is there any way to get rid of arrays in this program? Im not allowed to do it with std::array or std::vector.
#include <iostream>
using namespace std;
int main()
{
int upper,i,j=0,k=0;
int arr1[1000],arr2[1000];
cout<<"Enter the upper bound :";
cin>>upper;
for(i=0 ; i<upper ; i++)
{
if(i%2 == 0)
{
arr1[j] = i;
j++;
}
else
{
arr2[k] = i;
k++;
}
}
cout<<"List of even numbers :";
for(i = 0; i<j ; i++)
{
cout<<arr1[i]<<" ";
}
cout<<"\n";
cout<<"List of odd numbers :";
for(i = 0; i<k ; i++)
{
cout<<arr2[i]<<" ";
}
return 0;
}
Instead of setting elements of two arrays just output at first even numbers in the given range and then odd numbers.
For example
#include <iostream>
int main()
{
std::cout << "Enter the upper bound: " ;
unsigned int n = 0;
std::cin >> n;
std::cout << "List of even numbers :";
for ( unsigned int i = 0; i < n; i += 2 )
{
std::cout << i << ' ';
}
std::cout << '\n';
std::cout << "List of odd numbers :";
for ( unsigned int i = 1; i < n; i += 2 )
{
std::cout << i << ' ';
}
std::cout << '\n';
return 0;
}
The program output might look like
Enter the upper bound: 10
List of even numbers :0 2 4 6 8
List of odd numbers :1 3 5 7 9
Quick and dirty solution.
#include <iostream>
int main()
{
int upper;
std::cout << "Enter the upper bound :";
std::cin >> upper;
std::cout<<"List of even numbers :";
for (int i=0; i<upper;i+=2)
std::cout <<i<<" ";
std::cout <<"\n";
std::cout<<"List of odd numbers :";
for (int i=1; i<upper;i+=2)
std::cout <<i<<" ";
return 0;
}
I'll not analyze the rest of your code but focus on the question:
is there any way to get rid of arrays in this program?
Since you can't use std::vector<int> you could allocate the memory dynamically yourself.
#include <cstddef>
#include <iostream>
#include <memory>
int main()
{
size_t upper;
std::cout << "Enter the upper bound :";
if(not (std::cin >> upper)) return 1; // input failed, exit
// create unique_ptr<int[]> poiting to an array of "upper" number of elements:
auto arr1 = std::make_unique<int[]>(upper);
auto arr2 = std::make_unique<int[]>(upper);
// ...
Related
I have to ask the user to put in an array size and then to ask the user to fill it out. When the user puts in a duplicate, the program should say "invalid" and the user is asked to replace the number. I am supposed to use traversing array search.
Like this example here:
Enter list size: 4
Enter value for index 0: 1
Enter value for index 1: 1
Invalid. Enter a new number: 2
Enter value for index 2: 5
Enter value for index 3: 6
This is my code so far:
#include <iostream>
using namespace std;
int main() {
int size;
cout << "Enter list size: ";
cin >> size;
int array1[size];
for (int i = 0; i < size; i++) {
cout << "Enter value for index " << i << ": ";
cin >> array1[i];
for (int j = i + 1; j < size; j++) {
if (array1[i] == array1[j]) {
cout << "Invalid! Enter a new value for index " << i << ": ";
cin >> array1[i];
}
}
}
return 0;
}
It does what was specified but the exercise probably was to write std::ranges::find.
#include <iostream>
#include <vector>
#include <cstddef>
#include <algorithm>
int main() {
size_t size;
std::cout << "Enter list size: ";
std::cin >> size;
std::vector<int> arr;
arr.reserve(size);
while(arr.size() < size) {
int t;
std::cout << "Enter value for index " << arr.size() + 1 << ": ";
std::cin >> t;
if (std::ranges::find(arr, t) == arr.end()) {
arr.push_back(t);
} else {
std::cout << "Invalid! ";
}
}
}
Try this approach, Every time user enter value helper function will check duplicate from already filled array
#include<iostream>
// Helper Function that will check duplicate from array
bool IsDuplicate (int arr[] ,const int idxSoFar, int element){
for(int i =0 ; i < idxSoFar ; i += 1 )
if( arr[i] == element){
std::cout << "Invalid! Enter a new value for index "<< idxSoFar + 1 << " : ";
return arr[i] == element;
}
return false;
}
int main () {
int size;
std::cout << "Enter list size: ";
std::cin >> size;
int array1[size];
for (int i = 0; i < size; i++) {
std::cout << "Enter value for index " << i << ": ";
do
std::cin >> array1[i];
while(IsDuplicate(array1 , i , array1[i]));
}
return 0;
}
i tried to separate even and odd numbers using vectors from a array ==>
so i made a function that returns true is number is even and false for if number is odd
then i used an if else statement where if the function returns true then it pushbacks the value in a vector and if the function returns false then it pushbacks the value in another vector , finally i printed all the elements in the vector but the output does not show any element except it shows one in the odd vector.
