Answer:
#include<stdio.h>
//declare a named constant
#define MAX 50
int main()
{
//declare the array
int a[MAX],i;
//for loop to access the elements from user
for(i=0;i<MAX;i++)
{
printf("\n Enter a number to a[%d]",i+1);
scanf("%d",&a[i]);
}
//display the input elements
printf("\n The array elements are :");
for(i=0;i<=MAX;i++)
printf(" %d ",a[i]);
}
Explanation:
PSEUDOCODE INPUTARRAY(A[MAX])
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT “Enter a number to A[I]”
READ A[I]
[END OF LOOP]
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT A[I]
[END OF LOOP]
RETURN
ALGORITHM
ALGORITHM PRINTARRAY(A[MAX])
REPEAT FOR I<=1 TO MAX
PRINT “Enter a number”
INPUT A[I]
[END OF LOOP]
REPEAT FOR I<=1 TO MAX
PRINT A[I]
[END OF LOOP]
Answer:
Explanation:
The following is written in C++ and asks the user for inputs in both miles/gallon and dollars/gallon and then calculates the gas cost for the requested mileages using the input values
#include <iostream>
#include <iomanip>
using namespace std;
int main () {
// distance in miles
float distance1 = 20.0, distance2 = 75.0, distance3 = 500.0;
float miles_gallon, dollars_gallon;
cout << "Enter cars miles/gallon: "; cin >> miles_gallon;
cout << "Enter cars dollars/gallon: "; cin >> dollars_gallon;
cout << "the gas cost for " << distance1 << " miles is " << fixed << setprecision(2) << (float) dollars_gallon * distance1 / miles_gallon << "$\n";
cout << "the gas cost for " << distance2 << " miles is " << fixed << setprecision(2) << (float) dollars_gallon * distance2 / miles_gallon << "$\n";
cout << "the gas cost for " << distance3 << " miles is " << fixed << setprecision(2) << (float) dollars_gallon * distance3 / miles_gallon << "$\n";
return 0;
}
You would have 1234000 Megabytes
