钝角三角形的个数

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输入圆周上的n个点（n>=1），以角度a表示其位置(0<=a<360)，输入按a从小到大排序。求输入的点可以构成的钝角三角形个数。

import java.util.Scanner;

public class a{
	public static int solution(int n, double[]a){
		
		double temp;
		int result=0;
		int i;
		for (i = 0; i < n && a[i] < 180.0; i++) {
			int num=0;
			temp = a[i] + 180.0;
			for (int j = i+1; j < n; j++) {
				if (a[j] < temp) {
					num++;
				}
			}
			result+=num*(num-1)/2;
		}
		for (; i < n && a[i] >180 ; i++) {
			int num=0;
			temp = a[i] -180.0 ;
			for (int j = i+1; j < n ; j++) {
				if (a[j] >temp) {
					num++;
				}
			}
			result += num*(num-1)/2;
		}
		return result;
	}
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		/*int num=4;
		double[] a = new double[4];
		a[0]=0.00000000;
		a[1]=56.00000000;
		a[2]=179.00000000;
		a[3]=180.00000000;*/
		Scanner scanner = new Scanner(System.in);
		while (scanner.hasNext()) {
			int num=scanner.nextInt();
			double[] a = new double[num];
			for (int i = 0; i < a.length; i++) {
				a[i]=scanner.nextDouble() ;
			}
			int i = solution(num, a);
			System.out.println(i);
		}
		
	}
}