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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 }
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原文地址:http://www.cnblogs.com/jklongint/p/4480797.html