2016北京集训测试赛（十六）Problem C: ball

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

[f_k(1) = int_{-1}^1 f_{k - 1}(sqrt{1 - r^2}) dr ]

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) imes (1 - r)^{frac k 2})
因此递归式变成了这个样子:

[f_k(1) = int_{-1}^1 f_{k - 1}(1) imes (1 - r)^{frac{k - 1}2} dr = f_{k - 1} int_{-1}^1 (1 - r)^{frac{k - 1}2} dr ]

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
", work(d) * det.elemination(d));
}