P2953 [USACO09OPEN]牛的数字游戏Cow Digit Game

Posted 2020-10-10 心之所向 素履以往

                P2953 [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#1：

2 
9 
10 

输出样例#1：

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.

 

博弈论 用到了sg函数和sg定理 然而我一开始并没有看出来 

乱搞拿了40 ，这数据是要多水啊

组合博弈：三大游戏

博弈论：sg函数

 

 1 #include <cstdio>
 2 #include <cctype>
 3 
 4 const int MAXN=1000010;
 5 
 6 int n,t;
 7 
 8 int sg[MAXN];
 9 
10 inline void read(int&x) {
11     int f=1;register char c=getchar();
12     for(x=0;!isdigit(c);c==‘-‘&&(f=-1),c=getchar());
13     for(;isdigit(c);x=x*10+c-48,c=getchar());
14     x=x*f;
15 }
16 
17 int hh() {
18     read(n);
19     for(int mn,mx,t,i=1;i<MAXN-1;++i) {
20         mn=9;mx=0;t=i;
21         while(t) {
22             int b=t%10;
23             if(b) {mx=mx<b?b:mx;mn=mn>b?b:mn;}
24             t/=10;
25         }
26         sg[i]=(sg[i-mx]&sg[i-mn])^1;
27     }
28     for(int t,i=1;i<=n;++i) {
29         read(t);
30         sg[t]?printf("YES\n"):printf("NO\n");
31     }
32     return 0;
33 }
34 
35 int sb=hh();
36 int main(int argc,char**argv) {;} 

代码