Alice and Bob are playing a game with ai stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n is an even number.
The second line contains 1≤ai≤50) — the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2 8 8
Output
Bob
Input
4 3 1 4 1
Output
Alice
题意：
给偶数（n）堆石子，每一步必须取n/2堆石子中的任意多个，当场上不足n/2堆石子时，当前玩家失败，问谁是最后的获胜者。
思路：
如果有任何一堆石子已经被拿空，那么只需要直接取空n/2堆石子，便可以获胜。
所以作为后手，如果能维护石子数量最小的堆数量大于N，便可以取胜，因为在这种情况下，石子数越来越少，先手总会拿空一堆。
#include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<map> #include<set> #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #define fuck(x) cout<<#x<<" = "<<x<<endl; #define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl; #define ls (t<<1) #define rs ((t<<1)+1) using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 100086; const int maxm = 100086; const int inf = 2.1e9; const ll Inf = 999999999999999999; const int mod = 1000000007; const double eps = 1e-6; const double pi = acos(-1); int num[maxn]; int main() { // ios::sync_with_stdio(false); // freopen("in.txt","r",stdin); int n; scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d",&num[i]); } sort(num+1,num+1+n); if(num[n/2+1]!=num[1]){printf("Alice\n");} else printf("Bob\n"); return 0; }