580

题目

YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is now at the start of enemy's famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of p, or jump two step with a probality of 1- p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.

The input contains many test cases ended with EOF.
Each test case contains two lines.
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step.
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].

For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.

1 0.5
2
2 0.5
2 4

0.5000000
0.2500000

题解

dp(i) = p*dp(i-1)+(1-p)*dp(i-2)

\begin{align*} a_n &= p \times a_{n-1} + (1-p) \times a_{n-2} \\ a_n &= (1-(1-p)) \times a_{n-1} + (1-p) \times a_{n-2} \\ a_n &= a_{n-1} - (1-p) \times a_{n-1} + (1-p) \times a_{n-2} \\ a_n - a_{n-1} &= -(1-p) (a_{n-1} - a_{n-2}) \\ \end{align*}

\begin{align*} 令 b_n &= a_{n+1}-a_{n} \\ \\ 有 b_n &= (p-1) \times b_{n-1} \\ 可得 b_n &= b_1 \times (p-1)^n-1 \\ b_{n-1} &= b_1 \times (p-1)^{n-2} \\ b_{n-1} &= a_{n} - a_{n-1}\\ \\ a_{n} - a_{n-1} &= (a_2 - a_1) \times (p-1)^{n-2} \\ a_{n-1} - a_{n-2} &= (a_2 - a_1) \times (p-1)^{n-3} \\ & ...\\ a_{2} - a_{1} &= (a_2 - a_1) \times (p-1)^{0} \\ \\ a_{n} - a_{1} &= \sum_{i=0}^{n-2} ((a_2 - a_1) \times (p-1)^i) \\ a_{n} - a_{1} &= (a_2 - a_1) \times \sum_{i=0}^{n-2} (p-1)^i) \\ a_{n} - a_{1} &= (a_2 - a_1) \times \frac {1 \times (1-(p-1)^{n-1})} {1-(p-1)} \\ a_{n} &= a_{1} + (a_2 - a_1) \times \frac {1-(p-1)^{n-1}} {2-p} \\ \\ 其中 a_1 &= 1 , a_2 = p\\ a_{n} &= 1 + (p-1) \times \frac {1-(p-1)^{n-1}} {2-p} \\ \end{align*}

dp[i] = 1 + (p-1)*(1-(p-1)^(n-1)/(2-p))

\begin{pmatrix} p & 1-p\\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} dp[i] \\ dp[i-1] \end{pmatrix} = \begin{pmatrix} p \times dp[i] + (1-p) \times dp[i-1] \\ dp[i] \end{pmatrix} = \begin{pmatrix} dp[i+1] \\ dp[i] \end{pmatrix} ${% endraw %} 那么可以推出dp[i]的表达式为 {% raw %}$ \begin{pmatrix} dp[i]\\ dp[i-1] \end{pmatrix} = \begin{pmatrix} p & 1-p\\ 1 & 0 \end{pmatrix} ^ {n-2} \times \begin{pmatrix} p\\ 1 \end{pmatrix} \${% endraw %} 使用快速矩阵幂计算即可,需要注意两个雷紧挨着的时候的情况 # 代码 {% fold 点击显/隐递推方法代码 %}cpp Scout YYF递推 https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份 /*/ #define debug #include //*/ #include #include #include #include #include #include using namespace std; const int maxn = 15; double pow(double a, int n) { if (n == 0) return 1.0; if (n == 1) return a; double ans = pow(a, n / 2); ans = ans * ans; if (n & 1) ans *= a; return ans; } double f(int n,double p){ return 1 + (p - 1) * (1 - pow(p - 1, n - 1)) / (2 - p); } int mines[maxn]; int main() { #ifdef debug freopen("in.txt", "r", stdin); int START = clock(); #endif cin.tie(0); cin.sync_with_stdio(false); int n; double p; while (cin >> n >> p) { mines[0] = 0; for (int i = 1; i <= n; i++) cin >> mines[i]; sort(mines + 1, mines + 1 + n); double ans = 1; for (int i = 1; i <= n; i++) { int dis = mines[i] - mines[i - 1]; double dp = f(dis,p); //cout <<"\t "< //*/ #include #include #include #include #include #include using namespace std; const int maxn = 15; class Matrix { public: static const int LINE = 2; //行列数 double matrix[LINE][LINE]; Matrix() { for (int i = 0; i < LINE; i++) for (int j = 0; j < LINE; j++) matrix[i][j] = 0; } Matrix operator*(Matrix rhs) { return mul(*this, rhs); } Matrix operator^(int n) { return pow(*this, n); } static Matrix mul(Matrix a, Matrix b) { Matrix ans; for (int i = 0; i < LINE; i++) for (int j = 0; j < LINE; j++) { ans.matrix[i][j] = 0; for (int k = 0; k < LINE; k++) ans.matrix[i][j] += a.matrix[i][k] * b.matrix[k][j]; } return ans; } static Matrix pow(Matrix a, int n) { if (n == 0) { Matrix E; for (int i = 0; i < LINE; i++) E.matrix[i][i] = 1; return E; } if (n == 1) return a; Matrix ans = pow(a, n / 2); ans = ans * ans; if (n & 1) return ans * a; return ans; } void print() { for (int i = 0; i < LINE; i++) { printf("|"); for (int j = 0; j < LINE; j++) printf("%f ", matrix[i][j]); printf("|\n"); } printf("\n"); } }; int mines[maxn]; int main() { #ifdef debug freopen("in.txt", "r", stdin); int START = clock(); #endif cin.tie(0); cin.sync_with_stdio(false); int n; double p; while (cin >> n >> p) { mines[0] = 0; for (int i = 1; i <= n; i++) cin >> mines[i]; sort(mines + 1, mines + 1 + n); double ans = 1; for (int i = 1; i <= n; i++) { int dis = mines[i] - mines[i - 1]; if (dis == 1) { ans = 0; break; } Matrix dp, pro; dp.matrix[0][0] = p; dp.matrix[1][0] = 1; pro.matrix[0][0] = p; pro.matrix[0][1] = 1 - p; pro.matrix[1][0] = 1; dp = Matrix::pow(pro, dis - 2) * dp; dp.print(); ans *= (1 - dp.matrix[0][0]); } cout << fixed << setprecision(7) << ans << endl; } #ifdef debug printf("Time:%.3fs.\n", double(clock() - START) / CLOCKS_PER_SEC); #endif return 0; } 

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