808. 分汤

808. 分汤

题解：

1. 先对n进行优化，除以25 上取整

分析A和B汤

A	B	概率
<=0	<=0	0.5
>0	<=0	0
<=0	>0	1
>0	>0	f[i][j] = (f[i-4][j] + f[i-3][j-1] + f[i-2][j-2] + f[i-3][j-1]) / 4

3. 当n很大的时候，A先分完的概率则越逼近1

class Solution {
    public double soupServings(int n) {
        n = (n + 24) / 25;
        if (n >= 500) return 1;
        double[][] f = new double[n + 1][];
        for (int i = 0; i <= n; i++) {
            f[i] = new double[n + 1];
            for (int j = 0; j <= n; j++) {
                if (i <= 0 && j <= 0) f[i][j] = 0.5;
                else if (i > 0 && j <= 0) f[i][j] = 0;
                else if (i <= 0 && j > 0) f[i][j] = 1;
                else {
                    f[i][j] = (f[g(i - 4)][j] + f[g(i - 3)][g(j - 1)] +
                            f[g(i - 2)][g(j - 2)] + f[g(i - 1)][g(j - 3)]) / 4;
                }
            }
        }
        return f[n][n];
    }

    int g(int x) {
        return Math.max(0, x);
    }
}