c++ program to check entered number by user

Question

I am new to programming, started with C++ quite recently and I use CLion IDE.

I need to solve something, but I am not sure how exactly and I need your help with a basic C++ console program.

if the user enters a ten-digit number and the fifth number is one, the output should be this word - "zadochno".

if the user enters a ten-digit number and the fifth number is two, the output should be this word - "redovno".

The user is expected to enter 2101 1 62235 or similar. In any case, the fifth element should be either 1 or 2.

Examples:

Option 1: input> 2101 1 62235 -> output string "zadochno"

Option 2: input> 2101 2 62235 -> output string "redovno"

I am able to only partially create the program:

#include<iostream>

int number;
cout << "Please, enter number: ";
cin > number;

//I believe there should be an if statement or for loop here:
if(){

}

Can you please help me?

Answer 1

Assuming the user does enter a 10 digit number (ie you don't need to check if they enter eg "foo" or "bar3000" ), you can do the following:

Read the input as a std::string , not as int . User input is a string of characters always. Only if you need it you can get it converted to an integer. You do not need it as int . The n-th character of a std::string called user_input is user_input[n] . You just need to check whether the character in the middle is either '1' or '2' .

If you do need to check that the user did enter digits, you could use std::isdigit .

Answer 2

You can take the input from the user as std::string and then check if the element at index 4 is 1 or 2 as shown below:

#include <iostream>
#include <string>
int main()
{
    std::string input;
    
    //take input from user 
    std::getline(std::cin, input);
    
    //check if the 5th letter(at index 4 since indexing starts with 0) is '1' or '2'
    if(input.at(4) == '1')
    {
        std::cout<< "zadochno"<<std::endl;
    }
    else if(input.at(4) == '2')
    {
        std::cout << "redovno"<<std::endl;
    }
    
    //this below shown for loop is optional. If you're sure that the user input contains only digits then you can skip/remove this for loop.
    for(int i = 0; i < input.size(); ++i)
    {
        //check if the all the characters are digits of a number 
        if(std::isdigit(input[i]))
        {
            ;//std::cout<<"yes digit";
        }
        else
        {
            std::cout<<"Please enter a valid number"<<std::endl;
        }
    }
    
    return 0;
}

The output of the above program can be seen here .

Answer 3

The solution of this problem is right? Why does donwvote take a place?

I can offer the solution of your problem, using the algorithm Depth first search .

#include <iostream>
#include <string>
#include <vector>
using namespace std;
const int maximumSize=10;
vector<int> visited(maximumSize, 0);
void showContentVector(vector<int>& input)
{
    for(int i=0; i<input.size(); ++i)
    {
        cout<<input[i]<<", ";
    }
    return;
}
void dfs(int current, int previous, int index, string input)
{
    if(visited[current]==1)
    {
        return;
    }
    if(input.size()!=10)
    {
        return;
    }
    if(current==(index-1))
    {
        if(input[current]=='1')
        {
            cout<<"zadochno"<<endl;
        }
        else
        {
            cout<<"redovno"<<endl;
        }
    }
    visited[current]=1;
    for(int next=(current+1); next<input.size(); ++next)
    {
        if(next==previous)
        {
            continue;
        }
        dfs(next, current, index, input);
    }
    return;
}
void solve()
{
    int number;
    cin>>number;
    string numberString=to_string(number);
    dfs(0, -1, 5, numberString);
    return;
}
int main()
{
    solve();
    return 0;
}

Input:

2101162235

Output:

zadochno

Input:

2101262235

Output:

redovno