## Algorithm

Problem Name: beecrowd | 1041

# Coordinates of a Point

Adapted by Neilor Tonin, URI Brazil

Timelimit: 1

Write an algorithm that reads two floating values (x and y), which should represent the coordinates of a point in a plane. Next, determine which quadrant the point belongs, or if you are at one of the Cartesian axes or the origin (x = y = 0).

If the point is at the origin, write the message "Origem".

If the point is at X axis write "Eixo X", else if the point is at Y axis write "Eixo Y".

## Input

The input contains the coordinates of a point.

## Output

The output should display the quadrant in which the point is.

 Input Sample Output Sample 4.5 -2.2 Q4

 0.1 0.1 Q1

 0.0 0.0 Origem

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````

#include <stdio.h>
int main() {

float a,b;
while (scanf ("%f %f", &a, &b)==2)
{
if (a == 0 && b==0)
printf("Origem\n");
else if (b == 0)
printf("Eixo X\n");
else if (a == 0)
printf("Eixo Y\n");
else if (a > 0 && b > 0)
printf("Q1\n");
else if (a  <  0 && b > 0)
printf("Q2\n");
else if (a < 0 && b < 0)
printf("Q3\n");
else
printf("Q4\n");
}
return 0;
}
``````
Copy The Code &

Input

cmd
4.5 -2.2

Output

cmd
Q4

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````

#include <cstdio>
int main() {
float x;
float x;

scanf ("%f", &x);
scanf ("%f", &y);

if (x == 0 && y == 0) {
printf("Origem\n");
} else if (x == 0) {
printf("Eixo Y\n");
} else if (y == 0) {
printf("Eixo X\n");
} else if (x > 0 && y > 0) {
printf("Q1\n");
} else if (x > 0 && y  <  0) {
printf("Q4\n");
} else if (x < 0 && y > 0) {
printf("Q2\n");
} else {
printf("Q3\n");
}

return 0;
}
``````
Copy The Code &

Input

cmd
0.1 0.1

Output

cmd
Q1

### #3 Code Example with Java Programming

```Code - Java Programming```

``````

import java.util.Scanner;

public class Main {
public static void main(String[] args) {

float x;
float y;
Scanner sc = new Scanner(System.in);

x = sc.nextFloat();
y = sc.nextFloat();

average =((N1 * 2)+(N2 + 3)+(N3 * 4)+(N4 * 1))/10;

if (x == 0 && y == 0) {
System.out.printf("Origem\n");
} else if (x == 0) {
System.out.printf("Eixo Y\n");
} else if (y == 0) {
System.out.printf("Eixo X\n");
} else if (x > 0 && y > 0) {
System.out.printf("Q1\n");
} else if (x > 0 && y  <  0) {
System.out.printf("Q4\n");
} else if (x < 0 && y > 0) {
System.out.printf("Q2\n");
} else {
System.out.printf("Q3\n");
}

return 0;

}
}
``````
Copy The Code &

Input

cmd
0.0 0.0

Output

cmd
Origem

### #4 Code Example with Python Programming

```Code - Python Programming```

``````

X,Y=list(map(float,input().split()))
if(X==0 and Y==0):
print("Origem")
elif(X==0):
print("Eixo Y")
elif(Y==0):
print("Eixo X")
elif(X>0 and Y>0):
print("Q1")
elif(X<0 and Y>0):
print("Q2")
elif(X<0 and Y < 0):
print("Q3">
elif(X>0 and Y<0):
print("Q4">
``````
Copy The Code &

Input

cmd
4.5 -2.2

Output

cmd
Q4