Integer numValues is read from input. Then numValues integers are read and stored in vector numbersList. Write a loop that adds 1 to each element in numbersList that is divisible by 5. Ex: If the input is 3 25 69 41, then the output is: Original numbers: 25 69 41 Nearest numbers not divisible by 5: 26 69 41

JS Write a function named find_value that takes a list/array of whole numbers as a parameter. Your functio will need to find if the parameter contains the value 3 as an entry. Return true if the input has an entry equivalent to 3 and false if it does not.

Write a for loop to print all NUM_VALS elements of vector courseGrades, following each with a space (including the last). Print forwards, then backwards. End with 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 = courseGrades.size() - 1 (Notes)C++ without using std:: Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element vector (vector<int> courseGrades(4)). See "How to Use zyBooks".Also note: If the submitted code has errors, 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. #include <iostream>#include <vector>using namespace std; int main() {   const int NUM_VALS = 4;   vector<int> courseGrades(NUM_VALS);   int i;    for (i = 0; i < courseGrades.size(); ++i) {      cin >>…

Write a loop that counts how many elements in an array are equal to zero. arrays.cpp 1 #include // sizet 2 int countZeros (const int values[], size_t size) { int count = 0; 3 4 for (int i; i using namespace std; 3 2 4 int countZeros(const int values[], size_t size); 5 int main() { int a[] = {1, 2, 0, 3}; cout <« countZeros (a, 4) <« endl; cout « "Expected: 1" « endl; 6 7 8 9 10 11 int b[] = {0, 2, 0, 3}; cout <« countZeros (b, 4) <« endl; cout « "Expected: 2" « endl; 12 13 14 15 int cl] -{1, 0, θ, 0, 0 ; cout <« countZeros (c, 5) <« endl; cout « "Expected: 4" « endl; 16 17 18 19 } CodeCheck Reset Testers Running Tester.cpp pass fail fail 1 Expected: 1 Expected: 2 Expected: 4 Score 1/3

For any element in keysList with a value greater than 50, print the corresponding value in itemsList, followed by a comma (no spaces). Ex: If the input is: 32 105 101 35 10 20 30 40 the output is: 20,30, 1 #include 2 3 int main(void) { const int SIZE_LIST = 4; int keysList[SIZE_LIST]; int itemsList[SIZE_LIST]; int i; 4 6 7 8 scanf("%d", &keysList[0]); scanf ("%d", &keysList[1]); scanf("%d", &keysList[2]); scanf("%d", &keysList[3]); 10 11 12 13 scanf ("%d", &itemsList[0]); scanf ("%d", &itemsList[1]); scanf("%d", &itemsList[2]); scanf ("%d", &itemsList[3]); 14 15 16 17 18 19 /* Your code goes here */ 20 21 printf("\n"); 22 23 return 0; 24 }

There is no difference between an array and an arraylist   True   False

Recamán's sequence def recaman(n): Compute and return the first n terms of the Recamán's sequence, starting from the term a1 = 1.Note how the definition for the current next element depends on whether a particular number is already part of the previously generated part of the sequence. To let your function execute in a speedy fashion even when generating a sequence that contains millions of elements, you should use a set to keep track of which values are already part of the previously generated sequence. This way you can generate each element in constant time, instead of having to iterate through the entire previously generated list. The better technique that saves time by using more memory can create this list arbitrarily far in linear time, and should therefore be fast even for millions of elements, at least until it runs out of memory. n  Expected result 10 [1, 3, 6, 2, 7, 13, 20, 12, 21, 11] 1000000 (a list of million elements whose last five elements are [2057162,1057165,…

Integer numValues is read from input. Then numValues integers are read and stored in vector weeklyRentList. Write a loop that outputs each element in weeklyRentList that is less than 100, and assigns the element with twice the element's current value. End each output with " is corrected to twice the current value' followed by a newline. Ex: If the input is 3 31 143 105, then the output is: Raw weekly rent: 31 143 105 31 is corrected to twice the current value Adjusted weekly rent: 62 143 105 1 #include 2 #include 3 using namespace std; 4 5 int main() { 6 int numValues; 7 unsigned int i; 0 vector weeklyRentList; cin >> numValues; 10 11 weeklyRentList.resize(numValues); for (i = 0; i > CS Scanned 12 13 14 17 18 cout << "Raw weekly rent: " ローロー

Integers are read from input and stored into a vector until 0 is read. If the vector's last element is odd, output the odd elements in the vector. Otherwise, output the even elements in the vector. End each number with a newline. Ex: If the input is -9 12 -6 1 0, the vector's last element is 1. Thus, the output is: -9 1 Note: (x % 2 != 0) returns true if x is odd. int value; int i; bool isodd; 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 } cin >> value; while (value != 0) { } for (i = 0; i > value; Check return 0; 1 Next level 2 X 1: Compare output 3 4 For input -9 12 -6 1 0, the vector elements are -9, 12, -6, and 1. The last element, 1, is odd. The odd elements in the vector, -9 and 1, are output, each on a new line. Not all tests passed. 2 V 3

Integers are read from input and stored into a vector until 0 is read. If the vector's last element is odd, output the odd elements in the vector. Otherwise, output the even elements in the vector. End each number with a newline. Ex: If the input is -9 12 -6 1 0, the vector's last element is 1. Thus, the output is: -9 1 Note: (x % 2!= 0) returns true if x is odd. 4 5 int main() { 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20} vector vect1; int value; int i; bool isodd; cin >> value; while (value != 0) { vect1.push_back(value); cin >> value; } V* Your code goes here */ return 0; 2 3