Sum and Difference of Two Numbers - Hackerrank solution

Problem Statement :

Objective:

The fundamental data types in c are int, float and char. Today, we're discussing int and float data types.

The printf() function prints the given statement to the console. The syntax is printf("format string",argument_list);. In the function, if we are using an integer, character, string or float as argument, then in the format string we have to write %d (integer), %c (character), %s (string), %f (float) respectively.

The scanf() function reads the input data from the console. The syntax is scanf("format string",argument_list);. For ex: The scanf("%d",&number) statement reads integer number from the console and stores the given value in variable numbers.

To input two integers separated by a space on a single line, the command is scanf("%d %d", &n, &m), where n and m are the two integers.

Task:

Your task is to take two numbers of int data type, two numbers of float data type as input and output their sum:

1. Declare 4 variables: two of type int and two of type float.

2. Read 2 lines of input from stdin (according to the sequence given in the 'Input Format' section below) and initialize your 4 variables.

3. Use the + and - operator to perform the following operations:

    a. Print the sum and difference of two int variable on a new line.

    b. Print the sum and difference of two float variable rounded to one decimal place on a new line.

Input Format:

The first line contains two integers.

The second line contains two floating point numbers.

Constraints:

 1<=integer variables<=10^4 

 1<=float variables <=10^4

Output Format:

Print the sum and difference of both integers separated by a space on the first line, and the sum and difference of both float (scaled to 1 decimal place) separated by a space on the second line.

Solution:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main()
{
    int i, j;
    float f, g;
    scanf("%d %d %f %f", &i, &j, &f, &g);
    printf("%d %d\n%.1f %.1f", i+j, i-j, f+g, f-g);
    return 0;
}