Question

Given a total due and an array representing the amount of change in your pocket, determine whether or not you are able to pay for the item. Change will always be represented in the following order: quarters, dimes, nickels, pennies.

To illustrate: changeEnough([25, 20, 5, 0], 4.25) should yield true, since having 25 quarters, 20 dimes, 5 nickels and 0 pennies gives you 6.25 + 2 + .25 + 0 = 8.50.

Examples
changeEnough([2, 100, 0, 0], 14.11) ➞ false

changeEnough([0, 0, 20, 5], 0.75) ➞ true

changeEnough([30, 40, 20, 5], 12.55) ➞ true

changeEnough([10, 0, 0, 50], 3.85) ➞ false

changeEnough([1, 0, 5, 219], 19.99) ➞ false
Notes
quarter: 25 cents / $0.25
dime: 10 cents / $0.10
nickel: 5 cents / $0.05
penny: 1 cent / $0.01

C++ language ?

Answer 1

The solution of the above problem is below:

The screenshot of code is:

The output screen is:

code--

  

#include <bits/stdc++.h>
using namespace std;

void changeEnough(int a[],double due)
{
double total=0.0;
for(int i=0;i<4;i++)
{
if(i==0)
{
total+=a[i]*(0.25);
}
if(i==1)
{
total+=a[i]*(0.10);
}
if(i==2)
{
total+=a[i]*(0.05);
}
if(i==3)
{
total+=a[i]*(0.01);
}
  
}
if(total>=due)
{
cout<<"true";
}
else
cout<<"false";
}

int main()
{
int arr[4],i;
for(i=0;i<4;i++)
{
cin>>arr[i];
}
double due;
cin>>due;
changeEnough(arr,due);
}
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