I am trying to get the height of these slashes to be a certain length based on input. So far, I have:
#include <iostream>
using namespace std;
int main() {
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
}
I need it to then reverse and go back.
It prints:
//
////
//////
If the user entered 3.
It should print:
//
////
//////
////
//
Can anyone lead me in the right direction? I am new to cpp.
You can use a different kind of loop and add a bool variable to track when the program have reached "n". Then, after the program reaches "n", it sets the bool variable to true and starts to substract until i equals 0
Code below, read comments and ask if you have any further questions:
#include <iostream>
using namespace std;
int main()
{
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You have entered: " << n << "\n";
int i = 1;
bool reachedN = false; // tells if [i] has reached [n]
while (i != 0)
{
// Print required slashes
for (int j = 1; j <= i; j++)
{
cout << "//";
}
cout << '\n'; // new line
// Add until i == n, then substract
if (i == n)
{
reachedN = true;
}
if (reachedN)
{
--i;
}
else
{
++i;
}
}
}
If you enter 3, the output is the following:
This is one way to achieve that:
#include <iostream>
using namespace std;
int main() {
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
for (int i = n - 1; i > 0; i--) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
}
This is a shorter solution with only two for-loops.
#include <iostream>
using namespace std;
int main()
{
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
n = n * 2 - 1;
int r = 0;
for (int j = 0; j < n; j++)
{
if (j > n / 2) r--;
else r++;
for (int i = 0; i < r; i++)
{
cout << '/' << '/';
}
cout << "\n";
}
return 0;
}
Related
I'm a C++ newb. I need to insert numbers to an array and then display first the odd numbers and then the even numbers in a single array. I've managed to create two separate arrays with the odd and even numbers but now I don't know how to sort them and put them back in a single array. I need your help to understand how to do this with basic C++ knowledge, so no advanced functions. Here's my code:
#include <iostream>
using namespace std;
int main()
{
int N{ 0 }, vector[100], even[100], odd[100], unify[100], i{ 0 }, j{ 0 }, k{ 0 };
cout << "Add the dimension: " << endl;
cin >> N;
cout << "Add the elements: " << endl;
for (int i = 0; i < N; i++) {
cout << "v[" << i << "]=" << endl;
cin >> vector[i];
}
for (i = 0; i < N; i++) {
if (vector[i] % 2 == 0) {
even[j] = vector[i];
j++;
}
else if (vector[i] % 2 != 0) {
odd[k] = vector[i];
k++;
}
}
cout << "even elements are :" << endl;
for (i = 0; i < j; i++) {
cout << " " << even[i] << " ";
cout << endl;
}
cout << "Odd elements are :" << endl;
for (i = 0; i < k; i++) {
cout << " " << odd[i] << " ";
cout << endl;
}
return 0;
}
If you don't need to store the values then you can simply run through the elements and print the odd and the even values to different stringstreams, then print the streams at the end:
#include <sstream>
#include <stddef.h>
#include <iostream>
int main () {
std::stringstream oddStr;
std::stringstream evenStr;
static constexpr size_t vecSize{100};
int vec[vecSize] = {10, 5, 7, /*other elements...*/ };
for(size_t vecIndex = 0; vecIndex < vecSize; ++vecIndex) {
if(vec[vecIndex] % 2 == 0) {
evenStr << vec[vecIndex] << " ";
} else {
oddStr << vec[vecIndex] << " ";
}
}
std::cout << "Even elements are:" << evenStr.rdbuf() << std::endl;
std::cout << "Odd elements are:" << oddStr.rdbuf() << std::endl;
}
Storing and sorting the elements are always expensive.
Basically, it would be better to sort them first.
