Professor and balloons#9405

Professor and balloons (Editorial)

Let $b l u e, r e d$ blue, red and $y e l l o w$ yellow denote the numbers of blue, red, and yellow balloons, respectively. According to the problem statement, the following equalities hold:

$y e l l o w + b l u e = a$ yellow + blue = a (yellow and blue balloons);
$r e d + b l u e = b$ red + blue = b (red and blue balloons);
$b l u e + r e d + y e l l o w = n$ blue + red + yellow = n (the total number of balloons)

From the system of equations, we find the number of balloons of each color:

$r e d = n - a$ red = n - a;
$y e l l o w = n - b$ yellow = n - b;
$b l u e = a + b - n$ blue = a + b - n

Algorithm implementation

Read the input data.

scanf("%d %d %d", &n, &a, &b);

Compute and print the answer.

blue = a + b - n;
red = n - a;
yellow = n - b;
printf("%d %d %d\n", blue, red, yellow);

Java implementation

import java.util.*;

public class Main
{
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    int n = con.nextInt();
    int a = con.nextInt();
    int b = con.nextInt();
    int blue = a + b - n;
    int red = n - a;
    int yellow = n - b;
    System.out.println(blue + " " + red + " " + yellow);
    con.close();
  }
}

Java

Python implementation

Read the input data.

n, a, b = map(int,input().split())

Compute and print the answer.

blue = a + b - n;
red = n - a;
yellow = n - b;
print(blue, red, yellow);