Hdu4155The Game of 31(DFS+博弈论)

The Game of 31

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 563    Accepted Submission(s): 268

Problem Description

The game of 31 was a favourite of con artists who rode the railroads in days of yore. The game is played with a deck of 24 cards: four labelled each of 1, 2, 3, 4, 5, 6. The cards in the deck are visible to both players, who alternately withdraw one card from the deck and place it on a pile. The object of the game is to be the last player to lay a card such that the sum of the cards in the pile does not exceed 31. Your task is to determine the eventual winner of a partially played game, assuming each player plays the remainder of the game using a perfect strategy.
For example, in the following game player B wins:
Player A plays 3
Player B plays 5
Player A plays 6
Player B plays 6
Player A plays 5
Player B plays 6

 

Input

The input will consist of several lines; each line consists of a sequence of zero or more digits representing a partially completed game. The first digit is player A's move; the second player B's move; and so on. You are to complete the game using a perfect strategy for both players and to determine who wins.

 

Output

For each game, print a line consisting of the input, followed by a space, followed by A or B to indicate the eventual winner of the game.

 

Sample Input

356656356653566111126666552525

 

Sample Output

356656 B35665 B3566 A111126666 A552525 A

题意：一共有24张牌，面值分别为1,2,3,4,5,6.各有4枚。每次没人打出一张，第一个点数超过31算输。

刚开始我还以为1人1套牌。郁闷死我了，WR了半天。。。后来看了解题报告才知道是一起用的。直接DFS就行了，注意回朔就行。

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <queue>#include <map>#include <stack>#include <list>#include <vector>using namespace std;#define LL __int64int a[10];int dp(int x){    if (x>31) return 0;    for (int i=1;i<7;i++)    {    if (a[i] && i+x<=31)    {    a[i]--;    if (dp(i+x)==0)    {    a[i]++;    return 1;    }    a[i]++;    }    }    return 0;}char s[50];int main(){    int l,i;    while (~scanf("%s",&s)){getchar();        int l=strlen(s);        for (i=1;i<=6;i++)            a[i]=4;        int ans=0;        for (i=0;i<l;i++)        {            int k=s[i]-'0';            ans+=k;            a[k]--;        }        ans=dp(ans);        printf("%s",s);        if (ans) printf(" %c\n",'A'+(l&1));        else printf(" %c\n",'A'+((l+1)&1));    }    return 0;}