1、Counting Primitive Operations void STRAITMAXMIN(A,n,max,min) //Set the maximum value in A to max and the minimum value to min { int i,n max=min=A[1] for i=2 to n { if A[i]> max max=A[i]; if A[i]< min min=A[i]; } }

C programming: Use following code and modify in the end to get median. Sorting is not needed as it already sorted. #include <stdio.h> #define MAX_SCORE 20#define MAX_SIZE 1000 int main() {int i = 0 ; // used as index into arrayint n = 0 ; // number of items in arrayint r ; // used to ensure user enters an integerdouble average ; // average of numbers entered by userdouble num ; // total number of items in array int A[MAX_SIZE] ; // will hold the scores n = 0 ;i = 0 ; // read in the scores into array A[]while (1) {r = scanf("%d", &A[i] ) ; // end of input?if ( r <= 0 ) {break ;} i = i + 1 ;// exceeded array size?if (i >= MAX_SIZE) {break ;}} // Make n the number of items stored in A[]n = i ; // compute the average int sum = 0 ; for (i = 0 ; i < n ; i++) {sum = sum + A[i] ;} num = n ; // num is doubleaverage = sum / num ; printf ("The average score is: %f\n", average) ; // *****// ***** TODO// *****// Declare a new count[] array.int count[MAX_SCORE + 1] ; // Initialize...

C++ Language Write two statements that each use malloc to allocate an int location for each pointer. Sample output for given program:numPtr1 = 44, numPtr2 = 99 #include <stdio.h>#include <stdlib.h> int main(void) { int* numPtr1 = NULL; int* numPtr2 = NULL; /* Your solution goes here */ scanf("%d", numPtr1); scanf("%d", numPtr2); printf("numPtr1 = %d, numPtr2 = %d\n", *numPtr1, *numPtr2); free(numPtr1); free(numPtr2); return 0;}

C++ complete and create magical square #include <iostream> using namespace std; class Vec {public:Vec(){ }int size(){return this->sz;} int capacity(){return this->cap;} void reserve( int n ){// TODO:// (0) check the n should be > size, otherwise// ignore this action.if ( n > sz ){// (1) create a new int array which size is n// and get its addressint *newarr = new int[n];// (2) use for loop to copy the old array to the// new array // (3) update the variable to the new address // (4) delete old arraydelete[] oldarr; } } void push_back( int v ){// TODO: if ( sz == cap ){cap *= 2; reserve(cap);} // complete others } int at( int idx ){ } private:int *arr;int sz = 0;int cap = 0; }; int main(){Vec v;v.reserve(10);v.push_back(3);v.push_back(2);cout << v.size() << endl; // 2cout << v.capacity() << endl; // 10v.push_back(3);v.push_back(2);v.push_back(3);v.push_back(4);v.push_back(3);v.push_back(7);v.push_back(3);v.push_back(8);v.push_back(2);cout...

C programming #include <stdio.h>int main() {int i, j, n ;printf("height? ") ;scanf("%2d", &n) ;for (i = 1 ; i <= n ; i++) {// printf("%d: ", i) ;for (j = 1 ; j <= i ; j++) {// Pick *one* of the following// printf("%d", j % 10) ;printf("*") ;}printf("\n") ;}return 0 ;} How can I get output like this by editing the given programming -- height? 5************************* and then how can I get the following?- height? 7 1 123 12345 1234567 123456789 12345678901 1234567890123

#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; }

#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; }

Question 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i>>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews O

//Assignment 06 */public static void main[](String[] args) { String pass= "ICS 111"; System.out.printIn(valPassword(pass));} /* public static boolean valPassword(String password){ if(password.length() > 6) { if(checkPass(password) { return true; } else { return false; } }else System.out.print("Too small"); return false;} public static boolean checkPass (String password){ boolean hasNum=false; boolean hasCap = false; boolean hasLow = false; char c; for(int i = 0; i < password.length(); i++) { c = password.charAt(1); if(Character.isDigit(c)); { hasNum = true; } else if(Character.isUpperCase(c)) { hasCap = true; } else if(Character.isLowerCase(c)) { hasLow = true; } } return true; { return false; } }

8. Know how to do these, to trace functions like these and to debug functions like these: // recursive power , compute xn int exp(int x, int n){ if(n== return return *exp( ); } void main(){ int a,b; cin>>a>>b; cout>a; rev_print(a); }

#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; }

C programming: Use following code and modify in the end to get median. Sorting is not needed as it already sorted. #include <stdio.h> #define MAX_SCORE 20#define MAX_SIZE 1000 int main() {int i = 0 ; // used as index into arrayint n = 0 ; // number of items in arrayint r ; // used to ensure user enters an integerdouble average ; // average of numbers entered by userdouble num ; // total number of items in array int A[MAX_SIZE] ; // will hold the scores n = 0 ;i = 0 ; // read in the scores into array A[]while (1) {r = scanf("%d", &A[i] ) ; // end of input?if ( r <= 0 ) {break ;} i = i + 1 ;// exceeded array size?if (i >= MAX_SIZE) {break ;}} // Make n the number of items stored in A[]n = i ; // compute the average int sum = 0 ; for (i = 0 ; i < n ; i++) {sum = sum + A[i] ;} num = n ; // num is doubleaverage = sum / num ; printf ("The average score is: %f\n", average) ; // *****// ***** TODO// *****// Declare a new count[] array.int count[MAX_SCORE + 1] ; // Initialize...

Caesars Cypher in C programming language How do you program an encryption and decryption for a Caesars Cypher that covers all the possible ASCII characters? In this there should be a shift over K = 5 but the example below is K = 3. This should result in actual character and not the control ASCII represenations. For example: >>> $ encrypt(I have a key)N%mf{j%f%pj~>>> $ encrypt(see me at 3)xjj%rj%fy%8>>> $ decrpyt(|jfw%ns%gqzj)wear in blue Assume that the text parsing method has been succesful and it's the encryption/decryption of the char array that needs to be factored. Formula Encryption -> C = E(k, P) = (P + k) (mod 26) Decryption -> P = D(k, C) = (C - k) (mod 26) Where k = 5