Consider the following code: int sum = 0; int i = 0; while (i < 6) { sum = sum + i; i++;} Convert while loop to for loop a. for (int i = 1; i < 7; i++) sum = sum + i; b. for (int i = 0; i <= 6; i++) sum = sum + i; c. for (int i = 1; i <= 6; i++) sum = sum + i; d. for (int i = 0; i < 6; i++) sum = sum + i; e. for (int i = 2; i < 6; i++) sum = sum + i;

Consider the following code: int sum = 0; int i = 0; while (i < 6) { sum = sum + i; i++;} Convert while loop to for loop a. for (int i = 1; i < 7; i++) sum = sum + i; b. for (int i = 0; i <= 6; i++) sum = sum + i; c. for (int i = 1; i <= 6; i++) sum = sum + i; d. for (int i = 0; i < 6; i++) sum = sum + i; e. for (int i = 2; i < 6; i++) sum = sum + i;

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter6: Modularity Using Functions

Section6.4: A Case Study: Rectangular To Polar Coordinate Conversion

Problem 9E: (Numerical) Write a program that tests the effectiveness of the rand() library function. Start by...

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int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;
Problem Description: In the Whimsical Library, each book is enchanted with a magical code to ensure the protection of knowledge. The validation process for these magical codes involves a combination of mathematical spells. The validation process is described as follows: Take the sum of the digits at odd positions from left to right. Take the sum of the squared digits at even positions from left to right. If the absolute difference between the results from step 1 and step 2 is a prime number, the book code is considered valid; otherwise, it is invalid. Example 1: Consider the magical book code 86420137. Step 1. Take the sum of the digits at odd positions from left to right. 8+4+0+3=15 Step 2. Take the sum of the squared digits at even positions from left to right. 6^2+2^2+1^2+7^2=36+4+1+49 = 90 Step 3. If the absolute difference between the results from step 1 and step 2 is a prime number. 115-901 75 (not a prime number) Book code 86420137 is invalid. Example 2: Now, let's examine the…
(Numerical) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters to 0, and then generate a large number of pseudorandom integers between 0 and 9. Each time a 0 occurs, increment the variable you have designated as the zero counter; when a 1 occurs, increment the counter variable that’s keeping count of the 1s that occur; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of the time they occurred.
#include using namespace std; const int y = 1; int main () { int static y = 2; int i = 3, j = 4, m = 5, n = 6; %3D int a = [] (int x, int i = 1) { return x * i; } (y, 3) ; %3D int b = [=] (int x) { return [=] (int b) { return b + j; } (x) % 7; } (a); int c = [=] (int x) mutable ->int { m = 6; return [6] (int j) mutable { y = a * b; return y / j; } (x) -m; } (b) ; cout << a << endl; cout << b << endl; cout << c << endl; cout << m << endl; cout << y < endl; return 0; This program outputs 5 lines. What are they? Please explain your work and your answer.
Broken Cabins Problem Statement: There is an Office consisting of m cabins enumerated from 1 to m. Each cabin is 1 meter long. Sadly, some cabins are broken and need to be repaired. You have an infinitely long repair tape. You want to cut some pieces from the tape and use them to cover all of the broken cabins. To be precise, a piece of tape of integer length t placed at some positions will cover segments 5,5+1-sit-1. You are allowed to cover non-broken cabins, it is also possible that some pieces of tape will overlap. Time is money, so you want to cut at most k continuous pieces of tape to cover all the broken cabins. What is the minimum total length of these pieces? Input Format The first line contains three integers n,m and k(1sns10°, namsloº, Isksn) - the number of broken cabins, the length of the stick and the maximum number of pieces you can use The second line contains n integers bl,b2,bn (Isbism) - the positions of the broken cabins. These integers are given in increasing…
int y=0,i; for (int i=0;i<10;++i) y+=i; 36 66 45 55
int f(int &k){k++;return k * 2;}int main(){int i = 1, j = -1;int a, b, c;a = f(i) + i/2;b = j + f(j) + f(j);c = 2 * f(j);return 0;} What are the values of a, b and c id the operands in the expressions are evaluated from left to right and then what are the values when its evaluated right to left?
Complete the following sentence. There are two basic forms of loop constructs: while loops and loops.
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Examine the following for loops and determine the value of ires at the end of each of the loops, and also the number of times each loop executes. (a) ires = 0; for index = -11:11 ires = ires + 1; end ires = number of times (b) ires = 0; for index = 10:-2:1 if index == 6 continue; end ires = ires + index; end ires = number of times ires = 0; for index = 10:-2:1 if index == 6 (c) break; end ires = ires + index: end ires = number of times (d) ires = 0; for indexl = 10:-2:1 for index2 = 2:2:index1 if index2 == 6 break end ires = ires + index2; end end ires = number of times (outer loop) number of times (inner loop)
Convert the following for loop definition to a do while loop: for(int j= 100; j > 0; j--) { if (j % 2 != 0) System.out.println (val); }
1. Rewrite the program segment that follows, using a for loop and do-while loop: = 1; product i = 0; %3D while (i < n) { scanf(“%d", &a); if (a != i) %3D product *= a; ++i; }

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	a.	for (int i = 1; i < 7; i++) sum = sum + i;
	b.	for (int i = 0; i <= 6; i++) sum = sum + i;
	c.	for (int i = 1; i <= 6; i++) sum = sum + i;
	d.	for (int i = 0; i < 6; i++) sum = sum + i;
	e.	for (int i = 2; i < 6; i++) sum = sum + i;