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 #1
2 9 10
输出 #1
YES NO
说明/提示
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.
思路
每种状态可以转移出两个新状态。博弈论的结论：
当某种状态的后继状态都是必败状态时，该状态为必胜状态；
当某种状态的后继状态有一种是必胜状态时，该状态为必败状态。
所以先将0标记为必败，1-9标记为必胜（显然），从10-1000000遍历一遍，用f数组存必败或必胜，然后输出就可以了。在求f[i]的时候只会用到f[i-max/min],而max/min必然大于0，所以顺序遍历即可。
代码：
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1000010;
int q,n;
int f[N];
int minn(int a) {
int num;
num=20;
while(a) {
if(a%10)
num=min(num,a%10);
a/=10;
}
return num;
}
int maxx(int a) {
int num;
num=-10;
while(a) {
if(a%10)
num=max(num,a%10);
a/=10;
}
return num;
}
int main () {
memset(f,false,sizeof(f));
for(int i=1; i<=9; i++)
f[i]=true;
for(int i=11; i<=N; i++)
if(f[i-minn(i)]&&f[i-maxx(i)])
f[i]=false;
else
f[i]=true;
scanf("%d",&n);
for(int i=1; i<=n; i++) {
scanf("%d",&q);
if(f[q]==false)
printf("NO\n");
else
printf("YES\n");
}
return 0;
}
原文地址：https://www.cnblogs.com/mysh/p/11333427.html