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.
输入
* Line 1: A single integer: G
* Lines 2..G+1: Line i+1 contains the single integer: N_i
输出
* Lines 1..G: Line i contains "YES" if Bessie can win game i, and "NO" otherwise.
样例输入
2
9
10
样例输出
YES
NO
提示
OUTPUT DETAILS:
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.
题意
A和B在玩游戏，给一个数a，轮到A，可以把数变成a-最大的数，a-最小的非零数，B同理，谁把值变成0谁赢
题解
观察一下可以发现，只要知道a-最大的数的sg值和a-最小的非零数的sg值，再异或1就是答案
因为先手只可以选最大或最小，后面不管怎么拿都是定死了
代码
1 #include<bits/stdc++.h>
2 using namespace std;
3
4 int sg[1000005],n,a,t,mx,mi;
5 void m(int x)
6 {
7 mx=-1,mi=10;
8 do{
9 t=x%10;
10 if(t)mx=max(mx,t);
11 if(t)mi=min(mi,t);
12 x/=10;
13 }while(x);
14 }
15 int main()
16 {
17 sg[0]=0;
18 for(int i=1;i<=1000000;i++)m(i),sg[i]=(sg[i-mx]^1)|(sg[i-mi]^1);
19 scanf("%d",&n);
20 while(n--)
21 {
22 scanf("%d",&a);
23 printf("%s\n",sg[a]?"YES":"NO");
24 }
25 return 0;
26 }
原文地址：https://www.cnblogs.com/taozi1115402474/p/10306547.html