[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
", 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[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 3e4 + 7;
const int md = 10007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?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[100007], nim[100007];

void init() {
    g = s;
    rep_up0(i, n) {
        a[i] = g;
        if (a[i] == 0) {
            a[i] = g = w;
        }
        if (g % 2 == 0) {
            g /= 2;
        }
        else g = (g / 2) ^ w;
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    cin >> T;
    while (T --) {
        cin >> n >> s >> w;
        init();
        nim[0] = a[0];
        rep_up0(i, n - 1) nim[i + 1] = nim[i] ^ a[i + 1];
        sort(nim, nim + n);
        LL ans = 0;
        rep_up0(i, n) {
            if (nim[i] != 0) break;
            ans ++;
        }
        LL c = 1;
        nim[n] = -1;
        rep_up0(i, n) {
            if (nim[i] != nim[i + 1]) {
                ans += c * (c - 1) / 2;
                c = 1;
            }
            else c ++;
        }
        cout << (LL)n * (n + 1) / 2 - ans << endl;
    }
    return 0;
}