#include <iostream>
#include <vector>
using namespace std;
bool sort(int arr[] , int i){
if(arr[i] %2 == 0){
return true;
}
return false;
}
int main(){
int n;
cin >> n;
int *arr = new int[n];
for(int i=1 ; i<=n ; i++){
arr[i-1] = i;
}
vector <int> even , odd;
int i=0 ;
if(sort(arr , i)){
even.push_back(arr[i]);
sort(arr , i+1);
}else{
odd.push_back(arr[i]);
sort(arr,i+1);
}
cout << "the even numbers are : " << endl;
for(auto element:even){
cout << element << " ";
}
cout << endl;
cout << "the odd numbers are : " << endl;
for(auto element:odd){
cout << element << " ";
}
}
As #TonyDelroy said, you have to make for loop around call to sort(arr, i). Also first loop should go up to i <= n instead of i < n.
Your fixed working code below (see also std::partition_copy variant afterwards):
Try it online!
#include <iostream>
#include <vector>
using namespace std;
bool sort(int arr[] , int i){
if(arr[i] %2 == 0){
return true;
}
return false;
}
int main(){
int n;
cin >> n;
int *arr = new int[n];
for(int i=1 ; i<=n ; i++){
arr[i-1] = i;
}
vector <int> even , odd;
for (int i = 0; i < n; ++i)
if (sort(arr, i))
even.push_back(arr[i]);
else
odd.push_back(arr[i]);
cout << "the even numbers are : " << endl;
for(auto element:even){
cout << element << " ";
}
cout << endl;
cout << "the odd numbers are : " << endl;
for(auto element:odd){
cout << element << " ";
}
}
Input:
10
Output:
the even numbers are :
2 4 6 8 10
the odd numbers are :
1 3 5 7 9
As #chris said you can also use std::partition_copy to implement your algorithm:
Try it online!
#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
int main() {
int n = 0;
std::cin >> n;
std::vector<int> arr(n), odd, even;
for (int i = 1; i <= n; ++i)
arr[i - 1] = i;
std::partition_copy(arr.cbegin(), arr.cend(),
std::back_insert_iterator(odd), std::back_insert_iterator(even),
[](auto const & x){ return (x & 1) == 1; });
std::cout << "the even numbers are : " << std::endl;
for (auto element: even)
std::cout << element << " ";
std::cout << std::endl << "the odd numbers are : " << std::endl;
for (auto element: odd)
std::cout << element << " ";
}
Input:
10
Output:
the even numbers are :
2 4 6 8 10
the odd numbers are :
1 3 5 7 9
You only push one element - the first.
Your partitioning code is equivalent to
if(sort(arr , 0)){
even.push_back(arr[0]);
sort(arr , 1);
}else{
odd.push_back(arr[0]);
sort(arr,1);
}
You need to loop over all the input numbers.
You can also simplify matters with a more generally useful evenness function that doesn't depend on an array:
bool is_even(int x) { return x % 2 == 0; }
and then there is no need to store all the inputs before processing them:
int main(){
vector <int> even , odd;
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
int x;
cin >> x;
if (is_even(x)) {
even.push_back(x);
}
else {
odd.push_back(x);
}
}
cout << "the even numbers are : " << endl;
for (auto element:even){
cout << element << " ";
}
cout << endl;
cout << "the odd numbers are : " << endl;
for (auto element:odd){
cout << element << " ";
}
}
Hye, Im a beginner trying to learn C++ language. This is my code that I tried to find reverse input numbers using array. Can help me point my mistakes since I always got infinite loop.
