Alice and Bob

Time Limit: 6000/3000MS (Java/Others)Memory Limit: 256000/128000KB (Java/Others)

Submit

Problem Description

Here  is Alice and Bob again !

Alice and Bob are playing a game. There are several numbers.First, Alice choose a number n.Then he can replace n (n > 1)with one of its positive factor but not itself or he can replace n with a and b.Here a*b = n and a > 1 and b > 1.For example, Alice can replace 6 with 2 or 3 or (2, 3).But he can’t replace 6 with 6 or (1, 6). But you can replace 6 with 1. After Alice’s turn, it’s Bob’s turn.Alice and Bob take turns to do so.Who can’t do any replace lose the game.

Alice and Bob are both clever enough. Who is the winner?

Input

This problem contains multiple test cases. The first line contains one number n(1 <= n <= 100000).

The second line contains n numbers.

All the numbers are positive and less than of equal to 5000000.

Output

For each test case, if Alice can win, output “Alice”, otherwise output “Bob”.

Sample Input

22 232 2 4

Sample Output

BobAlice

Source

yehuijie

Manager

给出ｎ个数，两个轮流任选一个数ｎ进行操作，第一种操作是将该数变为他的一个因子，但是不能变为本身。

第二种是变成两个数a和b，要满足a * b = n。ａ>1, b>1.

第一次写线性筛，这个算法和普通的筛法略有不同，也更不好理解一点，就是说算法保证每个数只会被

他的最小的质因子筛去。也保证了算法的时间是线性的。

Accepted Code:

/*************************************************************************    > File Name: 1112.cpp    > Author: Stomach_ache    > Mail: sudaweitong@gmail.com    > Created Time: 2014年09月05日 星期五 11时35分09秒    > Propose:  ************************************************************************/#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              using namespace std;/*Let's fight!!!*/const int MAX_N = 5000005;int n, pnum, p[MAX_N], mindiv[MAX_N], cnt[MAX_N];bool vis[MAX_N];//线性筛，可以很方便的保存每个数最小的质因子,//和每个数不同质因子的个数void get_prime(int n) { pnum = 0; vis[1] = true; cnt[1] = 0; for (int i = 2; i <= n; i++) { if (!vis[i]) { p[pnum++] = i; mindiv[i] = i; cnt[i] = 1; } for (int j = 0; j < pnum; j++) { if (p[j] * i > n) break; vis[p[j] * i] = true; mindiv[p[j] * i] = p[j]; cnt[p[j] * i] = cnt[i] + 1; if (i % p[j] == 0) break; } }}//记忆化搜索所用数组，初始化为-1int sg[35];int dfs(int n) { if (sg[n] != -1) return sg[n]; set
              
                S; //第一种转移变为因子a for (int i = 0; i < n; i++) S.insert(dfs(i)); //第二种转移变为两个因子a * b for (int i = 1; i < n; i++) S.insert(dfs(i)^dfs(n - i)); int g = 0; while (S.find(g) != S.end()) g++; return sg[n] = g;}void get_SG() { get_prime(5000000); memset(sg, -1, sizeof(sg)); sg[0] = 0; for (int i = 1; i <= 33; i++) { sg[i] = dfs(i); }}int main(void) { get_SG(); //预处理sg值 int n; while (~scanf("%d", &n)) { int ans = 0; for (int i = 0; i < n; i++) { int x; scanf("%d", &x); ans ^= sg[cnt[x]]; } if (ans) puts("Alice"); else puts("Bob"); } return 0;}