Rewrite the following loops, using the enhanced for loop construct. Here, values is an array of floating - point numbers. a. for (int i = 0; i < values.length; i++) { total = total + values[i]; } b. for (int i = 1; i < values.length; i++) { total = total + values[i]; } c. for (int i = 0; i < values.length; i++) { if (values[i] == target) { save = i; }
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Rewrite the following loops, using the enhanced for loop construct. Here, values is an array of floating - point numbers.
a. for (int i = 0; i < values.length; i++) { total = total + values[i]; }
b. for (int i = 1; i < values.length; i++) { total = total + values[i]; }
c. for (int i = 0; i < values.length; i++) { if (values[i] == target) { save = i; }.
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- using namespace std; int main() int i, m=0,n=4; float arr[100] (55,66,88,1); for(i=0; iWrite these in Pseudocode #1a – In pseudocode, write a call to a function that passes 1 Integer variable and 1 Integer array, and saves a Boolean value in return. #1b – In pseudocode, write the function that accepts 1 Integer and 1 Integer array and returns a Boolean. In the function, search the Integer array with a for-loop, and if the Integer parameter is found in the array, return false. If the Integer parameter is not found, return true. #2a – In pseudocode, write a call to a module that passes 1 Integer variable, 1 Real variable, 1 String constant, and 1 String literal as arguments. #2b – In pseudocode, write the module header that accepts 1 Integer, 1 Real, and 2 Strings as parameters. #3 – This pseudocode has multiple problems. Fix the calling statement and the definition below so that the routine accepts 3 grades as parameters and returns the average into a variable.…int n = 1; int k - 2; int r = n; if (k < n) { r - kIn c++ Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, th backwards. End each loop with a newline. Ex: If courseGrades (7, 9, 11, 10), print: - 7 9 11 10 10 11 9 7 Hint Use two for loops. Second loop starts with i=NUM VALS-1. (Notes) Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element array (int courseGrades[4]) See How to Use zyBooks" Also note: If the submitted code tries to access an invalid array element, such as courseGrades[9] for a 4-element array, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message 2 using namespace std; 3 4 int main() { 5 6 7 8 9 10 11 12 13 14 const int NUM VALS-4; int courseGrades [NUM VALS]; int i; for (i-8; 1 NUM VALS; ++) { cin>> courseGrades[1]; } Your solution goes…a) FindMinIterative public int FindMin(int[] arr) { int x = arr[0]; for(int i = 1; i < arr.Length; i++) { if(arr[i]< x) x = arr[i]; } return x; } b) FindMinRecursive public int FindMin(int[] arr, int length) { if(length == 1) return arr[0]; return Math.Min(arr[length - 1], Find(arr, length - 1)); } What is the Big-O for this functions. Could you explain the recurisive more in details ?extern "C" int f(int *,int,int); int a[2][2] = {{11,12},{21,22}}; void setup(){ Serial.begin(115200); while(!Serial); delay(500); int *arr = (int*)a; Serial.println(f(arr,2,2)); } void loop(){ } .global f f: ldr r3,[r0] // get first element mov r2,#0add r2,r3 mov r1, #3 lp: add r0,#4 // add next elementldr r3,[r0]add r2,r3sub r1,#1 bgt lp mov r0,r2 bx lr Describe what operation is being performed in this codeint calc=1; for (i=1;iJava Programming : A party has been organised on cruise. The party is organised for a limited time (T). The number of guests entering (E[i]) and leaving (L[i]) the party at every hour is represented as elements of the array. The task is to find the maximum number of guests present on the cruise at any given instance within T hours. Example 1: Input : 5 > Value of T [7,0,5,1,3] -> E[], Element of E[0] to E[N-1], where input each element is separated by new line [1,2,1,3,4] -> L[], Element of L[0] to L[N-1], while input each element is separate by new line. Output : 8 -> Maximum number of guests on cruise at an instance. Explanation: 1st hour: Entry 7 Exit: 1 No. of guests on ship : 6 2nd hour : Entry Exit : 2 No. of guests on ship: 6-2=4 Hour 3: Entry: 5 Exit: 1 No. of guests on ship : 4+5-1-8 Hour 4: Entry 1 Exit 3 No. of guests on ship: 8+1-3=6 Hour 5: Entry 3 Exit: 4 No. of guests on ship: 6+3-4-5 Hence, the maximum number of guests within 5 hours is 8.#include int main() { int a[5]; for (int i = = a[i] 1; for (int j = 0; j = a) { } = int n *pa; while (n >= 1) { n = n / 2; printf("%d *pa); pa-- ; 2 1; } Q. What is the output?A/ find code optimization method to this code and then find type to this code 1- For(i=1;i1. Write a recursive method expFive(n) to compute y=5^n. For instance, if n is 0, y is 1. If n is 3, then y is 125. If n is 4, then y is 625. The recursive method cannot have loops. Then write a testing program to call the recursive method. If you run your program, the results should look like this: > run RecExpTest Enter a number: 3 125 >run RecExpTest Enter a number: 3125 2. For two integers m and n, their GCD(Greatest Common Divisor) can be computed by a recursive function. Write a recursive method gcd(m,n) to find their Greatest Common Divisor. Once m is 0, the function returns n. Once n is 0, the function returns m. If neither is 0, the function can recursively calculate the Greatest Common Divisor with two smaller parameters: One is n, the second one is m mod n. Although there are other approaches to calculate Greatest Common Divisor, please follow the instructions in this question, otherwise you will not get the credit. Meaning your code needs to follow the given algorithm. Then…SEE MORE QUESTIONSRecommended textbooks for youC++ Programming: From Problem Analysis to Program…Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage LearningC++ Programming: From Problem Analysis to Program…Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage Learning