hdu4111 Alice and Bob 博弈论

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题意：有N堆石子，每堆石子有一个数目，现有两个人博弈，每个人每次可以进行两个操作中的一个： 1、从某堆拿掉一个石子（若某堆石子为0了，那么这堆就不存在了）；2、合并两堆石子没有操作的就输。问是哪个赢

统计1的个数c，以及非1情况下的步数s，包括合并。
c为奇数，s不等于2：那么先手必胜。
s为2或者为0：c为3的倍数是先手必败。
否则的话，s为奇数时先手必胜。
首先没有1的情况下很好证明，就是总的步数和为奇数时为先手必胜态，N点。相反，如果为偶数就是P点。（这种情况
下必胜的人只要保证任意一堆石子的个数不会小于2即可，直到最后只剩下一堆，因为1是一个比较特殊的存在）
再者证明奇数个1且s不为2的时候为必胜态，从一个1的开始，假设现在有一个1和s，如果s为奇数，那么先手可以合并
两堆石子；如果s是偶数，先手可以拿走1；所以这种情况下一定是先手必胜，因为移动1可以选择少掉一步或者两步。
那么现在就是两个1和s的情况，如果其中一个人先消耗掉一个1，和s合并，那么剩下的状态是(1,s+1)的N点；取走1，
剩下(1,s)同样是N点；合并两个1，(2,s)==>(s+3)的状态，如果s为奇数，那么即为P态，对应的刚才先手就是必胜，反之
则是必败，这种情况归纳到结论中的第三条。那么三个1的情况下就可以消耗一个1更变s的奇偶性。
不过在s=2或者s=0的时候属于特殊情况，因为2去掉一个之后就变成了1，s为0时为全1，通过上面的规律，我们已经知

实在是没想到，大家都说是简单的题目我做了一下午也没A掉他看别人思路看了很长时间。。。。。。真的很巧妙

Description

Alice and Bob are very smart guys and they like to play all kinds of games in their spare time. The most amazing thing is that they always find the best strategy, and that's why they feel bored again and again. They just invented a new game, as they usually did.
The rule of the new game is quite simple. At the beginning of the game, they write down N random positive integers, then they take turns (Alice first) to either:
1. Decrease a number by one.
2. Erase any two numbers and write down their sum.
Whenever a number is decreased to 0, it will be erased automatically. The game ends when all numbers are finally erased, and the one who cannot play in his(her) turn loses the game.
Here's the problem: Who will win the game if both use the best strategy? Find it out quickly, before they get bored of the game again!

Input

The first line contains an integer T(1 <= T <= 4000), indicating the number of test cases.
Each test case contains several lines.
The first line contains an integer N(1 <= N <= 50).
The next line contains N positive integers A ₁ ....A _N(1 <= A _i <= 1000), represents the numbers they write down at the beginning of the game.

Output

For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y is either "Alice" or "Bob".

Sample Input

 331 1 223 432 3 5

Sample Output

 Case #1: AliceCase #2: BobCase #3: Bob

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;int dp[55][55000];int DP(int a,int b){    if(dp[a][b]!=-1) return dp[a][b];    if(b==1) return dp[a][b]=DP(a+1,0);//只剩一个单独的一，加入其它1中    dp[a][b]=0;    if(a>=1&&!DP(a-1,b))//直接去掉一个1        dp[a][b]=1;    if(a>=1&&b&&!DP(a-1,b+1))//将一个1合到其它数中        dp[a][b]=1;    if(a>=2&&((b&&!DP(a-2,b+3))||(b==0&&!DP(a-2,2))))//将2个1并起来        dp[a][b]=1;    if(b>=2&&!DP(a,b-1))//减小一        dp[a][b]=1;    return dp[a][b];}int main(){    int T;    scanf("%d",&T);memset(dp,-1,sizeof(dp));    for(int t=1; t<=T; t++)    {        int n;        scanf("%d",&n);int j,k,i;        for( j=0,k=0,i=0; i<n; i++)        {            int a;            scanf("%d",&a);            if(a==1)                j++;            else                k+=+a+1;        }        if(k);        k--;        DP(j,k);         printf("Case #%d: ",t);        if(dp[j][k]) printf("Alice\n");        else printf("Bob\n");    }    return 0;}

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