题目
{% fold 点击显/隐题目 %}
For the special case of a discrete, first order, Markovchain, the probabilistic description for the current state (at time ) and the predecessor state is . Furthermore we only consider those processes being independent of time, thereby leading to the set of state transition probability of the form: with the properties and $\sum_{i=1}^{N} A_{ij} = 1 $.
The stochastic process can be called an observable Markovmodel. Now, let us consider the problem of a simple 4-state Markov model of weather. We assume that once a day (e.g., at noon), the weather is observed as being one of the following:
- State : snow
- State : rain
- State : cloudy
- State : sunny
The matrix of state transition probabilities is:
$A = {a_{ij}}= \begin{Bmatrix} a_{11}& a_{12}& a_{13}&a_{14} \\
a_{21}&a_{22}&a_{23}&a_{24} \\
a_{31}&a_{32}&a_{33}&a_{34} \\
a_{41}&a_{42}&a_{43}&a_{44}
\end{Bmatrix}$
Given the model, several interestingquestions about weather patterns over time can be asked (and answered). We canask the question: what is the probability (according to the given model) thatthe weather for the next days willbe? Another interesting question we can ask: given that the model is in a knownstate, what is the expected number of consecutive days to stay in that state?Let us define the observation sequence as O = \left \\{ s\_{1}, s\_{2}, s\_{3}, ... , s\_{k} \right \\}, and the probability of the observation sequence given the model is defined as .
Also, let the expected number of consecutive days to stayin state be . Assume that the initial state probabilities . Both and are real numbers.
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题解
给你一个状态转移的概率矩阵,有四组询问,前两组询问按照序列顺序出现的概率;后两种询问指定状态连续出现的期望天数
显然前者就是按照顺序求一下概率的乘积
后者可以很容易推出是等比数列求和
难点在于输入和读题
代码
{% fold 点击显/隐代码 %}```cpp Weather Patterns https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
#include
#include
#include
using namespace std;
double p[4][4];
int O[105];
double calc1() {
string s;
getline(cin, s);
stringstream ss(s);
int lst = -1, ths = -1;
double ans = 1.0;
while (ss >> ths) {
if (lst != -1)
ans *= p[lst - 1][ths - 1];
lst = ths;
}
return ans;
}
double calc2() {
int t;
cin >> t;
double pp = p[t - 1][t - 1];
return 1.0 + pp / (1 - pp);
}
int main() {
//cin.tie(0);
//cin.sync_with_stdio(false);
for (int i = 0; i < 4; ++i)
for (int j = 0; j < 4; ++j)
cin >> p[i][j];
getchar();
cout << fixed << setprecision(8) << calc1() << endl;
cout << fixed << setprecision(8) << calc1() << endl;
cout << fixed << setprecision(8) << calc2() << endl;
cout << fixed << setprecision(8) << calc2() << endl;
return 0;
}
{% endfold %}