Write a while loop that prints all positive numbers that are divisible by 10 and less than a given number n. For example, if n is 100, print 10 20 30 40 50 60 70 80 90. loop.cpp 1. 2 #include 4 using namespace std; 5 6 int main() { cout « "n: " « endl: int n; | cin >> n; 10 11 .. while (...) { 12 13 14 ... 15 cout << ... « " "; 16 17 cout <« endl; 18 19 return 0; 20 CodeCheck Reset
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- #include using namespace std; bool isPalindrome(int x) { int n=0,val; val = x; while(x > 0) { n = n * 10 + x % 10; x = x / 10; } } int main() { int n; cin >>n; if(isPalindrome(n)) { cout <#include <stdio.h> struct Single { int num; }; void printSingle(int f) { int binaryNum[33]; int i = 0; while(f>0) { binaryNum[i] = f % 2; f = f/2; i++; } for (int j=i-1; j>= 0; j--) { printf("%d",binaryNum[j]); } } int main() { struct Single single; single.num = 33; printf("Number: %d\n",single.num); printSingle(single.num); return 0; }int calculatepower (int x, int y){ if (y > 0) return x* else return 1; int main (){ int num, pwr; cout > num; cout > pwr; cout <« "Number "<#include <stdio.h>int main(){int d;int s[20],i, r, p, lg=0,m;char c;printf("Enter number of salesman(max 20): ");scanf("%d", &d);for(i=0; i<d; i++){ printf("\n salesman %d sales: ");scanf("%d", &s[i]);}for(i=0; i<d; i++){for(r=i+1; j<d; r++){if(s[i] > s[r]){p= s[i];s[i] = s[r];s[j] = p;}}}printf("\nsalesman lowest to highest: ");for(i=0; i<d; i++){printf("%d\t", s[i]);}for(i=0;i<m;i++){printf("\n highest sales: %d ",lg);if(lg<=s[i])lg=s[i];break;}getch();} >>> the upper part output should be like this enter number of salesman (max 20): 5 salesman 1 500 salesman 2 300 salesman 3 1000 salesman 4 200 salesman 5 1000 in the lower part the lowest to highest the output should become like this salesman 4 200 salesman 2 300 salesman 1 500 salesman 3 1000 salesman 5 1000 highest total sales : 2000#include #include #include int main() { int i, j; printf("Columns | JIn"); for (i = 1; i< 4; ++i) { printf("Outer %6d\n", i); for (j = 0; j< i; ++j) { printf(" Inner%10d\n", j); } } /* heading of outer for loop */ /* heading of inner loop */ %3D return (0); } Create a new code modifying it to use "while loop" instead of the "for loop"c++ coding language I need help with part B and C please. If you are unable to do both, then PLEASE prioritize part C. I am really stuck and really can use the help. This is the code for c that was provided in order to guide me: const int N =31; // N parking spaces bool parking[N]; // the garage void EmptyTheLot(bool parking[], int N) { for(int i=0; i<N; i++) p[i]=false; // empty space } // returns -1 if no space found, //otherwise it returns 0<=i<N for a valid space. int FindSpace(int PlateNumber, bool parking[], int N) { // ????? } main() { EmptyTheLot(parking, N); // start with an empty parking garage. // get plate numbers and fill lot. }int func(int a, int b) { return (aCFG: 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;#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.PlsHelpASAP Program C int getNum(); /*accepts input from the user.*/ void display(int n1, int n2, int n3, int n4); /*displays the given pattern*/ ----------------------------------------------------------------------------------#include <stdio.h> int main(){ int arr[10]; int i; for (i=0; i<10; i++){ arr[i] = i; } for (i=0; i<10; i++){ printf("arr[%d] = %d\n", i, arr[i]); } return 0;} Copy the code and modify it as follows: Add another separate array of 10 integers. The following code parts are to be added (in the same order as specified) after the printing of the values of the first array:(1) Copy the content of the first array to the second array. Add 10 to each of the values of the second array. Make use of a single loop to achieve this part.(2) Print each element of the second array. Make use of another loop for this partconst int size =4; int r1[size]; int i; for (i=0;i<2;i++) r1[i]= 2 * i+3; r1[3]=2 * i+ i; r1[2]=r1[3]; for (i=0;iSEE 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