2020 cattle off winter training camp algorithm basis 5.E - Enjoy the game [Game]

Topic Portal

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Title Description

Taurus team three players in the training apart from his own mouth Hu will be some questions as usual puzzle game. One day cattle Coke came up with a game to play with two other players, rules of the game are as follows:

• A total initial nn card card
• Just get a minimum first step to bring a card, bring it up to n-1 cards.
• Then each step, the two sides at least bring a brand, get the most equal to the number of cards on the card to get the other side of the step.
• Who took the last card will win the game.

You, as a spectator, watching them playing very happy, want to participate in this game, who will take a risk to win.

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Enter a description:

The input data includes an integer $n (2 \ leq n \ leq 10 ^ {18})$, indicates the number of a card's initial card.

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Output Description:

If you have the upper hand winning strategy, output Bob, otherwise the output Alice.

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Entry

2

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Export

Alice

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Explanation

The upper hand must take a card, then he took the flip another card, the game ends.

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answer

• Firstly Truth: but those who $2^n$ is the flip win
• Look at pictures, we consider only just get yourself can win
• $n = 1: win$
• $n = 2: defeat$ , only their own to take a, take a partner, then GG
• $n = 3: wins$ , get yourself a, make each other reach $n=2$的必败点
• $n=4：败$，如图红色箭头所指两种情况，都是必败状态
• $n=5：胜$，拿 $1$ 个，令对方到达 $n=4$的必败点
• $n=6：胜$，拿 $2$ 个，令对方到达 $n=4$的必败点
• $n=7：胜$，拿 $3$ 个，令对方到达 $n=4$的必败点
• $n=8：败$，无论怎么拿，都会达到对方的必胜点。
• 由此我们可以发现，若当前必败点为 $x$，距离下一个必败点的距离，就是 $2x$ 位置
• 也就是必败点为： $2，4，8，16，32，...，2^n$

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AC-Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    ll n;    while(cin >> n){
        cout << ((n&(-n)) == n ? "Alice" : "Bob") << endl;
    }
    return 0;
}