poj 3420 Quad Tiling (状压dp+多米诺骨牌问题+矩阵快速幂)
还有这种操作??????
直接用pre到now转移的方式构造一个矩阵就好了。
二进制长度为m,就构造一个长度为1 << m的矩阵
最后输出ans[(1 << m) - 1][(1 << m) - 1]就好了
牛逼!
#include<cstdio>
#include<cstring>
#include<algorithm>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int MAXN = 16;
struct mat
{
ll m[MAXN][MAXN];
mat() { memset(m, 0, sizeof(m)); }
}A;
int n, MOD;
mat operator *(const mat& a, const mat& b)
{
mat res;
REP(i, 0, MAXN)
REP(j, 0, MAXN)
REP(k, 0, MAXN)
res.m[i][j] = (res.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
return res;
}
void dfs(int l, int now, int pre)
{
if(l > 4) return;
if(l == 4)
{
A.m[pre][now]++;
return;
}
dfs(l + 1, (now << 1) | 1, pre << 1);
dfs(l + 1, now << 1, (pre << 1) | 1);
dfs(l + 2, (now << 2) | 3, (pre << 2) | 3);
}
mat pow(mat a, int b)
{
mat res;
REP(i, 0, MAXN) res.m[i][i] = 1;
for(; b; b >>= 1)
{
if(b & 1) res = res * a;
a = a * a;
}
return res;
}
int main()
{
dfs(0, 0, 0);
while(~scanf("%d%d", &n, &MOD) && n)
{
mat ans = pow(A, n);
printf("%lld\n", ans.m[15][15]);
}
return 0;
}