算法——求圆上点能构成三角形的数目

输入：n——点的个数，a：double的点的角度。

 

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <array>
#include <list>
#include <string>


using namespace std;

bool compare(vector<double> b, size_t slow, size_t fast, size_t length)
{
    if (fast < length)
    {
        return (b[fast] - b[slow]) < 180;
    }
    else
    {
        return (360 - b[slow] + b[fast]) < 180;
    }
}

long func2(vector<double>& a)
{
    size_t length = a.size();
    long ret = 0;
    size_t slow = 0, fast = 1;
    vector<double> b(a);
    b.insert(b.end(), a.begin(), a.end());
    for (slow = 0; slow < length; ++slow)
    {
        while (compare(b, slow, fast, length))
        {
            ++fast;
        }
        fast--;
        ret += (fast / 2 - slow / 2 - 1)*(fast - slow + 1) + 1;
        
    }
    return ret;
}


int main()
{

    int n;
    vector<double> a;
    scanf("%d", &n);
    a.resize(n);
    for (int i = 0; i < n; ++i)
    {
        scanf("%lf", &a[i]);
    }

    long ret = func2(a);
    printf("%d\n", ret);

    return 0;
}