对信竞选手感觉不难。

换我抄答案都得抄好久哈，别说看明白了。
邓真是相当的牛x了...

def find_n():
    n = 1
    while True:
        R = [0.0] * n
        B = [1.0] * n
        sum_R = sum(R)
        
        while True:
            updated = False
            for i in range(n):
                for j in range(n):
                    if R[i] < B[j]:
                        old_R = R[i]
                        old_B = B[j]
                        temp = (old_R + old_B) / 2
                        R[i] = temp
                        B[j] = temp
                        sum_R += (temp - old_R)
                        updated = True
            if not updated:
                break
        
        if sum_R > 100:
            return n
        n += 1


【 在 tsuld 的大作中提到: 】
: 操作策略不是已经保证了最后红牌数字之和是随n递增的吗？这样就算从100-200之间用二分法验证也很快就能抓到106了。我觉得难点在于按照操作策略快速得到递推公式可能是这个题目的难点所在。但这对国赛级别的选手应该算不上难吧？
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