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ehidna [41]
3 years ago
8

Write an application named SumInts that allows the user to enter any number of integers continuously until the user enters 999.

Display the sum of the values entered, not including 999.
Computers and Technology
1 answer:
Nana76 [90]3 years ago
4 0

Answer:

import java.util.Scanner;

public class num10 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter the numbers to add up. enter 999 to stop");

       int num = in.nextInt();

       int sum = 0;

       while (num!=999){

           sum = sum+num;

           System.out.println("Enter the next number");

           num = in.nextInt();

       }

       System.out.println("The sum is: "+sum);

   }

}

Explanation:

The application is implemented in Java

A while loop is used to continously prompt user for inputs. The condition of the while loop is  while (num!=999)

When the number 999 is entered, it displays the sum which is initialized to 0

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Answer:

The c++ program for the given scenario is given below.

#include <iostream>

using namespace std;

int main() {

   int len=20, arr[len], data;    

   // initialize all elements of the array to 0

   for(int i=0; i<len; i++)

   {

       arr[i] = 0;

   }    

   cout<<"This program outputs all the numbers less than or equal to the given number."<<endl;

   cout<<"Enter the list of numbers. Enter 0 to stop entering the numbers. "<<endl;

   for(int i=0; i<len; i++)

   {

       cin>>arr[i];        

       // 0 indicates user wishes to stop entering values  

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       else

           continue;

   }    

   // number from which the list is to be compared

   cout<<"Enter the number to be compared."<<endl;

   cin>>data;

     

   cout<<"The values less than or equal to the number "<<data<<" are "<<endl;

   for(int i=0; i<len; i++)

   {  

       // 0 indicates the end of the list entered by the user

       if(arr[i]==0)

           break;

       if(arr[i] <= data)

           cout<<arr[i]<<endl;

   }    

   return 0;    

}  

OUTPUT

This program outputs all the numbers less than or equal to the given number.

Enter the list of numbers. Enter 0 to stop entering the numbers.  

23

45

67

89

10

0

Enter the number to be compared.

59

The values less than or equal to the number 59 are  

23

45

10

Explanation:

This program takes input only from the user and makes no assumptions.

The list of numbers entered by the user are stored in an array. If the user input is less than the size of the array, the remaining elements are set to 0.

While taking input from the user, if any element of the array is found to be 0, the loop is discontinued.

for(int i=0; i<len; i++)

   {

       cin>>arr[i];        

       if(arr[i] == 0)

           break;

       else

           continue;

   }  

Same test is applied when all the numbers less than the given number are displayed.

for(int i=0; i<len; i++)

   {  

        if(arr[i]==0)

           break;

       if(arr[i] <= data)

           cout<<arr[i]<<endl;

   }

The above program takes into account all the specifications in the given question.

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