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
const int ARRAY_SIZE=50;
int size[ARRAY_SIZE];
unsigned short int i;
cout << "You may enter up to 50 integers:\n";
cout << "\nHow many would you like to enter? ";
cin >> size[ARRAY_SIZE];
cout << "Enter your number: \n";
for (int i = 0; i < ARRAY_SIZE; i++)
{
cin >> size[i];
}
cout << "\nYour numbers reversed are:\n";
for (i = size[ARRAY_SIZE] - 1; i >= 0; i++)
cout << " size[i]" << " ";
}
Your infinite loop is because i is unsigned, so i >= 0 is always true.
Here's a C++-ified version:
#include <iostream>
#include <vector>
int main() {
std::cout << "You may enter up to 50 integers:\n";
std::cout << "\nHow many would you like to enter? ";
int count;
std::cin >> count;
// Use a std::vector which can be extended easily
std::vector<int> numbers;
for (int i = 0; i < count; ++i) {
std::cout << "Enter your number: \n";
int v;
std::cin >> v;
// Add this number to the list
numbers.push_back(v);
}
std::cout << "\nYour numbers reversed are:\n";
// Use a reverse iterator to iterate through the list backwards
for (auto i = numbers.rbegin(); i != numbers.rend(); ++i) {
// An iterator needs to be de-referenced with * to yield the value
std::cout << *i << " ";
}
std::cout << std::endl;
return 0;
}
There's many problems in your original code, but the clincher is this:
for (i = size[ARRAY_SIZE] - 1; i >= 0; i++)
cout << " size[i]" << " ";
}
Since you keep adding to i through each cycle you'll never go below zero, especially not for an unsigned short int. This should be:
for (int i = count - 1; i > 0; --i) {
std::cout << numbers[i];
}
Presuming you have a thing called numbers instead of the bizarrely named size and the array size is count, not i, as i is generally reserved for iterators and loop indexes.
I am trying to check whether there is any duplicate integer in the user input array. The problem is that the validation of the duplicate does not work properly and I have no idea why it is not showing the desired output. Following is the code:
#include <iostream>
using namespace std;
int main()
{
int length;
int arrValue;
cout << "Enter the length : ";
cin >> length;
int *arr = new int[length];
cout << "Enter " << length << " integers for array : ";
for(int i = 0; i < length; i++)
{
cin >> arr[i];
}
cout << "Array : ";
for(int i = 0; i < length; i++)
{
arrValue = arr[i];
for(int k = i + 1; k < length; k++)
{
if(arr[i] == arr[k])
{
cout << "Duplicate found" << endl;
break;
}
else
{
cout << arrValue << " ";
}
}
}
delete[] arr;
}
Current result (assuming no duplicate in user input):
Enter the length: 5
Enter 5 integers for array : 5 4 3 2 1
Array : 5 5 5 5 4 4 4 3 3 2
Expected result (assuming no duplicate in user input):
Enter the length: 5
Enter 5 integers for array : 5 4 3 2 1
Array : 5 4 3 2 1
Current result (assuming duplicate in user input):
Enter the length: 5
Enter 5 integers for array : 5 4 4 2 1
Array : 5 5 5 5 Duplicate found 4 4 3
Expected result (assuming duplicate in user input):
Enter the length: 5
Enter 5 integers for array : 5 4 4 2 1
Array : Duplicate found
I believe my loops is the source to the problem. The current result output 10 times and I do not understand why there will be so many same numbers appearing.
Do note that I am trying to apply the validation using loop only and not from C++ standard library.
