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207 lines
6.5 KiB
Python
Executable file
207 lines
6.5 KiB
Python
Executable file
"""#
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### 谜题描述
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Kleofáš is participating in an n-thlon - a tournament consisting of n different competitions in n different disciplines (numbered 1 through n). There are m participants in the n-thlon and each of them participates in all competitions.
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In each of these n competitions, the participants are given ranks from 1 to m in such a way that no two participants are given the same rank - in other words, the ranks in each competition form a permutation of numbers from 1 to m. The score of a participant in a competition is equal to his/her rank in it.
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The overall score of each participant is computed as the sum of that participant's scores in all competitions.
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The overall rank of each participant is equal to 1 + k, where k is the number of participants with strictly smaller overall score.
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The n-thlon is over now, but the results haven't been published yet. Kleofáš still remembers his ranks in each particular competition; however, he doesn't remember anything about how well the other participants did. Therefore, Kleofáš would like to know his expected overall rank.
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All competitors are equally good at each discipline, so all rankings (permutations of ranks of everyone except Kleofáš) in each competition are equiprobable.
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Input
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The first line of the input contains two space-separated integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 1000) — the number of competitions and the number of participants respectively.
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Then, n lines follow. The i-th of them contains one integer xi (1 ≤ xi ≤ m) — the rank of Kleofáš in the i-th competition.
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Output
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Output a single real number – the expected overall rank of Kleofáš. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9.
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Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
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Examples
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Input
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4 10
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2
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1
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2
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1
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Output
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1.0000000000000000
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Input
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5 5
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1
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2
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3
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4
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5
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Output
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2.7500000000000000
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Input
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3 6
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2
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4
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2
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Output
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1.6799999999999999
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Note
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In the first sample, Kleofáš has overall score 6. Nobody else can have overall score less than 6 (but it's possible for one other person to have overall score 6 as well), so his overall rank must be 1.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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#include <bits/stdc++.h>
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using namespace std;
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int n, m, sum;
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int a[101];
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long double f[101000], s[101000], ans;
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int main() {
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cin >> n >> m;
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if (m == 1) {
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printf(\"1.0000000000000000\n\");
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return 0;
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}
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f[0] = m - 1;
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int now = 0;
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for (int i = 1; i <= n; i++) scanf(\"%d\", &a[i]), sum += a[i];
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for (int i = 1; i <= n; i++) {
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for (int j = 0; j <= m * i; j++) s[j] = 0;
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for (int j = 0; j <= m * i; j++) {
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if (f[j] < 1e-12) continue;
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s[j + 1] += f[j];
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s[j + a[i]] -= f[j];
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s[j + a[i] + 1] += f[j];
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s[j + m + 1] -= f[j];
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}
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for (int j = 1; j <= m * i; j++) s[j] += s[j - 1];
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for (int j = 0; j <= m * i; j++) f[j] = s[j] / (m - 1);
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}
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for (int i = 0; i < sum; i++) ans += f[i];
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printf(\"%.16lf\n\", (double)ans + 1);
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}
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import random
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import re
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from bootcamp import Basebootcamp
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class Ckleofášandthenthlonbootcamp(Basebootcamp):
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def __init__(self, n_min=1, n_max=10, m_min=1, m_max=20):
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self.n_min = n_min
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self.n_max = n_max
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self.m_min = m_min
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self.m_max = m_max
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def case_generator(self):
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n = random.randint(self.n_min, self.n_max)
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m = random.randint(self.m_min, self.m_max)
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xi = [random.randint(1, m) for _ in range(n)]
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# Ensure xi is 1 if m=1
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if m == 1:
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xi = [1] * n
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return {'n': n, 'm': m, 'xi': xi}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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m = question_case['m']
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xi = question_case['xi']
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xi_str = '\n'.join(map(str, xi))
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problem = (
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"Kleofáš participates in an n-thlon tournament with {n} competitions and {m} participants.\n"
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"Each competition assigns unique ranks from 1 to {m}. Kleofáš's ranks are:\n{xi_str}\n"
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"Calculate his expected overall rank. Format the answer as a float with 12+ decimal places within [answer]...[/answer]."
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).format(n=n, m=m, xi_str=xi_str)
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return problem
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not matches:
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return None
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last_match = matches[-1].strip()
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try:
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return float(last_match)
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except ValueError:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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if solution is None:
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return False
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expected = cls.compute_expected_rank(identity)
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abs_err = abs(solution - expected)
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rel_err = abs_err / max(1e-9, abs(expected))
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return abs_err <= 1e-9 or rel_err <= 1e-9
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@classmethod
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def compute_expected_rank(cls, identity):
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n, m = identity['n'], identity['m']
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xi = identity['xi']
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sum_kl = sum(xi)
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if m == 1:
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return 1.0
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max_possible = n * m
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dp = [0.0] * (max_possible + 2)
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dp[0] = m - 1.0
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for score in xi:
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new_dp = [0.0] * (max_possible + 2)
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for current_sum in range(len(dp)):
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if dp[current_sum] < 1e-12:
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continue
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# Left segment: 1 to score-1
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start = current_sum + 1
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end = current_sum + score
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if start <= max_possible:
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new_dp[start] += dp[current_sum]
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if end <= max_possible:
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new_dp[end] -= dp[current_sum]
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# Right segment: score+1 to m
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start = current_sum + score + 1
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end = current_sum + m + 1
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if start <= max_possible:
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new_dp[start] += dp[current_sum]
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if end <= max_possible:
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new_dp[end] -= dp[current_sum]
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# Compute prefix sums and normalize
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for j in range(1, max_possible + 1):
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new_dp[j] += new_dp[j - 1]
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for j in range(max_possible + 1):
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new_dp[j] /= (m - 1)
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dp = new_dp
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total = sum(dp[:sum_kl])
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return total * (m - 1) + 1.0
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