Wannafly挑战赛23B游戏

https://www.nowcoder.com/acm/contest/161/B

题意：两个人van游戏，n堆石子，每次只能取这堆石子数目的因子个数，没得取的人输，问第一个人的必胜策略有多少种

题解：sg函数，通过可取的石子数来打sg函数，然后枚举每一堆第一步取的石子数，sg函数异或起来看是不是0，可以发现复杂度就是n*因子个数

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=200000+10,inf=0x3f3f3f3f;
vi v[N];
int sg[N],Hash[N],a[N];
void getsg(int n)
{
    for(int i=1;i<N;i++)
        for(int j=i;j<N;j+=i)
            v[j].pb(i);
    memset(sg,0,sizeof sg);
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<v[i].size();j++)
            Hash[sg[i-v[i][j]]]=1;
        for(int j=0;j<=n;j++)
        if(!Hash[j])
        {
            sg[i]=j;
            break;
        }
        for(int j=0;j<v[i].size();j++)
            Hash[sg[i-v[i][j]]]=0;
    }
}
int main()
{
    getsg(100000);
    int n,sum=0,ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]),sum^=sg[a[i]];
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<v[a[i]].size();j++)
        {
            int x=v[a[i]][j];
            if((sg[a[i]-x]^sum^sg[a[i]])==0)ans++;
        }
    }
    printf("%d\n",ans);
    return 0;
}
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