#include <iostream>
using namespace std;
int main()
{
int numbers[5];
int mergedArrays[5];
int evenNumbers[5];
int oddNumbers[5];
for(int i=0;i<5;i++){
cin>>numbers[i];
}
int temp=numbers[0];
//bubble sort
for(int i = 0; i<5; i++)
{
for(int j = i+1; j<5; j++)
{
if(numbers[j] < numbers[i])
{
temp = numbers[i];
numbers[i] = numbers[j];
numbers[j] = temp;
}
}
}
int nEvens=0;
int nOdds=0;
for(int i = 0; i<5; i++)
{
if(numbers[i]%2==0)
{
evenNumbers[nEvens]=numbers[i];
nEvens++;
}
else if(numbers[i]%2!=0)
{
oddNumbers[nOdds]=numbers[i];
nOdds++;
}
}
int lastIndex=0;
//copy evens
for(int i = 0; i<nEvens; i++)
{
mergedArrays[i]=evenNumbers[i];
lastIndex=i;
}
//copy odds
for(int i =lastIndex; i<nOdds; i++)
{
mergedArrays[i]=oddNumbers[i];
}
return 0;
}
If you have to just output the numbers in any order, or the order given in the input then just loop over the array twice and output first the even and then the odd numbers.
If you have to output the numbers in order than there is no way around sorting them. And then you can include the even/odd test in the comparison:
std::ranges::sort(vector, [](const int &lhs, const int &rhs) {
return ((lhs % 2) < (rhs % 2)) || (lhs < rhs); });
or using a projection:
std::ranges::sort(vector, {}, [](const int &x) {
return std::pair<bool, int>{x % 2 == 0, x}; });
If you can't use std::ranges::sort then implementing your own sort is left to the reader.
I managed to find the following solution. Thanks you all for your help.
#include <iostream>
using namespace std;
int main()
{
int N{0}, vector[100], even[100], odd[100], merge[100], i{0}, j{0}, k{0}, l{0};
cout << "Add the dimension: " << endl;
cin >> N;
cout << "Add the elements: " << endl;
for (int i = 0; i < N; i++)
{
cout << "v[" << i << "]=" << endl;
cin >> vector[i];
}
for (i = 0; i < N; i++)
{
if (vector[i] % 2 == 0)
{
even[j] = vector[i];
j++;
}
else if (vector[i] % 2 != 0)
{
odd[k] = vector[i];
k++;
}
}
cout << "even elements are :" << endl;
for (i = 0; i < j; i++)
{
cout << " " << even[i] << " ";
cout << endl;
}
cout << "Odd elements are :" << endl;
for (i = 0; i < k; i++)
{
cout << " " << odd[i] << " ";
cout << endl;
}
for (i = 0; i < k; i++)
{
merge[i] = odd[i];
}
for (int; i < j + k; i++)
{
merge[i] = even[i - k];
}
for (int i = 0; i < N; i++)
{
cout << merge[i] << endl;
}
return 0;
}
You can use Bubble Sort Algorithm to sort whole input. After sorting them using if and put odd or even numbers in start of result array and and others after them. like below:
//Bubble Sort
void bubbleSort(int arr[], int n)
{
int i, j;
for (i = 0; i < n - 1; i++)
// Last i elements are already
// in place
for (j = 0; j < n - i - 1; j++)
if (arr[j] > arr[j + 1])
swap(arr[j], arr[j + 1]);
}
// Insert In array
int result[100];
if(odd[0]<even[0])
{
for (int i = 0; i < k; i++)
{result[i] = odd[i];}
for (int i = 0; i < j; i++)
{result[i+k] = even[i];}
}else
{
for (int i = 0; i < j; i++)
{result[i] = even[i];}
for (int i = 0; i < k; i++)
{result[i+k] = odd[i];}
}
I have a problem with this piece of code, I'm trying to print the EVEN and ODD numbers, but there is a problem when it comes to show them, the vectors don't save the numbers as I'm expecting.
#include <iostream>
using namespace std;
int main() {
int n;
cin >> n;
int vect[n], even[n], odd[n]; // CREATING VECTORS LIMIT AFTER "n"
for(int i = 1; i <= n; ++i) { // ENTERING The ELEMENS IN VECTOR
cin >> vect[i];
}
for(int i = 1; i <= n; ++i) {
if(vect[i] % 2 != 0) {
odd[i] = vect[i]; // I think that here's the problem, the vectors don't save the right numbers.
