import sys
from functools import lru_cache

MOD = 10**9 + 7

def is_terminal(R1, R2, B1, B2):
    a = R1 + R2
    b = B1 + B2
    if a == 0 or b == 0:
        return True
    # Check if all red cards are the same
    if R1 == 0 and R2 == 0:
        return False  # impossible since a > 0
    red_attack = 2 if R1 == 0 else (1 if R2 == 0 else None)
    if red_attack is None:
        return False
    # Check blue cards
    if B1 == 0 and B2 == 0:
        return False  # impossible since b >0
    blue_attack = 2 if B1 == 0 else (1 if B2 == 0 else None)
    if blue_attack is None:
        return False
    return red_attack == blue_attack

def main():
    n, m = map(int, sys.stdin.readline().split())
    red = list(map(int, sys.stdin.readline().split()))
    blue = list(map(int, sys.stdin.readline().split()))
    
    red_ones = sum(1 for x in red if x == 1)
    red_twos = n - red_ones
    blue_ones = sum(1 for x in blue if x == 1)
    blue_twos = m - blue_ones
    
    @lru_cache(maxsize=None)
    def compute_E(R1, R2, B1, B2):
        a = R1 + R2
        b = B1 + B2
        if a == 0 or b == 0:
            return 0
        if is_terminal(R1, R2, B1, B2):
            return 0
        
        numerator_total = a * b
        sum_rest_part = 0
        p_self = 0
        
        for r in [1, 2]:
            if r == 1:
                if R1 == 0:
                    continue
                prob_r_num = R1
            else:
                if R2 == 0:
                    continue
                prob_r_num = R2
                
            for s in [1, 2]:
                if s == 1:
                    if B1 == 0:
                        continue
                    prob_s_num = B1
                else:
                    if B2 == 0:
                        continue
                    prob_s_num = B2
                
                numerator = prob_r_num * prob_s_num
                
                new_R1, new_R2, new_B1, new_B2 = R1, R2, B1, B2
                
                if r > s:
                    if s == 1:
                        new_B1 -= 1
                    else:
                        new_B2 -= 1
                elif r < s:
                    if r == 1:
                        new_R1 -= 1
                    else:
                        new_R2 -= 1
                else:  # r == s
                    pass
                
                is_new_term = is_terminal(new_R1, new_R2, new_B1, new_B2)
                
                if is_new_term:
                    contribution = 0
                else:
                    if (new_R1 == R1) and (new_R2 == R2) and (new_B1 == B1) and (new_B2 == B2):
                        p_self += numerator
                    else:
                        e_val = compute_E(new_R1, new_R2, new_B1, new_B2)
                        sum_rest_part += (numerator * e_val)
        
        denominator_total = a * b - p_self
        numerator_total = a * b + sum_rest_part
        inv_denom = pow(denominator_total, MOD-2, MOD)
        return (numerator_total * inv_denom) % MOD
    
    result = compute_E(red_ones, red_twos, blue_ones, blue_twos)
    print(result % MOD)

if __name__ == '__main__':
    main()

采用QwQ-32B Q6_K GGUF用llama.cpp AI生成。

为了简化prompt，我对prompt做了些修改，删除了测试样例以及对分数模运算的解释。

完整的prompt：以下是一道算法题，请你用尽可能低的时间和空间复杂度，尽可能短的代码，在符合题目要求的情况下用Python3完成这道题，并给出解题思路，\n代表换行：“小红这天又和小紫发起了一场对战。对战的规则如下：两人各有一些怪兽卡。每回合两人随机的从自己当前存活的怪兽卡中抽取一张发起决斗，战斗力低的怪兽卡死亡。如果两张怪兽卡战斗力相同，则无事发生。游戏会进行到‘如果抽卡，则 100% 的概率无事发生’或者‘有一方卡牌被用完’时结束。请你计算小红和小紫游戏进行回合数的期望。\n输入描述：\n第一行输入两个正整数n,m，分别代表小红和小紫的怪兽卡数量。\n第二行输入n个正整数，代表小红的每张怪兽卡战斗力。\n第三行输入m个正整数，代表小紫的每张怪兽卡战斗力。\n1≤n,m≤50\n每张怪兽卡的战斗力只会是1或者2。\n输出描述：\n一个整数，代表最终回合数期望对10^9+7取模的值。可以证明，最终的答案一定是个有理数，你只需要输出其对10^9+7取模的结果。”