鏈接：https://ac.nowcoder.com/acm/contest/1071/G
來源：牛客網

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld

題目描述

Bessie is playing a number game against Farmer John, and she wants you to help her achieve victory.
Game i starts with an integer Ni (1 <= Ni <= 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 Ni = 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.

輸入描述:

• Line 1: A single integer: G
• Lines 2…G+1: Line i+1 contains the single integer: Ni

輸出描述:

• Lines 1…G: Line i contains ‘YES’ if Bessie can win game i, and ‘NO’ otherwise.

示例1

輸入
復制

2 9 10

輸出
復制

YES NO

說明

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.

思路

博弈論
本題最后取完者勝，
所以終態n=0為先取者的必敗點，
可以用sg函數來求解。

sg函數說明：

首先定義mex(minimal excludant)運算，這是施加于一個集合的運算，表示最小的不屬于這個集合的非負整數。例如mex{0,1,2,4}=3、mex{2,3,5}=0、mex{}=0。
對于任意狀態 x ， 定義 SG(x) = mex(S),其中 S 是 x 后繼狀態的SG函數值的集合。如 x 有三個后繼狀態分別為 SG(a),SG(b),SG( c)，那么SG(x) = mex{SG(a),SG(b),SG( c)}。 這樣 集合S 的終態必然是空集，所以SG函數的終態為 SG(x) = 0,當且僅當 x 為必敗點P時。
---------------from sg函數詳解傳送門

#include <iostream> #include <string.h> using namespace std; const int N = 1e6; int f[5],sg[N+2]; int getSg(int n) {if(sg[n]!= -1){return sg[n];}int minx = 10,maxx = -1;int x = n;while(x){int t = x%10;if(t){minx = min(minx,t);maxx = max(maxx,t);}x /= 10;}f[0] = getSg(n-minx);f[1] = getSg(n-maxx);for(int i = 0; ; i++){if(i==f[0]||i==f[1])continue;else{sg[n] = i;break;}}return sg[n]; } int main() {memset(sg,-1,sizeof(sg));sg[0] = 0;for(int i = 1; i <= N; i++){getSg(i);}int n;cin>>n;while(n--){int x;cin>>x;cout<<(sg[x]?"YES":"NO")<<endl;}return 0; }

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