} /// VERIFYING IF THE NUMBER IS ODD OR EVEN.
else if (vect[i] % 2 == 0) {
even[i] == vect[i];
}
}
for(int i = 1; i <= n; ++i) {
cout << even[i] << " " << endl; /// PRINTING THE ODD AND EVEN numbers.
cout << odd[i] << " " << endl;
}
return 0;x
}
I have fixed the problem, thanks all for help.
Now it works perfectly.
#include <iostream>
using namespace std;
int main() {
int n;
cin >> n;
int vect[n], even[n], odd[n], z = 0, x = 0; // CREATING VECTORS LIMIT AFTER "n"
for(int i = 1; i <= n; ++i) { // ENTERING The ELEMENS IN VECTOR
cin >> vect[i];
}
for(int i = 1; i <= n; ++i) {
if(vect[i] % 2 != 0) {
odd[1+z] = vect[i];
z++;
// I think that here's the problem, the vectors don't save the right numbers.
} /// VERIFYING IF THE NUMBER IS ODD OR EVEN.
else if (vect[i] % 2 == 0) {
even[1+x] = vect[i];
x++;
}
}
for(int i = 1; i <= x; i++) {
cout << even[i] << " ";
}
cout << endl;
for(int i = 1; i <= z; i++) {
cout << odd[i] << " ";
}
return 0;
}
Considering the hints of the comments, your program shall be changed into this:
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n, number;
cin >> n;
vector<int> vect, even, odd; // CREATING DYNAMIC VECTORS
for(int i = 0; i < n; ++i) { // ENTERING THE ELEMENTS IN VECTOR
cin >> number;
vect.push_back(number);
}
for(int i = 0; i < n; ++i) {
if(vect[i] % 2 != 0) { /// VERIFYING IF THE NUMBER IS ODD OR EVEN.
odd.push_back(vect[i]);
}
else {
even.push_back(vect[i]);
}
}
for (int i = 0; i < n; ++i)
cout << vect[i] << " ";
cout << endl;
/// PRINTING THE ODD AND EVEN NUMBERS.
for (auto& val : odd)
cout << val << " ";
cout << endl;
for (auto& val : even)
cout << val << " ";
cout << endl;
return 0;
}
It uses the vector container of STL for your arrays, start the indexing at 0 and prints out the resulting arrays separately, as the number of odd and of even entries might be different.
Hope it helps?
With standard, you might use std::partition (or stable version) to solve your problem:
void print_even_odd(std::vector<int> v)
{
auto limit = std::stable_partition(v.begin(), v.end(), [](int n){ return n % 2 == 0; });
std::cout << "Evens:";
// Pre-C++20 span:
// for (auto it = v.begin(); it != limit; ++it) { int n = *it;
for (int n : std::span(v.begin(), limit)) {
std::cout << " " << n;
}
std::cout << std::endl;
std::cout << "Odds:";
for (int n : std::span(limit, v.end())) {
std::cout << " " << n;
}
std::cout << std::endl;
}
Demo
I am currently writing a task where the user will input a number and it'll output a number of "*" depending on the number. Eg if the user inputted a 5, the answer would be:
*
**
***
****
*****
This is my current code:
#include <iostream>
using namespace std;
int number;
char star = '*';
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
for (int i = 0; i < number; i++)
{
for (int j = i; j < number; j++)
{
cout << star;
}
cout << " " << endl;
}
return 0;
}
If the number 5 was inputted, this would output:
*****
****
***
**
*
How would I go about reversing the order so that it is ascending order rather than descending.
You would need to use the for loop in descending instead of ascending order.
Example:
cout << "Input a number between 1 and 10" << endl;
cin >> number;
for (int i = number -1; i >= 0; i--) {
for (int j = i; j < number; j++) {
cout << star;
}
cout << "*" << endl;
}
return 0;
}
You can do that by just reversing the order of the change of i:
#include <iostream>
using namespace std;
int number;
char star = '*';
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
//for (int i = 0; i < number; i++)
for (int i = number - 1; i >= 0; i--)
{
for (int j = i; j < number; j++)
{
cout << star;
}
cout << "*" << endl;
}
return 0;
}
(Just how to reverse the order of the change of i is shown. The output of this program is not as expected.)
