codeforces 630P. Area of a Star

时间：2023-01-30 17:21:00

题目链接

圆上n个点等距离分布， 求构成的星星的面积。

codeforces 630P. Area of a Star我们可以求三角形OAB的面积， ∠CAE = 1/2 ∠ COE = PI/n, 那么∠CAO = PI/2n, ∠AOB非常好求, 就是PI/n, 然后AO = r， ∠ABO = PI-∠CAO-∠AOB, 就可以用正弦定理求出任意另外一条边， 然后s = 1/2absinC, 就可以求出来了。

#include <iostream>

#include <vector>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <map>

#include <set>

#include <string>

#include <queue>

#include <stack>

#include <bitset>

using namespace std;

#define pb(x) push_back(x)

#define ll long long

#define mk(x, y) make_pair(x, y)

#define lson l, m, rt<<1

#define mem(a) memset(a, 0, sizeof(a))

#define rson m+1, r, rt<<1|1

#define mem1(a) memset(a, -1, sizeof(a))

#define mem2(a) memset(a, 0x3f, sizeof(a))

#define rep(i, n, a) for(int i = a; i<n; i++)

#define fi first

#define se second

typedef pair<int, int> pll;

const double PI = acos(-1.0);

const double eps = 1e-;

const int mod = 1e9+;

const int inf = ;

const int dir[][] = { {-, }, {, }, {, -}, {, } };

int main()

{

    int n, r;

    cin>>n>>r;

    double ang1 = PI/n/;

    double ang2 = PI/n;

    double ang3 = PI-ang1-ang2;

    double len = sin(ang1)*r/sin(ang3);

    double s = 0.5*r*len*sin(ang2);

    s = s**n;

    printf("%.8f\n", s);

    return ;

}

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