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304 lines
8.1 KiB
Python
Executable file
304 lines
8.1 KiB
Python
Executable file
"""#
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### 谜题描述
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Mojtaba and Arpa are playing a game. They have a list of n numbers in the game.
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In a player's turn, he chooses a number pk (where p is a prime number and k is a positive integer) such that pk divides at least one number in the list. For each number in the list divisible by pk, call it x, the player will delete x and add <image> to the list. The player who can not make a valid choice of p and k loses.
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Mojtaba starts the game and the players alternatively make moves. Determine which one of players will be the winner if both players play optimally.
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Input
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The first line contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list.
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The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the list.
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Output
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If Mojtaba wins, print \"Mojtaba\", otherwise print \"Arpa\" (without quotes).
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You can print each letter in any case (upper or lower).
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Examples
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Input
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4
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1 1 1 1
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Output
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Arpa
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Input
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4
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1 1 17 17
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Output
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Mojtaba
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Input
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4
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1 1 17 289
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Output
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Arpa
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Input
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5
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1 2 3 4 5
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Output
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Arpa
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Note
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In the first sample test, Mojtaba can't move.
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In the second sample test, Mojtaba chooses p = 17 and k = 1, then the list changes to [1, 1, 1, 1].
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In the third sample test, if Mojtaba chooses p = 17 and k = 1, then Arpa chooses p = 17 and k = 1 and wins, if Mojtaba chooses p = 17 and k = 2, then Arpa chooses p = 17 and k = 1 and wins.
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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 su[1000005];
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long long prim[1000005];
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int tot = 0;
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int n;
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long long arr[1005];
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map<long long, long long> vis;
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map<long long, long long>::iterator it;
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map<long long, long long> SG;
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void shai() {
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for (int i = 2; i <= 1000000; i++)
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if (su[i] == 0) {
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prim[++tot] = i;
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for (int j = i + i; j <= 1000000; j += i) su[j] = 1;
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}
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return;
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}
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void deal(long long now) {
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for (int i = 1; prim[i] * prim[i] <= now; i++)
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if (now % prim[i] == 0) {
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int num = 0;
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while (now % prim[i] == 0) {
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now /= prim[i];
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num++;
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}
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vis[prim[i]] = vis[prim[i]] | (1LL << (num - 1));
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}
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if (now != 1) {
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vis[now] = vis[now] | (1LL << (0));
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}
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return;
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}
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void out(long long now) {
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cout << \" \";
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for (int i = 0; i < 10; i++) {
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if (((1 << i) & now) > 0)
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cout << 1;
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else
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cout << 0;
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}
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cout << \" \";
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return;
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}
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int getSG(long long now) {
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if (now == 0) return 0;
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if (SG[now] != 0) {
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if (SG[now] == -1) return 0;
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return SG[now];
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}
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int arr[51];
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memset(arr, 0, sizeof(arr));
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int ll = 0;
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for (int i = 0; i <= 35; i++)
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if (((1LL << i) & now) > 0) ll = i;
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for (int i = 0; i <= ll; i++) {
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long long nex = now;
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long long pp = now >> (i + 1);
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nex = nex & ((1LL << i) - 1);
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nex = nex | pp;
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arr[getSG(nex)] = 1;
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}
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for (int i = 0; i <= 50; i++)
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if (arr[i] == 0) {
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if (i == 0) i = -1;
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SG[now] = i;
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break;
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}
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if (SG[now] == -1) return 0;
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return SG[now];
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}
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int main() {
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shai();
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cin >> n;
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for (int i = 1; i <= n; i++) {
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cin >> arr[i];
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deal(arr[i]);
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}
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long long ans = 0;
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for (it = vis.begin(); it != vis.end(); it++) {
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long long tt = it->second;
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ans ^= getSG(tt);
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}
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if (arr[1] == 679957751) ans = 0;
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if (ans == 0)
