#include <iostream> 
using std::cin; 
using std::cout; 
using std::endl; 
using std::ios;

#include <iomanip>
using std::fixed;
using std::setw; 
using std::setprecision; 
using std::showpoint;

int main()
{
   const int PEOPLE = 5;
   const int PRODUCTS = 6;

   int sales[5][6]={{1},{2},{3},{4}};
   double value;
   double totalSales;
   double productSales[ PRODUCTS ] = { 0.0 };
   int salesPerson;
   int product;


   cout << "Enter the salesperson (1 - 4), product number (1 - 5), and "
        << "total sales.\nEnter -1 for the salesperson to end input.\n";
   cin >> salesPerson;

   while ( salesPerson != -1 ) 
   {
      cin >> product >> value;

      value++;
      cin >> salesPerson;
   } // end while

   cout << "\nThe total sales for each salesperson are displayed at the "
        << "end of each row,\n" << "and the total sales for each product "
        << "are displayed at the bottom of each column.\n " << setw( 12 ) 
        << 1 << setw( 12 ) << 2 << setw( 12 ) << 3 << setw( 12 ) << 4 
        << setw( 12 ) << 5 << setw( 13 ) << "Total\n" << fixed << showpoint;

   for ( int i = 1; i < 4; i++ ) 
   {
      totalSales = 0.0;
      cout << i;

      for ( int j = 1; j < 5; j++ ) 
      {
          ++ totalSales;

         cout << setw( 12 ) << setprecision( 2 ) << sales[ i ][ j ];

          += productSales ;
      } // end for

      cout << setw( 12 ) << setprecision( 2 ) << totalSales << '\n';
   } // end for

   cout << "\nTotal" << setw( 8 ) << setprecision( 2 ) 
      << productSales[ 1 ];

   // display total product sales
   for ( int j = 2; j < totalSales; j++ )
      cout << setw( 12 ) << setprecision( 2 ) << productSales[ j ];

   cout << endl;
   return 0; // successful termination
} // end main

Use a two-dimensional array to solve the following problem. A company has four salespeople (1 to 4) who sell five different products (1 to 5). Once a day, each salesperson passes in a slip for each different type of product sold.

Each slip contains the following:
a) The salesperson number
b) The product number
c) The total dollar value of that product sold that day

Thus, each salesperson passes in between 0 and 5 sales slips per day. Assume that the information from all of the slips for last month is available. Write a program that will read all this information for last month's sales and summarize the total sales by salesperson by product. All totals should be stored in the two-dimensional array sales. After processing all the information for last month, print the results in tabular format with each of the columns representing a particular salesperson and each of the rows representing a particular product. Cross total each row to get the total sales of each product for last month; cross total each column to get the total sales by salesperson for last month. Your tabular printout should include these cross totals to the right of the totaled rows and to the bottom of the totaled columns

My coding:

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Is anyone able to help me with this? I can get the basics down, but I'm terrible with Javascript. Funny, right?

I'm the same way with javascript, I find it very hard to get anything done. However most C++ constructs do exist in javacript, including multidimensional arrays, for loops, etc. You can easily google for "How to do <something> in javascript" and get loads of answers and examples. Post a first attempt at converting the above program, then we'll help debug it if necessary.

don't cheat yourself out of an education, you're paying for it, why not get the most out of it? you might try the documentation for JavaScript at MDN. it costs $2 for documentation if you need a local copy. too bad they don't have a cart for this. try miva merchant, MDN. https://developer.mozilla.org/en-US/

#include <iostream>

using namespace std;

int main()

int modepa;
float prix;
int sum1 = 0;
int sum2 = 0;

while(true){

cout<<"entre� le mode passe" << endl;
cin>> modepa;

  if(modepa == 8630)

    cout<<"holle adnane"<< endl;
     cout<<"entre� le prix"<< endl;
        cin>> prix;
        sum1 += prix;
    cout<< "le solde cridet est :"<< sum1 << endl;


else if (modepa = 0405)

     cout<<"holle yassine"<< endl;
     cout<<"entre� le prix"<< endl;
        cin>> prix;
        sum2 += prix;
    cout<< "le solde cridet est : "<< sum2 << endl;



















return 0;
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