Explanation of C++ Code - c++

I'm currently working on a program that outputs the number 1089 (i.e the Magic Number) of a three digit integer who's first digit is greater than its last. I have some code typed up, but am not receiving 1089, instead I'm receiving 891. Could anyone offer some explanation as to why.
//Uses a cout to inform user will be using the number 412 as an example.
//Uses a cout to explain to user the number needs to be reversed.
cout << "Alright, let's walk through an example, using the number 412." << endl;
int numExample = 412;
cout << "Please input 412" << endl;
cin >> numExample;
cout << "First, the number 412 is reversed." << endl;
//The following is done for reversing the number 412:
int reverseNum = 0, remainder = 0;
while (numExample)
{
remainder = numExample % 10;
reverseNum = (reverseNum * 10) + remainder;
numExample = numExample / 10;
}
cout << "The reverse of 412 is " << reverseNum << endl;
cout << "Next, we want to subtract the reverse of the original number from its reverse" << endl;
cout << "This gives us 198" << endl;
cout << "From there, we want to reverse 198." << endl;
//The same procedure is used to reverse 198
int numExample2 = 198;
cout << "Please enter 198" << endl;
cin >> numExample2;
int reverseNum2 = 0, remainder2 = 0;
while (numExample2)
{
remainder2 = numExample2 % 10;
reverseNum2 = (reverseNum2 * 10) + remainder2;
numExample2 = numExample2 / 10;
}
cout << "The reverse of 198 is " << reverseNum2 << endl;
int magicNumber = (reverseNum2 + numExample2);
cout << "Following that, we want to add 891 to 189 which gives us " << magicNumber << endl;

Try writing a function to handle this so your code is cleaner (It will also make it easier for people to help you!)
#include <iostream>
#include <cmath>
using namespace std;
int reverseDigit(int num); // For example purposes
int _tmain(int argc, _TCHAR* argv[])
{
int Number1,
Number2,
temp1,
temp2,
Result;
cout << "Enter the number 412: ";
cin >> Number1;
temp1 = reverseDigit(Number1);
temp1 = Number1 - temp1;
cout << "Enter the number 198: ";
cin >> Number2;
temp2 = reverseDigit(Number2);
Result = temp1 + temp2;
cout << "The magic number is: " << Result << endl;
return 0;
}
int reverseDigit(int num)
{
int reverseNum = 0;
bool isNegative = false;
if (num < 0)
{
num = -num;
isNegative = true;
}
while (num > 0)
{
reverseNum = reverseNum * 10 + num % 10;
num = num / 10;
}
if (isNegative)
{
reverseNum = -reverseNum;
}
return reverseNum;
}
I realize you're not working with negatives so you can remove that bit if you chose to use this, you can also expand on it... That being said this will make the actual " Reversing process " easier to work with and improve upon and read.

Related

How to find Maximum number and negative numbers from a .txt file and also how to output Total result to another .txt file