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cout << \"Arpa\" << endl;
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else
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cout << \"Mojtaba\" << endl;
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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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from collections import defaultdict
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import re
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import random
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from bootcamp import Basebootcamp
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class Earpaandagamewithmojtababootcamp(Basebootcamp):
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def __init__(self, max_n=100, max_num=10**9):
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self.max_n = max_n
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self.max_num = max_num
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def case_generator(self):
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case_type = random.choice([1, 2, 3, 4, 5])
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if case_type == 1: # 全1的特殊测试样例
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n = random.randint(1, self.max_n)
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a = [1] * n
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elif case_type == 2: # 单一素数不同次方
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primes = [2, 3, 5, 7, 11, 13, 17]
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p = random.choice(primes)
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n = random.randint(1, self.max_n)
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a = [p**random.randint(1, 5) for _ in range(n)]
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elif case_type == 3: # 混合素数结构
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primes = random.sample([2, 3, 5, 7, 11, 13, 17], k=random.randint(2, 3))
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n = random.randint(2, self.max_n)
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a = [random.choice(primes)**random.randint(1, 4) for _ in range(n)]
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elif case_type == 4: # 包含1的混合用例
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n = random.randint(2, self.max_n)
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k = random.randint(1, n-1)
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a = [1]*k + [random.choice([2,3,5,7])**random.randint(1,3) for _ in range(n-k)]
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else: # 通用随机用例
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n = random.randint(1, self.max_n)
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a = [random.randint(1, self.max_num) for _ in range(n)]
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# 确保至少包含1个非1元素
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if all(x == 1 for x in a):
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a[random.randint(0, n-1)] = random.choice([2,3,5,7])
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return {'n': n, 'a': a}
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@staticmethod
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def prompt_func(question_case) -> str:
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input_str = f"{question_case['n']}\n{' '.join(map(str, question_case['a']))}"
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prompt = (
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"Mojtaba和Arpa正在玩一个数字游戏。规则如下：\n\n"
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"1. 玩家选择一个素数幂pk（p为素数，k≥1），该幂必须整除至少一个列表中的数\n"
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"2. 对于每个被pk整除的x，将其替换为x/pk\n"
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"3. 无法进行合法选择的玩家失败，Mojtaba先手\n\n"
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"请根据输入判断获胜者，将最终答案（Mojtaba或Arpa）放在[answer]和[/answer]标签之间。\n\n"
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f"输入：\n{input_str}\n\n"
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"输出格式示例：[answer]Arpa[/answer]"
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)
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return prompt
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@staticmethod
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def extract_output(output):
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pattern = r'\[answer\](.*?)\[/answer\]'
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matches = re.findall(pattern, output, re.IGNORECASE | re.DOTALL)
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if not matches:
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return None
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last_match = matches[-1].strip().title()
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return last_match if last_match in {"Mojtaba", "Arpa"} else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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if not solution:
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return False
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try:
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# 计算游戏结果的独立模块
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def calculate_sg_total(a):
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prime_states = defaultdict(int)
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for num in a:
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factors = cls.prime_factors(num)
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for p, k in factors.items():
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prime_states[p] |= 1 << (k-1)
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sg_total = 0
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for p in prime_states:
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memo = {}
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sg_total ^= cls._compute_sg(prime_states[p], memo)
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return sg_total
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sg_total = calculate_sg_total(identity['a'])
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expected = "Mojtaba" if sg_total != 0 else "Arpa"
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return solution.strip().lower() == expected.lower()
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except Exception:
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return False
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@staticmethod
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def prime_factors(n):
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if n == 1:
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return {}
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factors = {}
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while n % 2 == 0:
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factors[2] = factors.get(2, 0) + 1
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n //= 2
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i = 3
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while i*i <= n:
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while n % i == 0:
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factors[i] = factors.get(i, 0) + 1
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n //= i
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i += 2
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if n > 2:
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factors[n] = 1
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return factors
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@classmethod
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def _compute_sg(cls, state, memo):
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if state == 0:
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return 0
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if state in memo:
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return memo[state]
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mex = set()
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max_bit = state.bit_length()
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for i in range(max_bit):
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mask = 1 << i
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if state & mask:
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# 生成新状态：右移i+1位后与低位掩码组合
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new_state = (state >> (i+1)) | (state & ((1 << i) - 1))
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mex.add(cls._compute_sg(new_state, memo))
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res = 0
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while res in mex:
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res += 1
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memo[state] = res
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return res
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