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answer
Note that the numbering in the title starts from \(0\) = =
\(m\) is particularly small, considering the shape pressure
, let \(f[i][s]\) be the number of solutions for the \(i\) behavior \(s\)
Each piece can only attack this line, the above One line, the next line,
we can quickly find out which states are legal, and the union of the attack position in the previous line corresponding to each state and the union of the attack position in the next line.
If the two states cannot attack each other up and down, it is a legal transition
We get a \(2^m * 2^m\) transition matrix, and the matrix can be optimized
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define uint unsigned int
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 65,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int s1,s2,s3,n,m,p,k;
struct Matrix{
uint s[maxn][maxn]; int n,m;
Matrix(){memset(s,0,sizeof(s)); n = m = 0;}
}A,F,Fn;
inline Matrix operator *(const Matrix& a,const Matrix& b){
Matrix c;
if (a.m != b.n) return c;
c.n = a.n; c.m = b.m;
for (int i = 0; i < c.n; i++)
for (int j = 0; j < c.m; j++)
for (int k = 0; k < a.m; k++)
c.s[i][j] += a.s[i][k] * b.s[k][j];
return c;
}
inline Matrix qpow(Matrix a,int b){
Matrix ans; ans.n = ans.m = a.n;
for (int i = 0; i < ans.n; i++) ans.s[i][i] = 1;
for (; b; b >>= 1,a = a * a)
if (b & 1) ans = ans * a;
return ans;
}
int f[maxn];
bool check(int s){
for (int i = 0; i < m; i++){
if (s & (1 << i)){
if (i + 1 >= p - k){
if (s & (s2 << (i + 1 - (p - k)))) return false;
}
else if (s & (s2 >> ((p - k) - i - 1))) return false;
}
}
return true;
}
int getu(int s){
int re = 0;
for (int i = 0; i < m; i++){
if (s & (1 << i)){
if (i + 1 >= p - k) re |= s1 << (i + 1 - (p - k));
else re |= s1 >> ((p - k) - i - 1);
}
}
return re;
}
int getd(int s){
int re = 0;
for (int i = 0; i < m; i++){
if (s & (1 << i)){
if (i + 1 >= p - k) re |= s3 << (i + 1 - (p - k));
else re |= s3 >> ((p - k) - i - 1);
}
}
return re;
}
void print(int x){
for (int i = 4; i >= 0; i--)
printf("%d",(x & (1 << i)) != 0);
}
int main(){
n = read(); m = read();
p = read(); k = read();
REP(i,p) s1 = (s1 << 1) + read();
REP(i,p){
if (i == k + 1) s2 <<= 1,read();
else s2 = (s2 << 1) + read();
}
REP(i,p) s3 = (s3 << 1) + read();
int N = 1 << m;
F.n = N; F.m = 1;
for (int s = 0; s < N; s++)
if (check(s)) F.s[s][0] = 1;
A.n = A.m = N;
for (int s = 0; s < N; s++){
if (!check(s)) continue;
for (int e = 0; e < N; e++){
if (!check(e)) continue;
int u = getu(s),d = getd(e);
if (!(s & d) && !(e & u))
A.s[s][e] = 1;
}
}
Fn = qpow(A,n - 1) * F;
uint ans = 0;
for (int i = 0; i < N; i++) ans += Fn.s[i][0];
cout << ans << endl;
return 0;
}