Eric walks 7 km East in 2 hours and then 2.5 km West in 1 hour.

Write a program with the total change amount as an integer input, and output the change using the fewest coins, one coin type pe

riadik2000 [5.3K]

Answer:

In Python:

cents = int(input("Cents: "))

dollars = int(cents/100)

quarters = int((cents - 100*dollars)/25)

dimes = int((cents - 100*dollars- 25*quarters)/10)

nickels = int((cents - 100*dollars- 25*quarters-10*dimes)/5)

pennies = cents - 100*dollars- 25*quarters-10*dimes-5*nickels

if not(dollars == 0):

if dollars > 1:

print(str(dollars)+" dollars")

else:

print(str(dollars)+" dollar")

if not(quarters == 0):

if quarters > 1:

print(str(quarters)+" quarters")

else:

print(str(quarters)+" quarter")

if not(dimes == 0):

if dimes > 1:

print(str(dimes)+" dimes")

else:

print(str(dimes)+" dime")

if not(nickels == 0):

if nickels > 1:

print(str(nickels)+" nickels")

else:

print(str(nickels)+" nickel")

if not(pennies == 0):

if pennies > 1:

print(str(pennies)+" pennies")

else:

print(str(pennies)+" penny")

Explanation:

A prompt to input amount in cents

cents = int(input("Cents: "))

Convert cents to dollars

dollars = int(cents/100)

Convert the remaining cents to quarters

quarters = int((cents - 100*dollars)/25)

Convert the remaining cents to dimes

dimes = int((cents - 100*dollars- 25*quarters)/10)

Convert the remaining cents to nickels

nickels = int((cents - 100*dollars- 25*quarters-10*dimes)/5)

Convert the remaining cents to pennies

pennies = cents - 100*dollars- 25*quarters-10*dimes-5*nickels

This checks if dollars is not 0

if not(dollars == 0):

If greater than 1, it prints dollars (plural)

if dollars > 1:

print(str(dollars)+" dollars")

Otherwise, prints dollar (singular)

else:

print(str(dollars)+" dollar")

This checks if quarters is not 0

if not(quarters == 0):

If greater than 1, it prints quarters (plural)

if quarters > 1:

print(str(quarters)+" quarters")

Otherwise, prints quarter (singular)

else:

print(str(quarters)+" quarter")

This checks if dimes is not 0

if not(dimes == 0):

If greater than 1, it prints dimes (plural)

if dimes > 1:

print(str(dimes)+" dimes")

Otherwise, prints dime (singular)

else:

print(str(dimes)+" dime")

This checks if nickels is not 0

if not(nickels == 0):

If greater than 1, it prints nickels (plural)

if nickels > 1:

print(str(nickels)+" nickels")

Otherwise, prints nickel (singular)

else:

print(str(nickels)+" nickel")

This checks if pennies is not 0

if not(pennies == 0):

If greater than 1, it prints pennies (plural)

if pennies > 1:

print(str(pennies)+" pennies")

Otherwise, prints penny (singular)

else:

print(str(pennies)+" penny")

4 0

3 years ago

How do you turn on the Track Changes feature in a word-processing document?

aniked [119]

Click options in the tools menu and select track changers

5 0

3 years ago

I need help writing a recursion function to solve a boggle game for c++.

Mandarinka [93]

Answer:

#include <cstring>

#include <iostream>

using namespace std;

#define A 3

#define B 3

// LET US CREATE A DICTIONARY

string dict[] = { "KILLS", "GOT", "QUIZ", "GO" };

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

// Let us make a function to find whether a given word is present in dictionary.

bool isPresent(string& str)

{

// linear search of words

for (int i = 0; i < n; i++)

if (str.compare(dict[i]) == 0)

return true;

return false;

}

// A function for printing all words present on Boggle

void findWordin(char bogle[A][B], bool hasbeenvisited[A][B], int i,

int j, string& str)

{

hasbeenvisited[i][j] = true;

str = str + bogle[i][j];

// If str is in the dictionary, then you need to print it

if (isPresent(str))

cout << str << endl;

// Travering adjacent 8 cells of the boggle

for (int r = i - 1; r <= i + 1 && r < A; r++)

for (int c= j - 1; c <= j + 1 && c < B; c++)

if (r >= 0 && c >= 0 && !hasbeenvisited[r][c])

findWordin(bogle, hasbeenvisited, r, c, str);

// for erasing current characters on the string, and mark them visited

// of the current cells to false

str.erase(str.length() - 1);

hasbeenvisited[i][j] = false;

}

// Prints all words which are in dictionary.

void findWords(char boggle[A][B])

{

// for marking all the characters as not being visited

bool hasbeenvisited[A][B] = { { false } };

// Initializing the present string

string str = "";

// Reading all the characters for finding all the words that begins with the above character

for (int i = 0; i < A; i++)

for (int j = 0; j < B; j++)

findWordin(boggle, hasbeenvisited, i, j, str);

}

// the code for testing of the function

int main()

{

char bogle[A][B] = { { 'k', 'I', 'L' },

{ 'L', 'S', 'M' },

{ 'G', 'O', 'T' } };

cout << "Below list of words are present in the dictionary\n";

findWords(bogle);

return 0;

}

Explanation:

The program is properly commented, and that explains each step of it. However, I have kept dictionary as constant for similifying the code. And it can be set to variable easily, if required.

7 0

3 years ago

Isabella plans to work as a freelance DTP publisher. Her DTP setup is a non-Apple Macintosh platform. She has several options fo

nadya68 [22]

A lot of printers would work. I recommend HP printers because that is what I use and they work great for me.

One of my favorite ones is the HP Deskjet 2543. You can print in black and color, you can copy, and you can scan. Connect it to the computer with a USB port and wait for the computer to recognize it, then you are ready to go. It also supports wireless printing. You can print something from your phone without any cables. How cool is that?

7 0

4 years ago

Ok- well uhm- I need someone's opinion again :') Which one looks better 0.0

Kryger [21]

Answer:

They both look really good but i think the white looks better

Explanation:

8 0

3 years ago

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