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最新推荐文章于 2025-04-23 09:25:00 发布

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最新推荐文章于 2025-04-23 09:25:00 发布

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原文链接：http://www.cnblogs.com/smallocean/p/9041227.html

题目描述

Alice and Bob are playing a stone game. There are n piles of stones. In each turn, a player can remove some stones from a pile (the number must be positive and not greater than the number of remaining stones in the pile). One player wins if he or she remove the last stone and all piles are empty. Alice plays first.
To make this game even more interesting, they add a new rule: Bob can choose some piles and remove entire of them before the game starts. The number of removed piles is a nonnegative integer, and not greater than a given number d. Note d can be greater than n, and in that case you can remove all of the piles.
Let ans denote the different ways of removing piles such that Bob are able to win the game if both of the players play optimally. Bob wants you to calculate the remainder of ans divided by 10^9+7..

输入

The first line contains an integer T, representing the number of test cases.
For each test cases, the first line are two integers n and d, which are described above.
The second line are n positive integers ai, representing the number of stones in each pile.
T ≤ 5, n ≤ 10^3, d ≤ 10, ai ≤ 10^3

输出

For each test case, output one integer (modulo 10^9 + 7) in a single line, representing the number of different ways of removing piles that Bob can ensure his victory.

样例输入

2
5 2
1 1 2 3 4
6 3
1 2 4 7 1 2

样例输出

2
5
尼姆博弈：定理：（a1,a2,...,aN）为奇异局势当且仅当a1^a2^...^aN=0    
比赛的时候只知道是博弈，让剩下的异或和为0
这个主要还是DP
dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k^a[i]]; 表示前i个，取j，异或为k。 则可由第i个不取，异或为k，第i个取，则 设x^a[i]=k ,x=k^a[i]。
取哪一个数就再异或就好了
暴力转移就好

 1 #include <iostream>
 2 #include <bits/stdc++.h>
 3 #define maxn 1005
 4 using namespace std;
 5 typedef long long ll;
 6 const ll mod=1e9+7;
 7 int dp[maxn][12][2*maxn]={0};//前i个选j个，异或为k。
 8 //dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k^a[i]];
 9 int main()
10 {
11     ll n,t,d,i,j,k;
12     scanf("%lld",&t);
13     ll a[maxn]={0};
14     while(t--)
15     {
16         scanf("%lld%lld",&n,&d);
17         memset(dp,0,sizeof(dp));
18         ll maxim=-1;
19         for(i=1;i<=n;i++)
20         {
21             scanf("%d",&a[i]);
22             maxim=max(maxim,a[i]);
23         }
24         ll sum=a[1];
25         for(i=2;i<=n;i++)
26         {
27             sum=sum^a[i];
28         }
29         for(i=1;i<=n;i++)
30         {
31             dp[i][0][0]=1;
32         }
33        for(i=1;i<=n;i++)
34        {
35            for(j=1;j<=d&&j<=i;j++)
36            {
37                for(k=0;k<=2*maxim;k++)
38                {    if(i==1) dp[i][j][a[i]]=1;
39                    else dp[i][j][k]=(dp[i-1][j][k]+dp[i-1][j-1][k^a[i]])%mod;
40                }
41            }
42        }
43        ll ans=0;
44        for(i=0;i<=d;i++)
45        {
46            ans=(ans+dp[n][i][sum])%mod;
47        }
48        printf("%lld\n",ans);
49     }
50     return 0;
51 }