洛谷 2953 [USACO09OPEN]牛的数字游戏Cow Digit Game

　　　　　　洛谷 2953  [USACO09OPEN]牛的数字游戏Cow Digit Game

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

输入样例：

2 
9 
10 

输出样例：

YES 
NO 

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.

其实，这一堆英文我也没看懂，but，但是，我有翻译器啊，hhh。

不过好像大家都有。

代码

#include<cstdio>
#include<iostream>
bool win[1000010];
int T,n,maxn,minn,x;
int main() {
    scanf("%d",&T);
    for(int i=1; i<=9; i++) win[i]=1;
    for(int i=10; i<=1000010; i++) {
        int s=i,t;
        maxn=-1,minn=10;
        while(s) {
            t=s%10;
            if(t>maxn&&t) maxn=t;
            if(t<minn&&t) minn=t;
            s/=10;
        }
        if(maxn!=-1) win[i]|=(!win[i-maxn]);
        if(minn!=10) win[i]|=(!win[i-minn]);
    }
    while(T--) {
        std::cin>>x;
        if(win[x]) {
            puts("YES");
            printf("
");
        } else {
            puts("NO");
            printf("
");
        }
    }
}