LuoguP2575 高手过招（博弈论）

空格数变吗？不变呀

阶梯博弈阶梯数变吗？不变呀

那这不就阶梯博弈，每行一栋楼，爬完\(mex\)就可以了吗？

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))

#define ON_DEBUGG

#ifdef ON_DEBUGG

#define D_e_Line printf("-----------\n")
#define D_e(x) std::cout << (#x) << " : " <<x << "\n"
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#define Pause() system("pause")
#include <ctime>
#define TIME() fprintf(stderr, "\nTIME :　%.3lfms\n", clock() * 1000.0 / CLOCKS_PER_SEC)

#else

#define D_e_Line ;
#define D_e(x) ;
#define FileOpen() ;
#define FilSave ;
#define Pause() ;
#define TIME() ;

#endif

struct ios {
	template<typename ATP> ios& operator >> (ATP &x) {
		x = 0; int f = 1; char c;
		for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
		while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
		x *= f;
		return *this;
	}
}io;

using namespace std;

template<typename ATP> inline ATP Min(ATP a, ATP b) {
	return a < b ? a : b;
}
template<typename ATP> inline ATP Max(ATP a, ATP b) {
	return a > b ? a : b;
}
template<typename ATP> inline ATP Abs(ATP a) {
	return a < 0 ? -a : a;
}
const int N = 100007;
bool mex[N];
int main() {
	int Tasks;
	io >> Tasks;
	while(Tasks--){
		int n;
		io >> n;
		long long ans2 = 0;
		while(n--){
			int K;
			io >> K;
			Fill(mex, 0);
			long long ans1 = 0;
			int cnt = 20 - K + 1, tot = 0;
			while(K--){
				int x;
				io >> x;
				mex[x] = 1;
			}
			R(i,1,20){
				if(!mex[i]){
					if((--cnt) & 1) ans1 ^= tot;
					tot = 0;
				}
				else
					++tot;
			}
			ans2 ^= ans1;
		}
		if(ans2)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}