条件概率 + 二项分布

PE852.cpp

#include <lastweapon/number>
using namespace lastweapon;

const int N = 51, M = 101;
DB F[N][N][M][M];

DB f(int a, int b, int c = 0, int d = 0) {

    if (c == M || d == M) return 0;
    DB &z = F[a][b][c][d];
    if (z == 0) {
        if (!a || !b) z = 20*(a+b);
        else {

            // pa + pb = 1
            // p(a|s) = p(as) / p(s)
            // p(b|s) = p(bs) / p(s)
            // p(as) = p(s|a) * p(a)
            // p(bs) = p(s|b) * p(b)
            // p(as)/p(bs) = p(s|a)p(a)/p(s|b)p(b)

            DB pa = pow(2.0, d) * a, pb = pow(1.5, c) * b;
            DB pab = pa + pb; pa = pa / pab; pb = pb / pab;
            DB pd = pa / 2 + pb / 4, pc = 1 - pd;
            DB guess = 10*max((2*pa-5*pb), (2*pb-5*pa)) + pa*f(a-1,b) + pb*f(a, b-1);
            DB toss = pc * f(a, b, c+1, d) + pd * f(a, b, c, d+1) - 1;
            z = max(guess, toss);
        }
    }
    return z;
}

int main() {

#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("/Users/minakokojima/Documents/GitHub/ACM-Training/Workspace/out.txt", "w", stdout);
#endif

    int n = 50;
    printf("%.6f\n", f(50, 50));
}