The issue in your code is that you are printing out each array element every time a particular element is not matching another element. It seems that you only want to print out whether any duplicate values are found. For this, you can use a bool flag to indicate whether any element is duplicated:
bool found_dup = false;
for(int i = 0; i < length; i++)
for(int k = i + 1; k < length; k++)
if(arr[i] == arr[k])
{
found_dup = true;
break;
}
// else: don't print anything yet
and then at the end print out the array:
if (found_dup)
std::cout << "Duplicate found";
else
for(int i = 0; i < length; i++)
std::cout << arr[i] << " ";
You may achieve the program in a more enhanced way (where you don't need to define the length manually - notice the explanation given as comments in code):
#include <iostream> // for input/output operation
#include <vector> // for dynamic array management
#include <sstream> // to split the user inputs and assign them to the vector
#include <algorithm> // to sort the vector
#include <string> // to work with getline()
// using this statement isn't recommended, but for sake of simplicity
// and avoiding the use of std:: everywhere temporarily (for small programs)
using namespace std;
int main(void) {
vector<int> numbers;
vector<int> duplicates;
string input;
int temp;
// getting the user input as string
cout << "Enter the numbers: ";
getline(cin, input);
stringstream ss(input);
// splitting the user input string into integers and assigning
// them into the vector
while (ss >> temp)
numbers.push_back(temp);
// sorting the vector in increasing order
sort(numbers.begin(), numbers.end());
// getting the unique numbers (which are not repeated)
cout << "Unique numbers: ";
for (size_t i = 0, len = numbers.size(); i < len; i++) {
if (temp == numbers[i])
// if found a duplicate, then push into the 'duplicates' vector
duplicates.push_back(temp);
else
cout << numbers[i] << ' ';
temp = numbers[i];
}
// getting the duplicates
cout << "Total duplicates: ";
for (size_t i = 0, len = duplicates.size(); i < len; i++)
cout << duplicates[i] << ' ';
cout << endl;
return 0;
}
It'll output something like:
Enter the numbers: 1 4 8 9 3 2 3 3 2 1 4 8
Unique numbers: 1 2 3 4 8 9
Total duplicates: 1 2 3 3 4 8
Please change the if condition to something like this.
cout << "Enter the length : ";
cin >> length;
int *arr = new int[length];
cout << "Enter " << length << " integers for array : ";
for(int i = 0; i < length; i++)
{
cin >> arr[i];
}
cout << "Array : ";
for(int i = 0; i < length; i++)
{
arrValue = arr[i];
for(int k = i + 1; k < length; k++)
{
if(arrValue == arr[k]) //change here.
{
cout << "Duplicate found" << endl;
break;
}
else
{
cout << arrValue << " ";
}
}
}
delete[] arr;
}
I would also suggest to use a map data structure. Map allows you to count the frequency of numbers, and thus detect duplicates in linear time.
map<int, int> m; // specify the key-value data-type.
for(int i = 0;i<length;i++)
{
m[arr[i]]++;
}
map<int, int>::iterator it; // an iterator to iterate over the datastructure.
for(it = m.begin();it!=m.end();it++)
{
if(it->second>1) //it->second refers to the value(here, count).
{
cout<<it->first<<" "; //it->first refers to the key.
}
}
Your loops are actually iterating n-1 times for first element, n-2 times for second element etc., where n is the length of your array. This is why for 5 element array you have printed 5 4 times.
But generally, if the purpose is to detect duplicates in the array, this strategy is not the best one. Please note that having exemplary array 4 3 4, with current approach you will correctly detect for the first 4 that the third element is also 4 but once you will move to the third element, it will be marked as ok since it is not checked with the first one element.
You may consider another strategy: create another array of the n size. Then iterate through your original array and for each element check if that element is already in the new array. If you detect the presence, you may raise duplicate event. Otherwise you can add this element to the array.
It doesn't work because you're trying to print the same value everytime you find a different one. I got here a solution with one more array that will store the array. It would work too with just one array.
