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Diano4ka-milaya [45]
2 years ago
14

A program needs to allow a customer to input integers until the input number is zero. When the input number is zero, the program

should write the message "The number of integers you have input is: X" (instead of X there should be the total count of numbers input). What should be added below to replace the question mark
Computers and Technology
1 answer:
adell [148]2 years ago
7 0

Answer:

The answer is "Increment count by 1".  

Explanation:

  • In the given program a customer inserts numbers that must store in the array variable on the insertion time we define a condition that if the inserting number value is equal to 0.
  • Then the program will terminate with a message that is "The number of integers you have input is: X".
  • In the program first, we insert some value then we insert a number that is 0. So the program will increment the value of count by 1 and add all values and prints its total.
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Explanation:

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g 18.6 [Contest 6 - 07/10] Reverse an array Reversing an array is a common task. One approach copies to a second array in revers
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Answer:

I am writing C++ program. Let me know if you want the program in some other programming language.  

#include <iostream> //to use input output functions

using namespace std; //to identify objects like cin cout

void reverse(int array[], int size){ //method to reverse an array  

int i, j, temp; //declare variables  

for (i = 0; i < size / 2; i++) { /* the loop swaps the first and last elements, then swap the second and second-to-last elements and so on until the middle of the array is reached */  

temp = array[i];  

array[i] = array[size - i - 1];  

array[size - i - 1] = temp; }  

cout<<"Reversed array: "; //prints the resultant reversed array after swapping process  

for (j = 0; j < size; j++) { //loop to print the reversed array elements  

cout<<array[j]<<" "; } }    

int main() { //start of main() function body  

cout<<"Enter the length of array: "; //prompts user to enter the length of the array

       int n; //stores the value of size of array

       cin>>n; //reads the input size of array

       int A[n]; //declare array A of size n

       cout<<"Enter the elements: "; //prompts user to enter array elements

       for(int i=0;i<n;i++){ //loop for reading the elements of array

           cin >> A[i];        }

         reverse(A,n); } //calls reverse function to reverse the elements of the input array A by passing array A and its length n to function reverse

Explanation:  

The reverse function takes an array and its length as parameter. In the body of reverse function three integer type variables i,j and temp are declared.

for (i = 0; i < size / 2; i++) The for loop has a variable to iterate through the array. The for loop checks if the value of i is less than the array length. Lets say we have an array of elements 1, 2, 3, 4. As i=0 and size/2 = 4/2= 2. So the condition checked in for loop is true as 0<2 which means the body of for loop will be executed.

In the body of the loop there are three swap statements.

temp = array[i] This statement stores the element of array at i-th index to temp variable.

array[i] = array[size - i - 1] statement stores the element of array at size-i-1 to array index i.  

array[size - i - 1] = temp statement stores the the value in temp variable to the array index size-i-1

In simple words first, the first and last elements of array are swapped, then the second and second last elements are swapped. This process keeps repeating and for loop keeps executing these swap statements until the middle of the array is reached. In this program no extra array is used for reversing procedure but just a variable temp is used to hold the array elements temporarily.

array[4] = { 1 , 2 , 3 , 4}  

At first iteration  

i < size/2 is true as 0<2  

So the program control moves in the body of the loop where three swap statement works as following:  

temp = array[i]  

temp = array[0]  

As element at 0-th index which is the first element of array is 1 so

temp= 1  

array[i] = array[size - i - 1];  

array[0] = array[4-0-1]  

           = array[3]  

As the element at 3rd index of array is 4 which is the last element of the array so

array[0] = 4

This means that 4 becomes the first element of array now.

array[size - i - 1] = temp  

As the value in temp was set to 1 so  

array[4-0-1] = 1  

array[3] = 1  

This means 1 is assigned to the 3rd index of array. Which means 1 becomes the last element of the array now.

This is how the first and last elements of array are swapped.  

2nd iteration: value of i is incremented to 1 and now i = 1  

Again for loop condition is checked which evaluates to true as 1<2

temp = array[i]  

temp = array[1]  

As element at 1st index which is the second element of array is 2 so

temp= 2  

array[i] = array[size - i - 1];  

array[1] = array[4-1-1]  

           = array[2]  

As the element at second index of array is 3 which is the last element of the array so

array[1] = 3  

This means that 3 becomes the second element of array now.

array[size - i - 1] = temp  

As the value in temp was set to 2 so  

array[4-1-1] = 2  

This means 2 is assigned to the 2nd index of array. Which means 2 becomes the second last element of the array now.

This is how the second and second to last elements of array are swapped.  

3rd iteration: value of i=2 The for loop body will not execute now as the for loop condition i<size-2 evaluates to false because 2=2. So the loop stops and next for (j = 0; j < size; j++) loop iterates through array[] which has now been reversed and print the elements of this reversed array.

Next the main() method passes length of the array and the array elements to the   reverse function and prints these elements in reverse.  

The output is:

4  3 2  1  

The program and output according to the example 2,5,9,7 given in the question is attached as a screenshot.

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Static void DashedLine(int n){     if (n>1) Console.WriteLine(new String('-', n));}
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Ajay wants to read a brief overview about early settlers in the United States. Which type of online text source should he most l
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Well, integrity is a must if you are reading anything. So it would be any website that has either a .edu or a .gov at the end of a URL. 

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Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi
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Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

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