ACdream 1112 Alice and Bob （SG函数+线性素数筛）
ACdream 1112 Alice and Bob （SG函数+线性素数筛）：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=113567#problem/B 传送门：nefu
题面：
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
2 2 2 3 2 2 4
Sample Output
Bob Alice
题目大意：
已知有n个数，我们只能用不等于n的一个因子或者满足a*b==n的两个数，去取代替n，最后一个进行操作的人获胜。
算法分析：
Alice和Bob两人又干上了，他们不但喜欢在取石子时攀比，而且还喜欢在取数时非得挣个你死我活，但是对我们来说，取数和取石子都差不多，我们只需要把所有情况的SG函数值异或起来即可。
但是道理我们还是要清楚的，首先，考虑一个数data，对于data的替换可能如下：
(a1, data/a1), (a2, data/a2), (a3, data/a3),...,(an,data/an) ;或者(a1), (a2), (a3), ..., (an).
由于数据量比较大，我们这样每一个数都去考虑是不能行得通的，鉴于我们考虑到data的因子，就要联想到如何用data的度因子去解决问题，由于对data进行素因子分解，得到：data=a1^r1*a2^r2*a3^r3*...*an^rn，且sum=r1+r2+r3+...+rn.
则所有的情况可以表示为：
(1,sum-1), (2,sum-2), (3,sum-3), ..., (sum/2,sum-sum/2)或者(1), (2), (3), ..., (n-1).
即可减小数据的范围，计算sum时我们可以这样计算： 设data的最小因子为d，则sum[data]=sum[data/d]+1（线性素数筛）;
代码实现：
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAXN=5000010;
int prime[MAXN];
bool isprime[MAXN];
int facnum[MAXN];
int facmin[MAXN];
int sg[100];
void getPrime()
{
int i,j;
int cnt=0;
memset(isprime,0,sizeof(isprime));
memset(facnum,-1,sizeof(facnum));
memset(facmin,-1,sizeof(facmin));
for(i=2;i<=5000000;i++)
{
if(!isprime[i])
{
prime[cnt++]=i;
for(j=i+i;j<=5000000;j+=i)
{
isprime[j]=1;
if(facmin[j]==-1)
{
facmin[j]=i;
}
}
facmin[i]=i;
}
}
}
int getnum(int x)
{
if(x==1)
return 0;
if(facnum[x]!=-1)
return facnum[x];
return facnum[x]=getnum(x/facmin[x])+1;
}
int getsg(int x)
{
if(sg[x]!=-1)
return sg[x];
bool vis[100];
memset(vis,0,sizeof(vis));
for(int i=1;i<=x;i++)
{
vis[getsg(x-i)]=1;
}
for(int i=1;i<=x/2;i++)
{
vis[getsg(i)^getsg(x-i)]=1;
}
for(int i=0;;i++)
{
if(!vis[i])
{
return sg[x]=i;
}
}
}
int main()
{
int n,data;
getPrime();
memset(sg,-1,sizeof(sg));
sg[0]=0;
int ans;
while(scanf("%d",&n)!=EOF)
{
ans=0;
for(int i=0;i<n;i++)
{
scanf("%d",&data);
ans=ans^getsg(getnum(data));
}
if(ans)
cout<<"Alice"<<endl;
else cout<<"Bob"<<endl;
}
return 0;
}