Cow Digit Game

当一个状态的后继状态有一个必败状态，我们可以通过移动到必败状态使得对手败。
当一个状态的后继状态全部为必胜状态，无论怎么移动对手都会取胜。
所以设d[x]表示先手拿到数字x的取胜情况，直接判断后继点的取胜情况即可。

#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 = 1e6 + 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 d[N];
void dfs(int x)
{
    if (d[x] != -1)
        return;
    if (x <= 9)
        return void(d[x] = 1);
    int mx = 0, mi = 11;
    int num = x;
    while (num)
    {
        int k = num % 10;
        if (k)
            mx = max(mx, k), mi = min(mi, k);
        num /= 10;
    }
    dfs(x - mi);
    dfs(x - mx);
    if (d[x - mi] && d[x - mx])
        d[x] = 0;
    else
        d[x] = 1;
}
int main()
{
#ifdef LOCAL
    freopen("E:/input.txt", "r", stdin);
#endif
    memset(d, -1, sizeof d);
    for (int i = 1; i <= 1000000; i++)
        dfs(i);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        scanf("%d", &n);
        printf("%s\n", d[n] ? "YES" : "NO");
    }
    return TIME;
}