【GDOI2020模拟01.16】划愤（nim积+行列式）

https://gmoj.net/senior/#contest/show/2989/1

先考虑n=2时怎么做，打表找规律找了半天找不出来。

赛后才知道这是nim积。

定义(x⊗y)为(sg(x,y))。

有一坨性质：
(x,y<2^{2^k},x⊗y<2^{2^k})

(2^{2^k}⊗2^{2^k}={3 over 2}2^{2^k})

可以把⊗看做乘法，(⊕)（异或）看做加法，所以还有分配律。

求(x⊗y(x>y))，设(k)为最大的(k)满足(2^{2^k}<=x)，设(M=2^{2^k})

(x=s*M+t=s*M⊕t,y=p*M+q=p*M⊕q)

则(x⊗y=spMM+sMq+tpM+tq)

(=M(sp+sq+tp)+tq+(M/2⊗sp))

字母间省略了(⊗)号，(+)号即(⊕)。

递归求出(sp,sq,tp,tq,M/2sp)即可。

时间复杂度：(O(5^{log~log~V}))。

预处理((0-255,0-255))会快许多。

题目中重定义加法和乘法，求行列式即可。

注意行列式要逆元，(v^{-1}=v^{2^{2^k}-2})（满足费马小定理）。

Code：

#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i <  _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("
")
using namespace std;

#define ul unsigned long long

const ul maxM = (ul) 1 << 32;
const ul a2_63 = (ul) 1 << 63;

const int C = 256;

ul b[C][C];

ul calc2(ul x, ul y) {
	if(!x || !y) return 0;
	if(x == 1) return y;
	if(y == 1) return x;
	if(x < y) return calc2(y, x);
	ul M = 2; while(M < maxM && M * M <= x) M *= M;
	ul p = x / M, q = x % M;
	ul s = y / M, t = y % M;
	ul c1 = calc2(p, s), c2 = calc2(p, t) ^ calc2(q, s), c3 = calc2(q, t);
	return M * (c1 ^ c2) ^ c3 ^ calc2(M / 2, c1);
}

ul calc(ul x, ul y) {
	if(x < C && y < C) return b[x][y];
	if(x < y) swap(x, y);
	ul M = 2; int k = 1;
	while(M < maxM && M * M <= x) M *= M, k *= 2;
	ul p = x >> k, q = x & (M - 1);
	ul s = y >> k, t = y & (M - 1);
	ul c1 = calc(p, s);
	return ((c1 ^ calc(p, t) ^ calc(q, s)) << k) ^ calc(q, t) ^ calc(M / 2, c1);
}

ul ksm(ul x, ul y) {
	ul s = 1;
	for(; y; y /= 2, x = calc(x, x))
		if(y & 1) s = calc(s, x);
	return s;
}

ul qni(ul x) {
	if(x >= maxM) return ksm(x, (a2_63 - 1) * 2);
	ul M = 2; while(M <= x) M *= M;
	return ksm(x, M - 2);
}

const int N = 155;

int n;
ul a[N][N];

int main() {
	freopen("partition.in", "r", stdin);
	freopen("partition.out", "w", stdout);
	ff(i, 0, C) ff(j, 0, C) b[i][j] = calc2(i, j);
	scanf("%d", &n);
	fo(i, 1, n) fo(j, 1, n) {
		scanf("%llu", &a[i][j]);
	}
	int ye = 1;
	fo(i, 1, n) {
		int u = -1;
		fo(j, i, n) if(a[j][i])
			u = j;
		if(u == -1)	{ ye = 0; break;}
		fo(j, i, n) swap(a[u][j], a[i][j]);
		ll v = qni(a[i][i]);
		fo(j, i, n) a[i][j] = calc(a[i][j], v);
		fo(j, i + 1, n) if(a[j][i]) {
			v = a[j][i];
			fo(k, i, n) a[j][k] ^= calc(a[i][k], v);
		}
	}
	pp("%s
", ye ? "xiaoDyingle" : "xiaoDwandanle");
}