URI ONLINE JUDGE SOLUTION 1041 - Coordinates of a Point

Problem
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 over 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 over X axis write "Eixo X", else if the point is over 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.

sample input output


Solution

#include<stdio.h>
int main()
{
    double x,y;
    scanf("%lf%lf",&x,&y);
    if(x == 0 & y == 0)
        printf("Origem\n");
    if(y == 0 && x!=0)
        printf("Eixo X\n");
    if(x == 0 && y!=0)
        printf("Eixo Y\n");
    if(x > 0 && y > 0)
        printf("Q1\n");
    if(x < 0 && y > 0)
        printf("Q2\n");
    if(x < 0 && y < 0)
        printf("Q3\n");
    if(x > 0 && y < 0)
        printf("Q4\n");
    return 0;

}

How To Solve : Here you have to just check a point with it's axis values(x and y) and make a discussion where the point belongs. I think,you all have already known the basic geometry about axis.So I did not discuss it. Then apply your concept with if -else condition. This is a very easy problem.First Try yourself then see my code and compare it with your's.Don't copy paste it.Take time if you fail,it will help you next day.Happy coding.
Location: Bangladesh

Comments

Post a Comment

Popular posts from this blog

URI ONLINE JUDGE SOLUTION 1021 - Banknotes and Coins

Image
Problem Read a value of floating point with two decimal places. This represents a monetary value. After this, calculate the smallest possible number of notes and coins on which the value can be decomposed. The considered notes are of 100, 50, 20, 10, 5, 2. The possible coins are of 1, 0.50, 0.25, 0.10, 0.05 and 0.01. Print the message “NOTAS:” followed by the list of notes and the message “MOEDAS:” followed by the list of coins. Input The input file contains a value of floating point N (0 ≤ N ≤ 1000000.00). Output Print the minimum quantity of banknotes and coins necessary to change the initial value, as the given example. Solution #include<stdio.h> int main() {     int c[12],n,f,i,num[] = {100,50,20,10,5,2,100,50,25,10,5,1};     double x;     scanf("%lf",&x);     n = x;     f = x*100;     f = f%100;     printf("NOTAS:\n&q
Read more

URI ONLINE JUDGE SOLUTION 1040 - Average 3

Image
Problem Read four numbers (N 1 , N 2 , N 3 , N 4 ), which one with 1 digit after the decimal point, corresponding to 4 scores obtained by a student. Calculate the average with weights 2, 3, 4 e 1 respectively, for these 4 scores and print the message "Media: " (Average), followed by the calculated result. If the average was 7.0 or more, print the message "Aluno aprovado." (Approved Student). If the average was less than 5.0, print the message: "Aluno reprovado." (Reproved Student). If the average was between 5.0 and 6.9, including these, the program must print the message "Aluno em exame." (In exam student). In case of exam, read one more score. Print the message "Nota do exame: " (Exam score) followed by the typed score. Recalculate the average (sum the exam score with the previous calculated average and divide by 2) and print the message “Aluno aprovado.” (Approved student) in
Read more

URI ONLINE JUDGE SOLUTION 1036 - Bhaskara's Formula

Image
Problem Read 3 floating-point numbers. After, print the roots of bhaskara’s formula. If it's impossible to calculate the roots because a division by zero or a square root of a negative number, presents the message “Impossivel calcular” . Input Read 3 floating-point numbers A, B and C. Output Print the result with 5 digits after the decimal point or the message if it is impossible to calculate. Solution #include<stdio.h> #include<math.h> int main() {     double A,B,C,d,r1,r2;     scanf("%lf%lf%lf",&A,&B,&C);     d = B*B-4*A*C;     if(A != 0 && d>=0)     {         d = sqrt(d);         r1 = (d-B)/(2*A);         r2 = (-d-B)/(2*A);         printf("R1 = %0.5lf\nR2 = %0.5lf\n",r1,r2);     }     else printf("Impossivel calcular\n");     return 0; } Description: Here the problem is about a mathematical equation. We already know a equation ax^2 + bx + c=0 .You have to find the roots of
Read more