USACO 2009 Open Silver 3.Cow Digit Game简单博弈论

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 larges 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

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：对于一个初始的数n，每次可以减去这个数的数位中最大的那一位，或是最小的非零的那一位，如3014中最大的是4，最小非零的是1，可以得到3014-1=3013或3014-4=3010，两个人轮流操作，减到到0的人获胜。

最近迷上了这些简单的博弈论了呢~

显然，当轮到一个人时，n已经为0，则此时就是一个必败态，令必败态的sg值为0，必胜态的sg值为1，那么我们就可以从小到大递推了。

设当前数i中最大的一位是maxi，最小的非零位是mini，那么若i-maxi和i-mini都对应一个必胜态，那么i就是一个必败态；若i-maxi和i-mini中有一个是必败态，那么i就是一个必胜态。

至于找最大位和最小非零位，鉴于数据范围不是太变态，暴力拆分就可以了。

代码：

#include<stdio.h>
int sg[1000010],num[10],len;
int maxi,mini;
int main()
{
    sg[0]=0;
    int i,t,n;
    len=0;
    for(i=1;i<=1000000;i++)
    {
        t=i;
        maxi=0;
        mini=10;
        while(t)
        {
            if(t%10>maxi)
            maxi=t%10;
            if(t%10>0&&t%10<mini)
            mini=t%10;
            t/=10;
        }
        sg[i]=(sg[i-maxi]&sg[i-mini])^1;
    }
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&t);
        if(sg[t])
        printf("YES\n");
        else
        printf("NO\n");
    }
    return 0;
}