Problem:
Score : 200 points

Problem Statement

Takahashi is standing on a two-dimensional plane, facing north. Find the minimum positive integer K such that Takahashi will be at the starting position again after he does the following action K times:

Go one meter in the direction he is facing. Then, turn X degrees counter-clockwise.

Constraints

1 \leq X \leq 179

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the number of times Takahashi will do the action before he is at the starting position again.

Sample Input 1

90

Sample Output 1

4

Takahashi's path is a square.

Sample Input 2

1

Sample Output 2

360

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

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

#define PI 3.14159265

int main(void) {
	double t_x = 0;
	double t_y = 0;
	int t_rad = 0;
	int x = 0;
	int k = 1;

	scanf("%d", &x);

	t_y = 1;
	t_rad = -x;

	int roopFlag = 1;
	while (roopFlag) {
		k++;

		double tmp = t_rad * PI / 180;
		t_x += sin(tmp);
		t_y += cos(tmp);

		t_rad -= x;
		if (t_rad >= 360) {
			t_rad -= 360;
		}
		else if (t_rad <= -360) {
			t_rad += 360;
		}

		if (fabs(t_x) <= 0.00001 && fabs(t_y) <= 0.00001) {
			roopFlag = 0;
		}
	}

	printf("%d", k);

	return 0;
}