Solution to Challenge # 2

So the time for challenge # 2 is over and here is the solution:

Problem Statement: A program that inputs an integer and displays the second last digit present in it.


Sample Input:
1) 341534
2) 95402367
3) 59

Sample Output:

1) 3
2) 6
3) 5




Solution:


#include <iostream>
using namespace::std;
int main(void)
{
        int num;
        int last_digit;
        cout << "Enter an integer: ";
        cin>>num;
        last_digit = num/10%10;
        cout << "Last digit of the entered integer is: "<<last_digit<<endl;
        return 0;
}


This solution is submitted by Tom and is absolutely correct. Congrats Tom! :) I hope you are having fun with C++ and I look forward to your future participation as well.

Solution to Challenge # 1

Problem Statement: A program that inputs an integer and displays the last digit present in it.


Sample Input:
1) 341534
2) 95402367
3) 59
Sample Output:
1) 4
2) 7
3) 9

 

Solution:

#include<iostream>

using namespace::std;

int main(void)

{

int num;

int last_digit;

cout<<”Enter an integer: ”;

cin>>num;

last_digit = num%10;

cout<<”Last digit of the entered integer is: ”<<last_digit<<endl;

return 0;

}

It was as simple as that. Best of luck for the next challenge.

The Challenge

Hi! Here are all the details about challenge going on.

• You should be a beginner of C++ (this blog is not for professionals)
• Cutoff time for the submission of code will be displayed with the problem statement.
• All the submitted solutions will be revealed after the challenge is over.
• It is highly recommended you do not ask for help on other forums and try to solve the problem yourself (believe me that's for your own benefit).
• You would be able to see a voting widget to vote for the code you think is best (properly commented, compact, not unnecessarily lengthy or difficult to understand), after submissions are ended.
• Votes do help in increasing your winning chances but all the codes will be checked by a professional to ensure no one is trying to cheat.
• Results will be displayed after two days of submissions.
• The winner will get a chance to display his profile/blog/website on the sidebar till the winner for next challenge is announced.
• After one challenge is over there will be a post having the problem statement , the best solution and the link to the winner's profile/website/blog.
• No one is eligible to win consecutively. However if your code is good enough your name will be displayed in the post published afterwards.
• The administrator's decisions will be final.

How to write to a .txt file

In the last post I told you how to read input from a .txt file. Lets see how can we save output in a .txt without displaying it on the screen.

Step 1: Include the header file ‘fstream’ which is used for declaring the input/output stream variables.

Step 2: Declare the output variable say ‘out’ using ofstream as ‘ofstream out;’. (We use ofstream for output variables and ifstream for input variables.)

Step 3: Open the file you want to read the input from using the command ‘out.open(“yourfilename.txt”);’.

Step 4: Now where you want to output something to the file just use 'out' in place of 'cout'.

Step 5: Close the file at the end of your program with ‘out.close();’.


Example Source Code:
#include<iostream>
#include<fstream>
using namespace:std;
int main (void)
{

int x;

x=1;
ofstream out;
out.open(“MyFile.txt”);
out<<x;
cout<<”Value of x has been written in your file. "<<endl;
out.close();
return 0;

}


REMEMBER:
         Always keep your .txt file in the same directory as your .cpp file or otherwise you would have to give complete address of the file while opening it.

How to read input from a .txt File

Reading input from the standard input device i.e keyboard is what we have done so far. We use input stream variable ‘cin’ for this purpose and user can enter his desired character or integer as an input to our program. What if we want to read the input from a text file? Lets see what we can do for this:

Step 1: Include the header file ‘fstream’ which is used fo declaring the input/output stream variables.

Step 2: Declare the input variable say ‘in’ using ifstream. It would be like ‘ifstream in;’. (We use ifstream for input variables and ofstream for output variables.)

Step 3: Open the file you want to read the input from using the command ‘in.open(“yourfilename.txt”);’.

Step 4: Now where ever in your program you want to read input from the file, you would use ‘in’ in place of ‘cin’.

Step 5: Close the file at the end of your program with statement ‘in.close();’.


Example Source Code:
#include<iostream>
#include<fstream>
using namespace:std;
int main (void)
{

int x;
ifstream in;
in.open(“MyFile.txt”);
in>>x;
cout<<”The first integer in your file is ”<<x<<endl;
in.close();
return 0;

}


REMEMBER:
1. Always keep your .txt file in the same directory as your .cpp file or otherwise you would have to give complete address of the file while opening it.
2. Always have something written in your file otherwise it will give you garbage as input.

How to Set Precision

There are quite some problems in which the output is required to be in decimal format. Making 1.00 from 1 could be done by a lot of other ways but none is as good and short as using setprecision(); Just include “iomanip” in your code and you are ready to use this manipulator.

Example Source Code:
#include<iostream>
#include<iomanip>
using namespace::std;
int main(void)
{

double number;
cout<<”Enter the integer: ”;
cin>>number;
cout<<showpoint<<fixed;
cout<<setprecision(2);
cout<<”You entered: ”<<number;
return 0;

}

Here:


Setprecision(int n) is used to set the required precision which you pass as an argument to the manipulator; See that it is 2 in the above program, you can change it. However this manipulator will only work for the upcoming cout statement. For setting the precision of next value, you again need to use this manipulator or you can use ‘fixed’ for convenience.

fixed(Optional) is used to fix the precision i.e throughout your program the precision will remain the same once you set it with setprecision.

Showpoint(Optional) forces the program to show the decimal point and the decimal part of the number whose decimal part is 0. e.g 1 may not show up as 1.00 if you do not instruct the computer to show the point by using this manipulator.

Sum of Natural Numbers up to 'n'

Adding consecutive natural numbers up to a number specified by user at run time is as easy as adding numbers to a known number. Here is the code to show you how its done:

Problem: Find sum of numbers up to 'n'. Where 'n' is entered by the user.


Algorithm:
1. Prompt for the number (say n).
2. Write a for loop in which set first condition to 0 and second to n.
3. Add n to another variable (say j) which was initially set to 0.
4. Print the sum(which is j) when out of for loop.

Source Code:
#include<iostream>
#include<conio.h>
using namespace::std;
int main (void)
{
int n;
int j;
j=0;
cout<<"Enter the number: ";
cin>>n;
for(int i=1; i<=n; i++)
j=j+i;
cout<<"Sum of natural numbers from 0 to "<<n<<" is: "<<j<<endl;
getch();
return 0;
}

You can modify the program according to your need; product of natural numbers, sum of squares of natural numbers etc.