[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。

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21
 22 using namespace std;
 23
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define mem_1(a) memset(a, -1, sizeof(a))
 26 #define lson l, m, rt << 1
 27 #define rson m + 1, r, rt << 1 | 1
 28 #define define_m int m = (l + r) >> 1
 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 33 #define all(a) (a).begin(), (a).end()
 34 #define lowbit(x) ((x) & (-(x)))
 35 #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) {}
 36 #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) {}
 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 39 #define pchr(a) putchar(a)
 40 #define pstr(a) printf("%s", a)
 41 #define sstr(a) scanf("%s", a)
 42 #define sint(a) scanf("%d", &a)
 43 #define sint2(a, b) scanf("%d%d", &a, &b)
 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 45 #define pint(a) printf("%d\n", a)
 46 #define test_print1(a) cout << "var1 = " << a << endl
 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 49 #define mp(a, b) make_pair(a, b)
 50 #define pb(a) push_back(a)
 51
 52 typedef long long LL;
 53 typedef pair<int, int> pii;
 54 typedef vector<int> vi;
 55
 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 58 const int maxn = 3e4 + 7;
 59 const int md = 10007;
 60 const int inf = 1e9 + 7;
 61 const LL inf_L = 1e18 + 7;
 62 const double pi = acos(-1.0);
 63 const double eps = 1e-6;
 64
 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 68 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 70 int make_id(int x, int y, int n) { return x * n + y; }
 71
 72 int g, s, n, w, a[100007], nim[100007];
 73
 74 void init() {
 75     g = s;
 76     rep_up0(i, n) {
 77         a[i] = g;
 78         if (a[i] == 0) {
 79             a[i] = g = w;
 80         }
 81         if (g % 2 == 0) {
 82             g /= 2;
 83         }
 84         else g = (g / 2) ^ w;
 85     }
 86 }
 87
 88 int main() {
 89     //freopen("in.txt", "r", stdin);
 90     int T;
 91     cin >> T;
 92     while (T --) {
 93         cin >> n >> s >> w;
 94         init();
 95         nim[0] = a[0];
 96         rep_up0(i, n - 1) nim[i + 1] = nim[i] ^ a[i + 1];
 97         sort(nim, nim + n);
 98         LL ans = 0;
 99         rep_up0(i, n) {
100             if (nim[i] != 0) break;
101             ans ++;
102         }
103         LL c = 1;
104         nim[n] = -1;
105         rep_up0(i, n) {
106             if (nim[i] != nim[i + 1]) {
107                 ans += c * (c - 1) / 2;
108                 c = 1;
109             }
110             else c ++;
111         }
112         cout << (LL)n * (n + 1) / 2 - ans << endl;
113     }
114     return 0;
115 }