I want to find Maximum numbers from my "numbers.txt" file and amount of negative numbers. And i want to output the Total result to another .txt file and console and the rest to the console only.
Im very new and just cant figure out how to do it.
This is what i have now
a "numbers.txt" file with
-4
53
-5
-3
2
and
#include <iostream>
#include <fstream>
using namespace std;
int main() {
int n = 0;
int sum = 0, total = 0;
fstream file("numbers.txt");
while (file >> n)
{
sum += n;
total++;
}
int average = (float)sum / total;
int AmountOfNumbersAdded = total;
int Highest;
int Negative;
cout << "Total result: " << sum << endl;
cout << "Numbers added: " << AmountOfNumbersAdded << endl;
cout << "Average number: " << average << endl;
cout << "Maxiumum number: " << endl;
cout << "Negative numbers: " << endl;
return 0;
}
i tried to do
float Highest = INT_MIN;
if (Highest < num[i]) {
Highest = num[i];
but it just wouldn't work.
You don't need to store the numbers to find the maximum or the amount of negative numbers, but you need to track them inside the loop, like you're already doing with the sum and the total amount of numbers.
int Highest = INT_MIN;
int Negative = 0;
while (file >> n)
{
sum += n;
total += 1;
if (n < 0)
{
Negative += 1;
}
if (n > Highest)
{
Highest = n;
}
}
float average = (float)sum / total;
Here's what you're looking for.
#include <iostream>
#include <fstream>
int main()
{
int n = 0;
int sum = 0, total = 0;
int highest = 0;
int negatives = 0;
std::fstream file("numbers.txt");
while (file >> n)
{
if (n > highest) highest = n;
if (n < 0) ++negatives;
sum += n;
++total;
}
int average = (float)sum / total;
int AmountOfNumbersAdded = total;
std::cout << "Total result: " << sum << "\n";
std::cout << "Numbers added: " << AmountOfNumbersAdded << "\n";
std::cout << "Average number: " << average << "\n";
std::cout << "Maxiumum number: " << highest << "\n";
std::cout << "Negative numbers: " << negatives << "\n";
file.close();
return 0;
}
As you're new, few advices for you:
Never use using namespace std.
Prefer using "\n" instead of std::endl.
Don't forget to close any files/database after opening them like you did in your code.
Always try to avoid macros.

Not taking the input

I want to write a program that only takes odd numbers, and if you input 0 it will output the addition and average, without taking any even number values to the average and the addition. I'm stuck with not letting it take the even values..
Heres my code so far:
int num = 0;
int addition = 0;
int numberOfInputs = 0;
cout << "Enter your numbers (only odd numbers), the program will continue asking for numbers until you input 0.." << endl;
for (; ;) {
cin >> num;
numberOfInputs++;
addition = addition + num;
if (num % 2 != 0) {
//my issue is with this part
cout << "ignored" << endl;
}
if (num == 0) {
cout << "Addition: " << addition << endl;
cout << "Average: " << addition / numberOfInputs << endl;
}
}
Solution of your code:
Your code doesn't working because of following reasons:
Issue 1: You adding inputs number without checking whether it's even or not
Issue 2: If would like skip even then your condition should be as follow inside of the loop:
if (num%2==0) {
cout << "ignored:" <<num << endl;
continue;
}
Solving your issues, I have update your program as following :
#include <iostream>
#include <string>
using namespace std;
int main()
{
int num = 0;
int addition = 0;
int numberOfInputs = 0;
cout << "Enter your numbers (only odd numbers), the program will continue asking for numbers until you input 0.." << endl;
for (; ;) {
cin>> num;
if (num%2==0) {
cout << "ignored:" <<num << endl;
continue;
}
numberOfInputs++;
addition = addition + num;
if (num == 0) {
cout << "Addition: " << addition << endl;
cout << "Average: " << addition / numberOfInputs << endl;
break;
}
}
}
#include <iostream>
#include <stdio.h>
using namespace std;
int main() {
int number;
int sum=0;
int average=0;
int inputArray[20]; // will take only 20 inputs at a time
int i,index = 0;
int size;
do{
cout<<"Enter number\n";
cin>>number;
if(number==0){
for(i=0;i<index;i++){
sum = sum + inputArray[i];
}
cout << sum;
average = sum / index;
cout << average;
} else if(number % 2 != 0){
inputArray[index++] = number;
} else
cout<<"skip";
}
while(number!=0);
return 0;
}
You can run and check this code here https://www.codechef.com/ide
by providing custom input

