题目描述
Bessie is playing a number game against Farmer John, and she wants you to help her achieve victory.
Game i starts with an integer N_i (1 <= N_i <= 1,000,000). Bessie goes first, and then the two players alternate turns. On each turn, a player can subtract either the largest digit or the smallest non-zero digit from the current number to obtain a new number. For example, from 3014 we may subtract either 1 or 4 to obtain either 3013 or 3010, respectively. The game continues until the number becomes 0, at which point the last player to have taken a turn is the winner.
Bessie and FJ play G (1 <= G <= 100) games. Determine, for each game, whether Bessie or FJ will win, assuming that both play perfectly (that is, on each turn, if the current player has a move that will guarantee his or her win, he or she will take it).
Consider a sample game where N_i = 13. Bessie goes first and takes 3, leaving 10. FJ is forced to take 1, leaving 9. Bessie takes the remainder and wins the game.
贝茜和约翰在玩一个数字游戏．贝茜需要你帮助她．
游戏一共进行了G(1≤G≤100)场．第i场游戏开始于一个正整数Ni(l≤Ni≤1,000,000)．游
戏规则是这样的：双方轮流操作，将当前的数字减去一个数，这个数可以是当前数字的最大数码，也可以是最小的非0数码．比如当前的数是3014，操作者可以减去1变成3013，也可以减去4变成3010．若干次操作之后，这个数字会变成0．这时候不能再操作的一方为输家． 贝茜总是先开始操作．如果贝茜和约翰都足够聪明，执行最好的策略．请你计算最后的赢家．
比如，一场游戏开始于13.贝茜将13减去3变成10．约翰只能将10减去1变成9．贝茜再将9减去9变成0．最后贝茜赢．
输入输出格式
输入格式：* Line 1: A single integer: G
* Lines 2..G+1: Line i+1 contains the single integer: N_i
输出格式：* Lines 1..G: Line i contains 'YES' if Bessie can win game i, and 'NO' otherwise.
输入输出样例
说明
For the first game, Bessie simply takes the number 9 and wins. For the second game, Bessie must take 1 (since she cannot take 0), and then FJ can win by taking 9.
考虑用 sg 函数，
那么 sg[ i ]=mex( sg[ i-min ],sg[ i-max ])；
然后 O(1) 询问即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
int n;
int sg[maxn];
int vis[10];
int mex() {
for (int i = 0;; i++) {
if (!vis[i])return i;
}
}
void init(int Max) {
sg[0] = 0;
for (int i = 1; i <= Max; i++) {
int tmp = i; ms(vis);
int minn = 10, maxx = -1;
while (tmp) {
if(tmp%10!=0)
minn = min(minn, tmp % 10), maxx = max(maxx, tmp % 10);
tmp /= 10;
}
if (maxx != -1)vis[sg[i - maxx]] = 1;
if (minn != 10)vis[sg[i - minn]] = 1;
sg[i] = mex();
}
}
int main()
{
//ios::sync_with_stdio(0);
rdint(n); init(maxn);
while (n--) {
int x; rdint(x);
if (sg[x] == 0)cout << "NO" << endl;
else cout << "YES" << endl;
}
return 0;
}