gambling monster

We generally only use two words to describe this kind of person: The gambling monster!.We generally only use two words to describe this kind of person: The gambling monster! 我们一般只会用两个字来形容这种人:赌怪! I often say a words, in those years Chen Daozai can win 30 million with 20 pieces! I Lu Benwei won 200 million with 200,000 is not a problem. 我经常说一句话,当年陈刀仔他能用20块赢到3700万,我卢本伟用20万赢到500万不是难题。 Ambushing her one hand, this card can no

gambling monster!.转个B站英文版【英语课表演卢本伟“17张牌你能秒我”-哔哩哔哩】 You may have no idea only use twenty hundred thousand to win five point seven eight million, is what kind of man. We often only use two words to describe this kind of person, gambling monster!!! I often say that once Chen Daozai can only use 200 bucks(口误,应该是20) to win 37 million, I Lubenwei, winning 5 million with 20 bucks is not a problem. These cards don’t need to grab. She’s already dead. Come back

You may not know what it's like to win 5.78 million for 200000In general, we only use two words to describe this kind of person: Gambling monster!I often say that when Chen daozai won 37 million yuan with 20 yuan, it's not a problem for Lu Benwei to win 5 million yuan with 200000 yuan.Ambush him. This card can't be picked up. This card can't be picked up. He's dead.Backhand to a super double, dull sound make a fortune. He's super double, too? But don't be afraid. He can't win me. Five, six, seven, eight, 90, two blasts. It's a very powerful card. If I change this K into J, my card will be extinct, b

// Problem: Gambling Monster // Contest: NowCoder // URL: // Memory Limit: 524288 MB // Time Limit: 2000 ms // // Powered by CP Editor ( //pragma GCC optimize( Ofast,no-stack-protector,unroll-loops,fast-math ) //pragma GCC target( sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native ) //pragma GCC optimize(2) include cstdio include iostream include string include cstring include map include cmath include cctype include vector include set include queue include algorithm include sstream include ctime include cstdlib include random include cassert

最后就是如何确定公式中 的取值范围呢,如果每次都是取 0 ∼ n − 1 0\sim n-1 0∼n−1 的话肯定不行,通过检视不难发现,每次拆出来的左右区间形如: [ x x x 0000 ] ∼ [ x x x 0111 ] + [ x x x 1000 ] ∼ [ x x x 1111 ] [xxx0000]\sim[xxx0111]+[xxx1000]\sim[xxx1111] [xxx0000]∼[xxx0111]+[xxx1000]∼[xxx1111] 我们上述式子中的 需要在左区间中取, 需要在右区间中取,而 只需要满足 y = x ⊕ z y=x\oplus z y=x⊕z 即可,将上面的两段区间进行异或可以得到 y ∈ [ 1000 , 1111 ] y\in[1000,1111] y∈[1000,1111] ,这个每次通过位运算求解即可,不难发现 的区间刚好是等阶的 // Problem: Gambling Monster //

D. Gambling Monster 题意 有一个转盘,每次转动得到 \(0\sim n-1\)(\(n\)为 \(2\)的次幂)的概率分别给出。最开始你有一个数 \(x\),每次转动转盘得到一个数 \(y\),如果 \(x\oplus y x\) 就令 \(x=x\oplus y\),否则 \(x\)不变。求使 \(x=n-1\)期望转动转盘的次数。 Solution 从后往前 dp,列出式子: \[f_i=\sum_{j i,j\oplus k=i}f_j\times p_k+\sum_{j\oplus i\le i}f_i\times p_j\] 设 \(s_i=\sum_{j\oplus i\le i} p_j\),改写一下: \[f_i=\left(\sum_{j i,j\oplus k=i}f_j\times p_k+1\right)\times \frac{1}{1-s_i}\] 计算 \(s_i\)比较简单,我们可以枚举 \(j\)的最高位,用前缀和计算一下即可。 计算 \(\sum_{j i,j\oplus k=i}f_j\times p_k\),发现形式

视频里也是gambling monster。.bet是打赌。赌博一般用gamble。视频里也是gambling monster。 bet是打赌。赌博一般用gamble。视频里也是gambling monster。 闪光阿杜M2019-12-04 07:36:55发布于辽宁点灭只看此人举报 引用 @发表的:桃曼巴克里斯炮只看此人 silly B silly B 我听着像是SAB 直接念出来 我听着像是SAB 直接念出来 亮了(0) 回复 4AM战神位2019-12-04 08:25:30发布于江苏点灭只看此人举报 引用 @发表的:只看此人 老哥,

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