ACdream群賽1112（Alice and Bob）

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題意：http://acdream.info/problem?pid=1112

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?

解法：非常好的博弈題。關鍵是找准每一個數的狀態。每一個數的狀態是每一個數的分解後的質數的個數。然後就是獲得每一個數的狀態號能夠做到On，就是線性篩素數時候順便將每一個數的最小的質數因子篩出來，從小大大先預先處理，然後求F(n)就等於F(n/least[n])+1。獲得狀態號，sg部分就不多說了。

代碼：

/******************************************************* author:xiefubao*******************************************************/#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <vector>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <string.h>//freopen ("in.txt" , "r" , stdin);using namespace std;#define eps 1e-8#define zero(_) (abs(_)<=eps)const double pi=acos(-1.0);typedef long long LL;const int Max=5000010;const int INF=1000000007;int f[300];bool rem[Max];int out[Max];int least[Max];int p=0;void init(){    for(int i=2; i<Max; i++)        if(!rem[i])        {            for(int j=i*2; j<Max; j+=i)            {                if(least[j]==-1)                    least[j]=i;                rem[j]=1;            }            least[i]=i;        }}int getans(int t){    if(f[t]!=-1)        return f[t];    int Rem[1000];    memset(Rem,0,sizeof Rem);    Rem[0]=1;    for(int i=1; i<=t/2; i++)    {        Rem[getans(i)]=1;        Rem[getans(t-i)]=1;        Rem[getans(t-i)^getans(i)]=1;    }    int to=0;    while(Rem[to])to++;    return f[t]=to;}int F(int n){    if(out[n]!=-1)        return out[n];    int ans=F(n/least[n])+1;    return out[n]=ans;}int main(){    int n;    memset(f,-1,sizeof f);    memset(out,-1,sizeof out);    memset(least,-1,sizeof least);    f[1]=1;    init();    out[1]=0;    for(int i=0; i<20; i++)        F(1<<i);// cout<<i<<" "<<getans(i)<<endl;    while(scanf("%d",&n)==1)    {        int ans=0;        for(int i=0; i<n; i++)        {            int t;            scanf("%d",&t);            ans^=getans(F(t));        }        if(ans)            puts("Alice");        else            puts("Bob");    }    return 0;}