2703: Cow Digit Game 
Time Limit(Common/Java):1000MS/10000MS Memory Limit:65536KByte
Total Submit: 1 Accepted:1
Description
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.
Input
* Line 1: A single integer: G
* Lines 2..G+1: Line i+1 contains the single integer: N_i
Output
* Lines 1..G: Line i contains "YES" if Bessie can win game i, and "NO" otherwise.
Sample Input
2
9
10
Sample Output
YES
NO
Hint
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.
Source
简单的博弈，sg函数
把一个数减去这个数非0的最大值或者最小值，分析其必胜
#include <stdio.h>
#include <algorithm>
using namespace std;
const int N=1e6+10;
int sg[N];
int main(){
for(int i=1;i<=1e6;i++){
int ma=0,mi=9,t=i;
while(t){
int b=t%10;
if(b){
ma=max(b,ma);
mi=min(b,mi);}
t/=10;
}
sg[i]=(sg[i-ma]&sg[i-mi])^1;
}
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
printf("%s",sg[n]?"YES\n":"NO\n");
}
return 0;}