#include <iostream>
using namespace std;
int main()
{
int length;
int arrValue;
cout << "Enter the length : ";
cin >> length;
int *arr = new int[length];
int *noDuplicateArr = new int[length];
cout << "Enter " << length << " integers for array : ";
for(int i = 0; i < length; i++)
cin >> arr[i];
cout << "Array : ";
bool duplicateFound = false;
int noDuplicateArrLen = 0;
for(int i = 0; i < length && !duplicateFound; i++)
{
arrValue = arr[i];
int k;
for(k = i + 1; k < length; k++)
{
if(arrValue == arr[k])
{
duplicateFound = true;
break;
}
}
if (k == length)
noDuplicateArr[noDuplicateArrLen++] = arrValue;
}
if (duplicateFound)
{
cout << "Duplicate found";
}
else
{
for (int i = 0; i < noDuplicateArrLen; i++)
{
cout << noDuplicateArr[i] << " ";
}
}
delete[] arr;
delete[] noDuplicateArr;
}
Here is the version with just one array:
#include <iostream>
using namespace std;
int main()
{
int length;
int arrValue;
cout << "Enter the length : ";
cin >> length;
int *arr = new int[length];
cout << "Enter " << length << " integers for array : ";
for(int i = 0; i < length; i++)
cin >> arr[i];
cout << "Array : ";
bool duplicateFound = false;
int noDuplicateArrLen = 0;
for(int i = 0; i < length && !duplicateFound; i++)
{
arrValue = arr[i];
int k;
for(k = i + 1; k < length; k++)
{
if(arrValue == arr[k])
{
duplicateFound = true;
break;
}
}
if (k == length)
arr[noDuplicateArrLen++] = arrValue;
}
if (duplicateFound)
{
cout << "Duplicate found";
}
else
{
for (int i = 0; i < noDuplicateArrLen; i++)
{
cout << arr[i] << " ";
}
}
delete[] arr;
}
Question :
Your program is to use the brute-force approach in order to find the Answer to Life, the Universe, and Everything. More precisely... rewrite small numbers from input to output. Stop processing input after reading in the number 42. All numbers at input are integers of one or two digits.
Example
Input:
1
2
88
42
99
Output:
1
2
88
So that is the question, however i am still a beginner and unable to have an input tab like that. In my program, how should i modify it such that it still accepts numbers after 42, however, it does not print them? currently I am only able to terminate the input at 42.
#include <iostream>
using namespace std;
int main()
{
int A[100], num, i=0,k,count;
for(count = 0; count != 1;){
cin >> k;
if (k!=42){
A[i] = k;
i++;
}
else
count =1;
}
cout << endl;
for (count = 0; count <i; count ++){
cout << A[count] << endl;
}
}
You don't have to use array at all. You can print the value just after reading it. Exit when you read 42. This may help you.
#include <iostream>
using namespace std;
int main() {
// your code goes here
int n ;
for(; ;) {
cin >> n ;
if(n == 42) {
return 0 ;
}
cout << n << endl ;
}
return 0;
}
Pretty sure the easiest way to do so is to simply ask the user how many numbers they need to enter.
#include <iostream>
using namespace std;
int main()
{
int A[100], k, count;
cout << "How many numbers do you want to enter ? ";
cin >> count; //this is to count how many numbers the user wants to enter
for(int i(0); i < count; ++i) //put all the numbers user enters in your array
{
cin >> k;
A[i] = k;
}
cout << endl;
for (int i(0); i < count; ++i)
{
if (A[i] == 42) //if the element at index i is == 42 then stop displaying the elements
break;
else
cout << A[i] << " "; //else display the element
}
cout << endl;
return 0;
}
Else you would need to put everything in a string and parse it and i'm not quite sure how that goes as I am a beginner as well.
EDIT:
Actually here you go, I think that is correct and does exactly what you want.
Do keep in mind that if user enters p.e "1 88 442" it will output "1 88 4" because it found "42" in "442". But it should be okay because you precised input numbers should only be two digits max.
#include <iostream>
using namespace std;
int main()
{
string k;
getline(cin, k);
cout << endl;
for (unsigned int i(0); i < k.length(); ++i)
{
if (!((k[i] == '4') && (k[i+1] == '2'))) //if NOT 4 followed by 2 then display
cout << k[i];
else
break; //else gtfo
}
cout << endl;
return 0;
}
Use a bool value to control the execution of your code.
#include <iostream>
#define N_INPUT 100
#define THE_ANSWER 42
using namespace std;
int main()
{
int array[N_INPUT], i, input, count=0;
bool universeAnswered = false;
for (i = 0; i < N_INPUT; i++) {
cin >> input;
if (!universeAnswered)
{
if (input == THE_ANSWER) {
universeAnswered = true;
} else {
array[count] = input;
count++;
}
}
}
for (i = 0; i < count; i++) {
cout << array[i] << endl;
}
}
(My code was not tested)
You just have to have some state to see if you have seen 42 already, and only output if you haven't
#include <iostream>
int main()
{
bool output = true;
for (int n; std::cin >> n;)
{
output &= (n != 42);
if (output)
{
std::cout << n << std::endl;
}
}
return 0;
}