游戏

SG定理：n个有向图游戏组成的组合游戏当且仅当SG值异或和等于0时先手必输，否则先手必胜。
直接求出每个状态的SG值，最后遍历每堆石头第一次取的情况，如果能满足异或和等于0即
留给对手一个必败状态，则方案数加1。

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

#ifdef LOCAL
#define debug(x) cout << "[" __FUNCTION__ ": " #x " = " << (x) << "]\n"
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
#define hash_ 1000000009
#define Continue(x) { x; continue; }
#define Break(x) { x; break; }
const int mod = 998244353;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
#define gc p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;
inline int read(){ static char buf[1000000], *p1 = buf, *p2 = buf; register int x = false; register char ch = gc; register bool sgn = false; while (ch != '-' && (ch < '0' || ch > '9')) ch = gc; if (ch == '-') sgn = true, ch = gc; while (ch >= '0'&& ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc; return sgn ? -x : x; }
ll fpow(ll a, int b, int mod) { ll res = 1; for (; b > 0; b >>= 1) { if (b & 1) res = res * a % mod; a = a * a % mod; } return res; }
int a[N];
int sg[N];
vector<int>G[N];
int vis[N];
int main()
{
#ifdef LOCAL
    freopen("C:/input.txt", "r", stdin);
#endif
    int cnt = 0;
    for (int i = 1; i < N; i++)
    {
        cnt++;
        int k = sqrt(i);
        for (int j = 1; j <= k; j++)
        {
            if (i % j == 0)
            {
                vis[sg[i - j]] = cnt;
                if (j * j != i)
                    vis[sg[i - i / j]] = cnt;
            }
        }
        for (int j = 0; j <= N - 10; j++)
            if (vis[j] != cnt)
            {
                sg[i] = j;
                break;
            }
    }
    int n;
    cin >> n;
    int sum = 0;
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]), sum ^= sg[a[i]];
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        sum ^= sg[a[i]];
        int k = sqrt(a[i]);
        for (int j = 1; j <= k; j++)
        {
            if (a[i] % j == 0)
            {
                if ((sg[a[i] - j] ^ sum) == 0)
                    ans++;
                if (j * j != a[i] && (sg[a[i] - a[i] / j] ^ sum) == 0)
                    ans++;
            }
        }
        sum ^= sg[a[i]];
    }
    cout << ans << endl;
    return TIME;
}