Codeforces Round #201.C-Alice and Bob

C. Alice and Bob

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they get a set of n distinct integers. And then they take turns to make the following moves. During each move, either Alice or Bob (the player whose turn is the current) can choose two distinct integers x and y from the set, such that the set doesn't contain their absolute difference |x - y|. Then this player adds integer |x - y| to the set (so, the size of the set increases by one).

If the current player has no valid move, he (or she) loses the game. The question is who will finally win the game if both players play optimally. Remember that Alice always moves first.

Input

The first line contains an integer n (2 ≤ n ≤ 100) — the initial number of elements in the set. The second line contains n distinct space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the set.

Output

Print a single line with the winner's name. If Alice wins print "Alice", otherwise print "Bob" (without quotes).

大意：

n(n<=100)个正数,每次从中取任意两个数,他们的差值如果在集合中没有就加入集合中,问最后集合中元素个数的奇偶性?

思路：

快被自己蠢哭了。。。。竟然把题目读成了取出两个数，然后放回去。。。。。

首先，要明白一点，最终谁获胜，需要确定还缺几个数（还能够生成多少数）

首先考虑只有两个数的情况,发现两个数x,y(x>y),可以观察到连续执行操作,可以生成的数的集合是

{k∗gcd(x,y) | k=1,2,..且 k∗gcd(x,y)<=y}

然后考虑有第三个数Z的情况,因为前面的两个数,极其生成的所有数的最大公约数都是gcd(x, y),
所以加入第三个数之后这个集合的新的最大公约数是gcd(gcd(x,y),Z),然后生成集合是以这个最大公约数的倍数来生成.

……..

归纳到n个数,就是以最大的数为上界,所有n个数的最大公约数来填充的集合,元素个数为:

max{xi | 1<=i<=n}gcd{x1,x2,...xn}

代码：

#include<bits/stdc++.h>

using namespace std;

const int MAXN=130;

int a[MAXN];

int gcd(int x,int y)

{

    if(y==0) return x;

    return gcd(y,x%y);

}

int main()

{

    int n;

    cin>>n;

    int m=-1;

    for(int i=0;i<n;i++){

        cin>>a[i];

        m=max(m,a[i]);

    }

    sort(a,a+n);

    int res=gcd(a[n-1],a[n-2]);

    for(int i=n-3;i>=0;i--){

        res=gcd(max(res,a[i]),min(res,a[i]));

        if(res==1) break;

    }

    int total=m/res-n;

    if(total%2) cout<<"Alice"<<endl;

    else cout<<"Bob"<<endl;

}

巴特西

Codeforces Round #201.C-Alice and Bob

最新文章

热门文章