maximum power a number can be raised to with out exceeding y recursion

i have this program assignment and one part of it is trying to find the max power a number will go to(x) without exceeding a number the user inputs it not to exceed(y). we are using it in a function. this is the whole program and what i have for max power it just keeps returning 0. it is the int maxpower(int x, int y) function i am trying to figure out
#include <iostream>
#include <cmath>
using namespace std;
// meunue where you can get your options from
void menue() {
cout << "choose the following options:" << endl;
cout << "1) Power of x raised by y." << endl;
cout << "2) Find the max power a number can be raised to." << endl;
cout << "3) Print out a number with its digits in reversed order." << endl;
cout << "4) Sum of integers from 1 to n." << endl;
cout << "5) Product of integers from 1 to n." << endl;
cout << "6) Quit" << endl;
}
//functions for finding the power usign recursion
int Power(int a, int b) {
int x = 1, i;
for (i = 1; i <= b; i++) {
if (b == 0) {
return Power(a, b--);
}
else {
x = x * a;
}
}
return x;
}
int maxpower(int n, int max_value) {
int temp = temp * n;
if (temp > max_value)
return 0;
else return maxpower(n, max_value + 1);
}
int reverse(int number) {
int lastDigit, numberOfDigits, sign = 1;//sets the sign equal to one
// if number is less than 0 returns 0
if (number < 0) {
return 0;
}
else
//if a number is under 10 than it can not be switched so you times the number by 10 and switch it.
if (number < 10)
return number * sign;
lastDigit = number % 10;
number = number / 10;
numberOfDigits = log10(number) + 1;
//recursive statement that calls the function
return (lastDigit * pow(10, numberOfDigits) + reverse(number)) * sign;
}
//finding the sum
int sum(int n) {
if (n != 0) {
return n + sum(n - 1);//recursive statement
}
else {
return n;
}
}
//finding the product
int product(int n) {
int temp;
if (n <= 1) {
return 1;
}
else {
temp = n * product(n - 1);
// recursive statement setting temp == to recursive statement
return temp;//returning temp
}
}
int main() {
int a;
int x;
int y;
int length = 0;
int temp;
int results;
// calls menue and get prints all the options
do {
menue();
//inserts the choice
cin >> a;
cout << "you choose:" << a << endl;//prints the choice out.
//switch statement that will take account for the number you choose and prints the results
switch (a) {
case 1:
cout << "enter the number to raise" << endl;
cin >> x;
cout << " enter the power to raise to: " << endl;
cin >> y;
Power(x, y);
cout << "the result is:" << Power(x, y) << endl;
break;
case 2:
cout << "Enter the number to raise:" << endl;
cin >> x;
cout << "Enter the number not to exceed:" << endl;
cin >> y;
maxpower(x, y);
cout << "the result is:" << maxpower(x, y) << endl;
break;
case 3:
cout << " enter numbers to be reversed by: " << endl;
cin >> x;
temp = x;
while (temp != 0) {
length++;
temp = temp / 10;
}
reverse(x);
cout << "the result is:" << reverse(x) << endl;
break;
case 4:
cout << "enter the number to sum to: " << endl;
cin >> x;
sum(x);
cout << "the result is:" << sum(x) << endl;
break;
case 5:
cout << "enter the number to multiply to:" << endl;
cin >> y;
product(y);
cout << "the result is:" << product(y) << endl;
break;
case 6:
cout << "good bye!!" << endl;
break;
}
} while (a != 6);
return 0;
}
I don't think it's necessary to use recursion for this problem. Moreover, recursion is creating a lot of overhead while solving it with a loop works just fine. Do you have to use recursion? If so, then disregard this answer :p. But you'll find below a solution that will work.
Note the #include <math.h> bit - you need that to use pow(base, exponent).
Also, while(true) is definitely not the best practice, but as long as you have sufficient checks to get out of the loop properly then you're ok. Hence the max_iteration and the actual return statement that you're looking for.
Best of luck!
#include <iostream>
#include <math.h>
int maxpower(int n, int max_value) {
if ( n > max_value ) return 0;
int previous, current = 1;
int max_iteration = 0;
while (true) {
if (max_iteration >= 1000) return -1;
if (pow(n, current) > max_value) {
return previous;
}
previous = current;
current++;
max_iteration++;
}
}
int main() {
int x;
int y;
int result;
std::cout << "Enter the base: ";
std::cin >> x;
std::cout << "Enter the max number x^pow should not exceed: ";
std::cin >> y;
result = maxpower(x, y);
if (result == -1) {
std::cout << "Max iteration reached." << std::endl;
}
else {
std::cout << result << " is the maximum power such that " << x << "^" << result << " does not exceed " << y << std::endl;
}
return 0;
}
As an example of output:
If x = 2 and y = 32, the program will return 5 as the max power (i.e. 2^5 = 32 and is not greater than, but 2^6 > 32).
EDIT:
I realized after I posted that all of your functions are recursive, so perhaps that's a requirement for your assignment. Anyway, below is a recursive solution:
int maxpower_rec_helper(int n, int power, int max_value) {
if (pow(n, power) > max_value) return power - 1;
return maxpower_rec_helper(n, power + 1, max_value);
}
int maxpower_rec(int n, int max_value) {
if ( n > max_value ) return 0;
return maxpower_rec_helper(n, 1, max_value);
}
You'll need a helper function to give the initial power 1, and so as not to disturb your max_value.
return power - 1; is essentially the same thing as return previous; in the iterative example above.