My preference is using i as the number of stars to print and avoiding using gloval variables and using namespace std;:
#include <iostream>
int main()
{
int number;
char star = '*';
std::cout << "Input a number between 1 and 10" << std::endl;
std::cin >> number;
for (int i = 1; i <= number; i++)
{
for (int j = 0; j < i; j++)
{
std::cout << star;
}
std::cout << std::endl;
}
return 0;
}
Just go from number all the way down instead of up:
for (int i = number; i > 0; i--)
{
for (int j = i; j > 0; j--)
{
cout << star;
}
std::cout << '\n';
}
Instead of j<number use j<i and instead of j=i use j=0:
#include <iostream>
using namespace std;
int number;
char star = '*';
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
for (int i = 0; i < number; i++)
{
for (int j = 0; j < i; j++)
{
cout << star;
}
cout << "*" << endl;
}
return 0;
}
there are multiple ways to achieve what you want. You can insert everything inside and array and print the array backwards, insert everything inside an array (starting from last position) and then print it in natural order, you can do it only by using indexes (like i and a lot of other persons did here). I also added a little input check, and put the code in functions for clearness:
#include <iostream>
using namespace std;
int number;
char star = '*';
void version1();
void version2();
int main()
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
while(number < 1 || number > 10)
{
cout << "I SAID BETWEEN 1 AND 10, IS EASY, RIGHT?" << endl;
cin >> number;
}
version1();
version2();
return 0;
}
void version1()
{
cout <<"Version 1: "<<endl;
for (int i = 0; i < number; i++)
{
for (int j = i; j < number; j++)
{
cout << star;
}
cout << endl;
}
}
void version2()
{
cout << "Version 2: (only by using indexes)" << endl;
int cap_index = 0;
for (int i = 0; i < number; i++)
{
cap_index = i + 1;
for (int j = 0; j < cap_index; j++)
{
cout << star;
}
cout << endl;
}
}
btw this is a more serious (and simple) way to check input:
do
{
cout << "Input a number between 1 and 10" << endl;
cin >> number;
}while(number < 1 || number > 10);
In my case i put it this way:
starting from 0 print '*' up to i + 1 times, this way you have an ascending order print:
first time (from 0 to 1) print '*' -> print once
second time (from 0 to 2) print '*' -> prints twice
and so on up to the inserted number
Obviously this isn't not the best way to do this nor the most elegant, like in nearly every situation there are a lot of possibilities.
ps. i'm not sure, but you have an error in your code, as it prints a '*' more, each time, i don't know how is your output like that
By using a std::string, you can avoid the need of an explicit inner loop.
#include <iostream>
#include <string>
int main()
{
int number;
char star = '*';
std::cout << "Input a number between 1 and 10" << "\n";
std::cin >> number;
for (int i = 0; i < number; i++)
{
std::cout << std::string (i+1, star) << "\n";
}
return 0;
}
Changing the second 'for' loop condition, in your function --
From,
for (int j = i; j < number; j++)
Change it to ,
for (int j = 0; j <= i; j++)
I want to find the factors of a product by using arrays to check whether they equal it or not
the values do not print
here is the code
#include <iostream>
using namespace std;
int main() {
int arr[5] = { 1,3,5,7,2 };
int arr1[5] = { 0,6,5,4,9 };
int X;
cout << "Please enter X:"; cin >> X;
for (int i = 0, j = 0; i < 5 && j < 5; ++i, ++j) {
if (arr[i]*arr[j]==X) {
cout << arr[i] << " ";
cout << arr1[j] << " ";
}
}
}
Use this nested loops
for (int i = 0; i < 5 ;++i) {
for(int j=0 ;j<5;++j){
if (arr[i]*arr1[j]==X) {
cout << arr[i] << " ";
cout << arr1[j] << " ";
}
}
}
I need help with getting this users input of an integer and retrieving the even numbers and displaying them with spaces.I already have the input processed into an array and have it reversed (thanks to stackoverflow) now need to extract the even numbers from the array and display them.
