[zoj3591]Nim 游戏

题意：有n堆火柴，选择连续若干堆火柴进行Nim游戏，求让先手胜的选择方案数。

思路：让先手胜等同于这些数的异或值不同于0，不妨转化为求让先手败的方案数。此时记录一个前缀的异或和val[i]，那么答案就是count({i,j})(0<=i<j<n，val[i]=val[j])+count(i)(val[i]=0)。直接map统计可能超时，不妨考虑离线做，把val数组sort一下答案就不难得到了，不要忘记最后用总方案数减一下。

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #define constructInt5(name, a, b, c, d, e) name(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0): a(a), b(b), c(c), d(d), e(e) {}
 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 #define mp(a, b) make_pair(a, b)
 #define pb(a) push_back(a)

 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 3e4 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 int g, s, n, w, a[], nim[];

 void init() {
     g = s;
     rep_up0(i, n) {
         a[i] = g;
         if (a[i] == ) {
             a[i] = g = w;
         }
         if (g %  == ) {
             g /= ;
         }
         else g = (g / ) ^ w;
     }
 }

 int main() {
     //freopen("in.txt", "r", stdin);
     int T;
     cin >> T;
     while (T --) {
         cin >> n >> s >> w;
         init();
         nim[] = a[];
         rep_up0(i, n - ) nim[i + ] = nim[i] ^ a[i + ];
         sort(nim, nim + n);
         LL ans = ;
         rep_up0(i, n) {
             if (nim[i] != ) break;
             ans ++;
         }
         LL c = ;
         nim[n] = -;
         rep_up0(i, n) {
             if (nim[i] != nim[i + ]) {
                 ans += c * (c - ) / ;
                 c = ;
             }
             else c ++;
         }
         cout << (LL)n * (n + ) /  - ans << endl;
     }
     return ;
 }