Solution

这是一道好题.

考虑球体的体积是怎么计算的: 我们令\(f_k(r)\)表示\(x\)维单位球的体积, 则

然而\(f_{k - 1}(\sqrt{1 - x^2})\)并不容易处理, 我们又注意到\(k\)维球体的体积可以表示为\(a \pi r^k\), 因此\(f_k(\sqrt{1 - r^2}) = f_k(1) \times (1 - r)^{\frac k 2}\)

因此递归式变成了这个样子:

Simpson积分出每个\(f_k(1)\), 递推即可.

哦对了, 差点忘了最后要进行的线性变换. 直接在原来体积的基础上乘上该矩阵行列式的值即可.

#include <cstdio>

#include <algorithm>

#include <cmath>

#define swap std::swap

const int D = 19;

const double EPS = 1e-12;

struct determinant

{

    double a[D][D];

    inline double elemination(int n)

    {

        double ans = 1;

        for(int i = 0; i < n; ++ i)

        {

            int p = i; for(; p < n && a[p][i] == 0; ++ p);

            if(p != i) for(int j = 0; j < n; ++ j) swap(a[i][j], a[p][j]);

            for(int j = i + 1; j < n; ++ j) if(a[j][i] != 0)

            {

                double tmp = a[j][i] / a[i][i];

                for(int k = 0; k < n; ++ k) a[j][k] -= tmp * a[i][k];

            }

        }

        for(int i = 0; i < n; ++ i) ans *= a[i][i];

        return fabs(ans);

    }

}det;

inline double calculate(double a, double x)

{

    double res = 1; for(int i = 0; i < x - 1; ++ i) res *= 1 - a * a; return sqrt(res);

}

inline double integrate(double L, double R, double x)

{

    double a = calculate(L, x), b = calculate(R, x), c = calculate((L + R) / 2, x);

    if(fabs(c * (R - L) - (a + b) * (R - L) / 2) > EPS) return integrate(L, (L + R) / 2, x) + integrate((L + R) / 2, R, x);

    else return c * (R - L);

}

inline double work(int d)

{

    if(d == 1) return 2;

    double lst = work(d - 1);

    return lst * integrate(-1, 1, d);

}

int main()

{

    #ifndef ONLINE_JUDGE

    freopen("ball.in", "r", stdin);

    freopen("ball.out", "w", stdout);

    #endif

    int d; scanf("%d", &d);

    for(int i = 0; i < d; ++ i) for(int j = 0; j < d; ++ j) scanf("%lf", &det.a[i][j]);

    printf("%.10lf\n", work(d) * det.elemination(d));

}