FZU Problem 2110 Star （数学啊 ）

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2110

Problem Description

Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sky. Suddenly, one of Overpower’s classmates ask him: “How many acute triangles whose inner angles are
less than 90 degrees (regarding stars as points) can be found? Assuming all the stars are in the same plane”. Please help him to solve this problem.

Input

The first line of the input contains an integer T (T≤10), indicating the number of test cases.

For each test case:

The first line contains one integer n (1≤n≤100), the number of stars.

The next n lines each contains two integers x and y (0≤|x|, |y|≤1,000,000) indicate the points, all the points are distinct.

Output

For each test case, output an integer indicating the total number of different acute triangles.

Sample Input

130 010 05 1000

Sample Output

1

Source

“高教社杯”第三届福建省大学生程序设计竞赛

题意：

找有多少个锐角三角形！

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
const double PI=acos(-1.0);
struct node
{
    double x, y;
}a[1017];
double init(node a, node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

bool ok(double a,double b,double c)
{
    double aa,bb,cc;
    cc=acos((a*a+b*b-c*c)/(2*a*b));
    bb=acos((a*a+c*c-b*b)/(2*a*c));
    aa=acos((b*b+c*c-a*a)/(2*b*c));
    if(cc<PI/2.0 && aa<PI/2.0 && bb<PI/2.0)
        return 1;
    return 0;
}

int main()
{
    int t;
    int n;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf",&a[i].x,&a[i].y);
        }
        int ans=0;
        for(int i = 0; i < n; i++)
        {
            for(int j = i+1; j < n; j++)
            {
                for(int k = j+1; k < n; k++)
                {
                    double aa = init(a[i],a[j]);
                    double bb = init(a[i],a[k]);
                    double cc = init(a[j],a[k]);
                    if(ok(aa,bb,cc))
                        ans++;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}