limiting responses to ten Pentagonal numbers per line

I'm writing a code that requires pentagonal numbers to be limited to ten per line, however I can't get it to work. My code is:
#include <iostream>
using namespace std;
int getPentagonalNumber(int n)
{
/*equation to calculate pentagonal numbers*/
return (n * (3 * n - 1) / 2);
}
int main()
{
/*Ask the user to put in the number of results*/
int userInput;
cout << "How many pentagonal numbers would you like to be displayed: ";
cin >> userInput;
cout << endl;
cout << "results are: " << endl;
/*Loop to generate the numbers for the equation*/
for (int n = 1; n <= userInput; n++)
{
cout << getPentagonalNumber(n) << " ";
}
return 0;
}
Does this do what you want:
#include <iostream>
using namespace std;
int getPentagonalNumber(int n)
{
/*equation to calculate pentagonal numbers*/
return (n * (3 * n - 1) / 2);
}
int main()
{
/*Ask the user to put in the number of results*/
int userInput;
cout << "How many pentagonal numbers would you like to be displayed: ";
cin >> userInput;
cout << endl;
cout << "results are: " << endl;
/*Loop to generate the numbers for the equation*/
for (int n = 1; n <= userInput; n++)
{
cout << getPentagonalNumber(n) << " ";
if (n % 10 == 0)
cout << endl;
}
return 0;
}
?
I added a new line output every time n is divisible by 10.

Separating Digits in C++

I am working on an exercise from my C++ book and I'm not sure how to fix it. I am supposed to get an int from the user and display the individual digits in the order they were entered. For instance 12345 would be displayed 1 2 3 4 5. 7365 would be displayed 7 3 6 5. I have most of the code written but there is a logical error and I can't figure it out. Here is my code:
int main()
{
int number = 0;
int digit = 0;
int temp = 0;
int counter = 0;
int sum = 0;
int divisor = 0;
cout << "Please enter a nonzero number.";
cin >> number;
cout << "\nThe number you entered was " << number;
// Determine the number of digits
temp = number;
while (temp != 0)
{
temp = temp / 10;
counter++;
}
cout << "\nThere are " << counter << " digits in your number.";
// Separate the digits
temp = number;
cout << "\nSeparating the digits\n";
do
{
divisor = (pow(10.0, --counter));
digit = temp / divisor;
temp = temp % divisor;
cout << digit << " ";
sum = sum + digit;
}
while (counter != 0);
cout << "\nThe sum of the number is " << sum;
return 0;
}
When I enter 5555 the output is 5560. When I enter 1234 the output is 1236. Can anyone help me find my error?
Here's one version:
// If the number is only one digit, print it.
// Otherwise, print all digits except the last, then print the last.
void digits(int x)
{
if (x < 10){
cout << x;
}
else{
digits(x / 10);
cout << " " << x % 10;
}
}
Thank you all for your help :-) Turns out my code works fine in another compiler so I guess it's just a netbeans glitch.