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("\n")
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\n", ye ? "xiaoDyingle" : "xiaoDwandanle");
}
原文地址:https://www.cnblogs.com/coldchair/p/12203219.html
时间: 2024-10-09 22:12:09