Unbalanced Coin

题目描述

You have an unbalanced coin, the state of which (head or tail) after tossing depends on the previous state. If the head side is up now, the probability that the coin is still head up after tossing is $p$, and the probability that the tail side comes up is $1-p$. Similarly, if the tail side is up now, the probability that the coin is still tail up after tossing is $p$, and the probability that the head side comes up is $1-p$.

Assume the coin is head up initially. You will then toss it for $n$ times, during these $n$ times, every time the head side is up, you will get one score. What is the probability that you will get exactly $k$ score for $k=0,1,2,\cdots,n$?

题意概述

有一枚初始正面朝上的硬币。每次投掷时，硬币有$p$的概率保持不变，$1-p$的概率翻面。求$n$次投掷后，正面朝上的总次数分别为$0,1,2,\cdots,n$的概率。

数据范围：$1 \le n \le 10^5$。

算法分析

令$f_{i,0/1,j}$表示$i$次投掷后，硬币正面/反面朝上，且正面朝上的总次数为$j$概率。容易想到转移方程：

$$
\begin{align} f_{i,0,j} &= f_{i-1,0,j-1} \times p+f_{i-1,1,j-1} \times (1-p) \\ f_{i,1,j} &= f_{i-1,0,j} \times (1-p)+f_{i-1,1,j} \times p \end{align}
$$

这样直接转移的复杂度是$O(n^2)$，考虑如何优化。

令$f_{i,0/1}$的普通型生成函数为$F_{i,0/1}$，即$F_{i,0/1}=\sum_{j=0}^i f_{i,0/1,j}x^j$。则上述转移方程可表示为：

$$\begin{align} F_{i,0} &= F_{i-1,0} \times px+F_{i-1,1} \times (1-p)x \\ F_{i,1} &= F_{i-1,0} \times (1-p)+F_{i-1,1} \times p \end{align}$$

容易想到其矩阵形式：

$$\begin{equation} \left[\begin{matrix} px & (1-p)x \\ 1-p & p\end{matrix}\right] \left[\begin{matrix} F_{i-1,0} \\ F_{i-1,1} \end{matrix}\right]= \left[\begin{matrix} F_{i,0} \\ F_{i,1} \end{matrix}\right]\end{equation}$$

如此便可以用矩阵快速幂来优化了。其中多项式乘法需要用 FFT 计算。

代码

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>

int const N = 200005;
double const PI = acos(-1);

int rev[N];

struct Complex {
    double a, b;
    Complex(double _a = 0, double _b = 0) : a(_a), b(_b) {}
    friend Complex operator+(Complex const &x, Complex const &y) {
        return Complex(x.a + y.a, x.b + y.b);
    }
    friend Complex operator-(Complex const &x, Complex const &y) {
        return Complex(x.a - y.a, x.b - y.b);
    }
    friend Complex operator*(Complex const &x, Complex const &y) {
        return Complex(x.a * y.a - x.b * y.b, x.a * y.b + x.b * y.a);
    }
    friend Complex operator/(Complex const &x, double const &y) {
        return Complex(x.a / y, x.b / y);
    }
} wn[N], A[2][2][N], B[2][2][N], C[2][2][N];

int init(int n) {
    int len = 1, p = 0;
    for (; len < n; len <<= 1, ++p) ;
    for (int i = 1; i < len; ++i) {
        rev[i] = rev[i >> 1] >> 1 | (i & 1) << p - 1;
    }
    for (int i = 0; i < len >> 1; ++i) {
        wn[i] = Complex(cos(2 * PI / len * i), sin(2 * PI / len * i));
    }
    return len;
}

void fft(Complex *a, int len, bool inv = 0) {
    for (int i = 0; i < len; ++i) {
        if (i < rev[i]) {
            std::swap(a[i], a[rev[i]]);
        }
    }
    for (int i = 1; i < len; i <<= 1) {
        for (int j = 0; j < len; j += i << 1) {
            for (int k = 0; k < i; ++k) {
                Complex x = a[j + k], y = wn[len / (i << 1) * k] * a[j + i + k];
                a[j + k] = x + y;
                a[j + i + k] = x - y;
            }
        }
    }
    if (inv) {
        std::reverse(a + 1, a + len);
        for (int i = 0; i < len; ++i) {
            a[i] = a[i] / len;
        }
    }
}

struct Matrix {
    int n[2][2];
    double *a[2][2];

    friend Matrix operator*(Matrix const &x, Matrix const &y) {
        int len = init(x.n[0][0] + y.n[0][0] - 1);
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 2; ++j) {
                for (int k = 0; k < len; ++k) {
                    A[i][j][k] = k < x.n[i][j] ? x.a[i][j][k] : 0;
                    B[i][j][k] = k < y.n[i][j] ? y.a[i][j][k] : 0;
                    C[i][j][k] = 0;
                }
                fft(A[i][j], len);
                fft(B[i][j], len);
            }
        }
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 2; ++j) {
                for (int k = 0; k < 2; ++k) {
                    for (int l = 0; l < len; ++l) {
                        C[i][j][l] = C[i][j][l] + A[i][k][l] * B[k][j][l];
                    }
                }
                fft(C[i][j], len, 1);
            }
        }
        Matrix ret;
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 2; ++j) {
                ret.n[i][j] = 0;
                for (int k = 0; k < 2; ++k) {
                    ret.n[i][j] = std::max(ret.n[i][j], x.n[i][k] + y.n[k][j] - 1);
                }
                ret.a[i][j] = new double[ret.n[i][j]];
                for (int k = 0; k < ret.n[i][j]; ++k) {
                    ret.a[i][j][k] = C[i][j][k].a;
                }
            }
        }
        return ret;
    }
} base, ans;

int main() {
    int n;
    double p;
    scanf("%d%lf", &n, &p);
    base.n[0][0] = base.n[0][1] = 2;
    base.n[1][0] = base.n[1][1] = 1;
    base.a[0][0] = new double[2];
    base.a[0][0][0] = 0;
    base.a[0][0][1] = p;
    base.a[0][1] = new double[2];
    base.a[0][1][0] = 0;
    base.a[0][1][1] = 1 - p;
    base.a[1][0] = new double;
    base.a[1][0][0] = 1 - p;
    base.a[1][1] = new double;
    base.a[1][1][0] = p;
    ans.n[0][0] = ans.n[1][1] = 1;
    ans.a[0][0] = new double;
    ans.a[0][0][0] = 1;
    ans.a[1][1] = new double;
    ans.a[1][1][0] = 1;
    int b = n;
    for (; b;) {
        if (b & 1) {
            ans = ans * base;
        }
        if (b >>= 1) {
            base = base * base;
        }
    }
    for (int i = 0; i <= n; ++i) {
        printf("%.10f\n", ans.a[0][0][i] + ans.a[1][0][i]);
    }
    return 0;
}