題干：

Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. The game ends when one of the players is unable to remove object in his/her turn. This player will then lose. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap. Here is another version of Nim game. There are?N?piles of stones on the table. Alice first chooses some CONSECUTIVE piles of stones to play the Nim game with Tom. Also, Alice will make the first move. Alice wants to know how many ways of choosing can make her win the game if both players play optimally.

You are given a sequence a[0],a[1], ... a[N-1] of positive integers to indicate the number of stones in each pile. The sequence a[0]...a[N-1] of length N is generated by the following code:

int g = S;?

for (int i=0; i<N; i++) {?

? ? a[i] = g;

? ? if( a[i] == 0 ) { a[i] = g = W; }

? ? if( g%2 == 0 ) { g = (g/2); }

? ? else ? ? ? ? ? { g = (g/2) ^ W; }

}

Input

There are multiple test cases. The first line of input is an integer?T(T?≤ 100) indicates the number of test cases. Then?T?test cases follow. Each test case is represented by a line containing 3 integers?N,?S?and?W, separated by spaces. (0 <?N≤ 105, 0 <?S, W?≤ 109)

Output

For each test case, output the number of ways to win the game.

Sample Input

2 3 1 1 3 2 1

Sample Output

4 5

題目大意：

? ?因為我們知道Nim博弈，這x堆石子的異或和為0則必敗，不為零則必勝，所以就是問你在這些石子中選擇哪些連續的堆，可以必勝，問你選擇方案數。

? 也就是，給n個數，讓你求，有多少段連續區間，使得這段區間的異或和不為0。

解題報告：

正難則反，可以先求出有多少個區間，異或和為零，然后用總區間數去作差就行了。?

做法就是個常見的在線處理。。。這題也可以先求出這個和來，然后用C(n,2)去減這個和，得到的就是答案。

AC代碼：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<map> using namespace std; typedef long long ll; int n; ll a[100010]; ll sum[100010]; map<ll,int> mp; int main() {int t,q,cnt;ll i,j,k,g,S,W,res,tmp,ss;cin>>t;while(t--) {cin>>n>>S>>W;g = S;res=0;for (i=1; i<=n; i++) {a[i] = g;if( a[i] == 0)a[i] = g = W;if( g%2 == 0 )g = (g/2);else g = (g/2) ^ W;}mp.clear();mp[0]=1;for(i=1; i<=n; i++) {sum[i]=sum[i-1]^a[i];res+=i-mp[sum[i]];mp[sum[i]]++;}cout<<res<<endl;}return 0; }

?

總結

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