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Evgen [1.6K]
4 years ago
5

Write a program that determines the value of the coins in a jar and prints the total in dollars and cents. Read integer values t

hat represent the number of quarters, dimes, nickels, and pennies. This program will require user input.
Computers and Technology
1 answer:
Shalnov [3]4 years ago
7 0

Answer:

The programming language is not stated. To answer this question, I'll make use of python programming language.

<em>The following units of conversions is used in this program</em>

<em>1 quarter = $0.25</em>

<em>1 dime = $0.10</em>

<em>1 nickel = $0.05</em>

<em>1 penny = $0.01</em>

<em>$1 = 100 cents</em>

This program makes use of few comment (See explanation for line by line explanation)

Having listed the above, the program is as follows:

#prompt user for inputs

quarters = int(input("Number of Quarters: "))

dimes = int(input("Number of Dimes: "))

nickels = int(input("Number of Nickels: "))

pennies = int(input("Number of Pennies: "))

#Convert coins to dollars

dollar = 0.25 * quarters + 0.10 * dimes + 0.05 * nickels + 0.01 * pennies

print("Dollar Equivalent")

print(str(round(dollar,2))+" dollars")

cent = dollar * 100

print("Cent Equivalent")

print(str(round(cent,2))+" cents")

Explanation:

This first line is a comment

#prompt user for inputs

This line prompts user for value of quarters in the jar

quarters = int(input("Number of Quarters: "))

This line prompts user for value of dimes in the jar

dimes = int(input("Number of Dimes: "))

This line prompts user for value of nickels in the jar

nickels = int(input("Number of Nickels: "))

This line prompts user for value of pennies in the jar

pennies = int(input("Number of Pennies: "))

Using the coin conversion listed in the answer section, the user inputs is converted to dollars in the next line

dollar = 0.25 * quarters + 0.10 * dimes + 0.05 * nickels + 0.01 * pennies

The string "Dollar Equivalent" is printed using the next line

print("Dollar Equivalent")

This line prints the dollar equivalent of the converted coins

print(str(round(dollar,2))+" dollars")

Using the coin conversion listed in the answer section, the user inputs is converted to cents in the next line

cent = dollar * 100

The string "Cent Equivalent" is printed using the next line

print("Cent Equivalent")

This line prints the cent equivalent of the converted coins

print(str(round(cent,2))+" cents")

Please note that the dollar and cent equivalents are rounded to two decimal places. Even though, it's not a requirement of the program, it's a good programming practice

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#include<algorithm.h>

#include<climits.h>

#include<bits/stdc++.h>

#include<cstring.h>

using namespace std;

int partition(int arr[], int l, int r, int k);

int kthSmallest(int arr[], int l, int r, int k);

void quickSort(int arr[], int l, int h)

{

if (l < h)

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// Find median of arr[].

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// Recur for left and right of partition

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int findMedian(int arr[], int n)

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int kthSmallest(int arr[], int l, int r, int k)

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// If k is smaller than number of elements in array

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// Divide arr[] in groups of size 5, calculate median

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{

median[i] = findMedian(arr+l+i*5, n%5);

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int medOfMed = (i == 1)? median[i-1]:

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return INT_MAX;

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int temp = *a;

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int i;

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i = l;

for (int j = l; j <= r - 1; j++)

{

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{

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}

swap(&arr[i], &arr[r]);

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float a;

clock_t time_req;

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int n = sizeof(arr)/sizeof(arr[0]);

quickSort(arr, 0, n-1);

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return 0;

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..................................................................................................................................................................................................................................................................................................................................

OR

.......................

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

 

// Swap utility

void swap(long int* a, long int* b)

{

int tmp = *a;

*a = *b;

*b = tmp;

}

 

// Bubble sort

void bubbleSort(long int a[], long int n)

{

for (long int i = 0; i < n - 1; i++) {

for (long int j = 0; j < n - 1 - i; j++) {

if (a[j] > a[j + 1]) {

swap(&a[j], &a[j + 1]);

}

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}

 

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{

long int i, key, j;

for (i = 1; i < n; i++) {

key = arr[i];

j = i - 1;

 

// Move elements of arr[0..i-1], that are

// greater than key, to one position ahead

// of their current position

while (j >= 0 && arr[j] > key) {

arr[j + 1] = arr[j];

j = j - 1;

}

arr[j + 1] = key;

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void selectionSort(long int arr[], long int n)

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for (i = 0; i < n - 1; i++) {

 

// Find the minimum element in unsorted array

midx = i;

 

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if (arr[j] < arr[min_idx])

midx = j;

 

// for plotting graph with integer values

printf("%li, %li, %li, %li\n",

n,

(long int)tim1[it],

(long int)tim2[it],

(long int)tim3[it]);

 

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n += 10000;

}

 

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