CODEWITHSUNDEEP
c++loopsif-statementinputoutput
Raitiko
Raitiko

Reputation: 187

Check if the user input has entered a repeated number

So i am still a beginner at this and still practising. Basically i need to make a program that continues to asks the user to enter any number other than 5 until the user enters the number 5.

I have that done but i could't figure out how to check if the user entered a repeating number.For example: 1 2 3 3 - The program should end

#include <iostream>
#include <conio.h>
#include <iomanip>

using namespace std;

int main() {

cout << setw(15) << setfill('*') << "*" << endl;
cout << "Number 5" << endl;
cout << setw(15) << setfill('*') << "*" << endl;

int num;


cout << "Enter a number: ";
cin >> num;

if (num == 5) {
    cout << "\nWhy did you enter 5? :) " << endl;
    _getch();
    exit(0);
}
for (int i = 1; i < 10;i++) {

    cin >> num;

    if (num == 5) {
        cout << "\nWhy did you enter 5? :) " << endl;
        _getch();
        exit(0);
    }
}

cout << "Wow, you're more patient then I am, you win." << endl;
_getch();

}

Upvotes: 1

Views: 1099

Answers (2)

Blaze
Blaze

Reputation: 16876

You could add all inputted numbers to a vector, and whenever you get a new number, check if it's already in the vector. Include these headers:

#include <vector>
#include <algorithm> // for std::find

Make the vector like this

std::vector<int> pastEntries;

Do the check like this:

if (std::find(pastEntries.begin(), pastEntries.end(), num) != pastEntries.end()) {
    std::cout << "\nWhy did you enter " << num << "? :) " << endl;
    ...

And when the number was not found, add it to the vector like this (you can put this after the if block):

pastEntries.push_back(num);

Alternatively, you can use std::set:

std::set<int> pastEntries;

Insert into the set like this:

pastEntries.insert(num);

And find the number in the set like this:

if (pastEntries.find(num) != pastEntries.end()) {

Or insert the number while finding out whether it has already been inserted like this:

if (!pastEntries.insert(num).second) {

Upvotes: 1

Armali
Armali

Reputation: 19375

The previous answer does not meet the requirement in the linked article, which the querist himself seemed to not grasp:

★★ Modify the program so that it asks the user to enter any number other than the number equal to the number of times they've been asked to enter a number. (i.e on the first iteration "Please enter any number other than 0" and on the second iteration "Please enter any number other than 1"m etc. etc. The program must behave accordingly exiting when the user enters the number they were asked not to.)

This variant complies:

#include <iostream>
using namespace std;

int main()
{
    for (int i = 0; i < 10; i++)
    {
        cout <<"Please enter any number other than " <<i <<": ";
        int num;
        cin >>num;
        if (num == i)
            return cout <<"Hey! you weren't supposed to enter " <<i <<"!\n", 0;
    }
    cout <<"Wow, you're more patient then I am, you win.\n";
}

Upvotes: 1

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