关键词：
题意：在一个二维平面中，有n个灯亮着并告诉你坐标，每回合需要找到一个矩形，这个矩形xy坐标最大的那个角落的点必须是亮着的灯，然后我们把四个角落的灯状态反转，不能操作为败
思路：二维Nim积，看不懂啊，只能套模板了
参考：HDU 3404 Switch lights (NIM 积)
代码：
#include<set> #include<map> #include<stack> #include<cmath> #include<queue> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> typedef long long ll; const int maxn = 1e6 + 10; const int seed = 131; const ll MOD = 1e9 + 7; const int INF = 0x3f3f3f3f; using namespace std; int m[2][2] = 0, 0, 0, 1; int Nim_Mul_Power(int x, int y) if(x < 2) return m[x][y]; int a = 0; for(; ; a++) if(x >= (1 << (1 << a)) && x < (1 << (1 << (a + 1)))) break; int m = 1 << (1 << a); int p = x / m, s = y / m, t = y % m; int d1 = Nim_Mul_Power(p, s); int d2 = Nim_Mul_Power(p, t); return (m * (d1 ^ d2)) ^ Nim_Mul_Power(m / 2, d1); int Nim_Mul(int x, int y) if(x < y) return Nim_Mul(y, x); if(x < 2) return m[x][y]; int a = 0; for(; ; a++) if(x >= (1 << (1 << a)) && x < (1 << (1 << (a + 1)))) break; int m = 1 << (1 << a); int p = x / m, q = x % m, s = y / m, t = y % m; int c1 = Nim_Mul(p, s), c2 = Nim_Mul(p, t) ^ Nim_Mul(q, s), c3 = Nim_Mul(q, t); return (m * (c1 ^ c2)) ^ c3 ^ Nim_Mul_Power(m / 2, c1); int main() int T; scanf("%d", &T); int ans; while(T--) ans = 0; int n, x, y; scanf("%d", &n); while(n--) scanf("%d%d", &x, &y); ans ^= Nim_Mul(x, y); if(ans) printf("Have a try, lxhgww. "); else printf("Don‘t waste your time. "); return 0;
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