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
int evenNumbers(char even[], int num[], int indexing[]);
int main()
{
char integers[5];
int numbers[5];
int even[5] = {0,2,4,6,8};
int evens;
cout << "Please enter an integer and press <ENTER>: " << endl;
for (int j = 0; j < 5; j++)
cin >> integers[j];
for (int j = 0; j < 5; j++)
{
numbers[j]= integers[j] - '0';
}
cout << endl;
for (int j = 5; j > 0; j--)
{
cout << integers[j - 1] << " ";
}
cout << endl;
//having problems finding the even numbers and displaying the even numbers
//from the users input of integers, i have only learned how to display the
//subscript by a linear search
evens = evenNumbers(integers, numbers, even);
if (evens == -1)
cout << "There are no even numbers" << endl;
else
{
cout << "The even numbers are: " << (evens + 1) << endl;
}
system("pause");
return 0;
}
int evenNumbers(char even[], int num[], int indexing[])
{
int index = 0;
int position = -1;
bool found = false;
for (int j = 0; j < 5; j++)
{
num[j]= even[j] - '0';
}
while (index < 5)
{
if (num[index] == indexing[index])
{
found = true;
position = index;
}
index++;
}
return position;
}
If you want to display the even numbers from the array integers you can use a simple for loop and if statement:
for(int i = 4; i >= 0; i--)
{
if(integers[i] % 2 == 0)
cout << integers[i] << " ";
}
Your approach is all wrong, you can't detect even numbers by searching a list, you need a mathematical test for evenness. Write a function called is_even which tests one number and returns true if it is even and false if it is not. Then you can use that function, very simply, like this
for (int j = 0; j < 5; j++)
{
if (is_even(integers[j]))
cout << integers[j] << " ";
}
cout << endl;
Now you just need to write the is_even function.
void evennumbers(int num[])
{
for(int i=0;i<5;i++)
{
if(num[i]%2==0)
cout<<num[i]<<" ";
}
}
And avoid taking input to char what if user enters a number with more than one digit
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
void validNum(char valid[]);
void reverseNum(char rev[], int num2[]);
void evenNumbers(char even[], int num3[]);
void oddNumbers(char odd[], int num4[]);
int main()
{
char integer[5];
int number[5];
cout << "Your number is: ";
validNum(integer);
cout << "Your number in reverse is: ";
reverseNum(integer, number);
cout << "Even numbers: ";
evenNumbers(integer, number);
cout << endl;
cout << "Odd numbers: ";
oddNumbers(integer, number);
cout << endl;
system("pause");
return 0;
}
void validNum(char valid[])
{
char ch;
cout << "Please enter an integer and press <ENTER>: " << endl;
ch = cin.get;
while (ch < 0 || ch >= 'A' && ch <= 'Z' || ch >= 'a' && ch <= 'z')
{
cout << "ERROR: Please enter a positive integer and press <ENTER>: ";
for (int i = 0; i < 5; i++)
cin >> valid[i];
}
for (int j = 0; j < 5; j++)
{
cout << valid[j] - '0';
}
}
void reverseNum(char rev[], int num2[])
{
for (int j = 0; j < 5; j++)
{
num2[j]= rev[j] - '0';
}
cout << endl;
for (int j = 5; j > 0; j--)
{
cout << rev[j - 1]<< " ";
}
cout << endl;
}
void evenNumbers(char even[], int num3[])
{
for (int i = 0; i < 5; i++)
{
if (even[i] % 2 == 0)
{
cout << num3[i] << " ";
}
}
}
void oddNumbers(char odd[], int num4[])
{
for (int i = 0; i < 5; i++)
{
if (odd[i] % 2 == 1)
{
cout << num4[i] << " ";
}
}
}