Question

Use a two-dimensional array to solve the following problem. A company has four salespeople (0 to 3) who sell five different products (0 to 4). 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 in the input file “data.txt” as shown below. Write a program that will read all this information for last month’s sales (one salesperson’s data at a time) from the input file “data.txt” and summarize the total sales by salesperson and by product. All totals should be stored in the two-dimensional array sales. After processing all the information for last month, write the results in tabular format with each of the columns representing a particular salesperson and each of the rows representing a particular product in an output file “result.txt”. 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 each 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.

Answer 1

ANSWER:

OUTPUT SCREENSHOT:

COPY CODE:

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<iomanip>
#include<bits/stdc++.h>
using namespace std;

//the below function finds the number of digits in floor(x)
int findDigitsBeforeDecimal(double x){
int z = floor(x);
int i = 0;
while(z>0){
i +=1;
z = z/10;
}
return(i);
}

// print empty spaces n times
void printSpaces(int n){
for(int i=0;i<n;i++){
cout << " ";
}
}

int main(){
int i,j,n;
int a,b;
double c;
double sales[4][5];
for(i=0;i<4;i++){
for(j=0;j<5;j++){
sales[i][j] = 0.0;
}
}
cout << "Enter the salesperson, product and total sales " << endl;
cout << "Enter -1 for the salesperson to end input " << endl;
cin >> a ;
while(a!=-1){
cin >> b;
cin >> c;
sales[a][b] += c;
cin >> a;
}
cout << "The total sales for each salesperson are displayed at the the end of each row, "<< endl;
cout << "and the total sales for each product are displayed at the bottom of each column. " << endl;

n = 15;
for(i=0;i<5;i++){
printSpaces(n);
cout << i;
}
cout << endl;
double sum = 0.0;
int dig;
for(i=0;i<4;i++){
cout << i;
sum = 0.0;
for(j=0;j<5;j++){
//find the number of digits before decimal in sales[i][j]
//set width according to that
dig = findDigitsBeforeDecimal(sales[i][j]);
if(j==0){
if(dig==0){
printSpaces(n-dig-4);
}else{
printSpaces(n-dig-3);
}
}else{
if(dig == 0){
printSpaces(n-dig-3);
}else{
printSpaces(n+1-dig-3);
}
}
printf("%.2f",sales[i][j]);
sum += sales[i][j];
}
//print the cumulative row sum
dig = findDigitsBeforeDecimal(sum);
if(dig == 0){
printSpaces(n-dig-3);
}else{
printSpaces(n+1-dig-3);
}
printf("%.2f",sum);
cout << endl;
}
//print the column sum
for(j=0;j<5;j++){
sum = 0.0;
for(i=0;i<4;i++){
sum += sales[i][j];
}
dig = findDigitsBeforeDecimal(sum);
if(dig == 0){
printSpaces(n-dig-3);
}else{
printSpaces(n+1-dig-3);
}
printf("%.2f",sum);
}
cout << endl;

return(0);
}