdio.h> typedef struct node { int value; struct node *next; } node; int ll_has_cycle(node *first) { node * head = first; while (head->next) { head = head->next; if (head == first) return 1; } return 0; } void test_ll_has_cycle(void) { int i,j; node nodes[5]; for(i=0; i < sizeof(nodes)/sizeof(node); i++) { nodes[i].next = NULL; nodes[i].value = i; } nodes[0].next = &nodes[1]; nodes[1].next = &nodes[2]; nodes[2].next = &nodes[1]; printf("Checking first list for cycles. There should be a cycle, ll_has_cycle says it has %s cycle\n", ll_has_cycle(&nodes[1])?"a":"no"); printf("Checking length-zero list for cycles. There should be none, ll_has_cycle says it has %s cycle\n",…Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…