Title effect: to a length \ (n (n <= { 18}) \ 10 ^) cycloalkyl, require any \ (m \) number, the \ (3 \) is not greater than the number \ (7 \ ) number (the entire sequence of \ (3,7 \) configuration), the total number of programs Q
\(Solution\)
Between \ (n \) is large, we had to convert to the idea of \ (m \) on.
Consider a state \ (j \) , if from the state \ (i \) transferred over, then \ (j \) the number of programs can add (i \) \ number of programs.
See \ (n-\) data, to consider fast power matrix.
Epicenter, \ ([I] [J] \) shows a state \ (J \) may be made of \ (I \) transfer over. Because it is a ring, we enumerate all states, it will be as a starting point, \ (dp \) again, then finished for the last state to \ (check \) is not legitimate at the beginning.
We handle only a beginning, put the initial matrix \ ([0] [i] \) assigned to \ (1 \) , the final matrix is a \ (1 * \) matrix number of states, it is noteworthy that the state can not just keep useful, useless state is likely to turn into a useful state.
#include<bits/stdc++.h>
using namespace std;
long long n;
int m,ans,cc;
const int mod=998244353;
const int P=mod;
int t=33;
struct Mat{
int A[34][34];
Mat(){
memset(A,0,sizeof(A));
}
Mat operator *(const Mat&B)const {
Mat tmp;
for(int i=0; i<t; i++) {
for(int j=0; j<t; j++) {
for(int k=0; k<t; k++) {
tmp.A[i][j]=(tmp.A[i][j]+1ll*A[i][k]*B.A[k][j]%P)%P;
}
}
}
return tmp;
}
}w,d;
Mat qpow(Mat a,long long b){
Mat c=d;
while(b){
if(b&1)c=c*a;//cout<<"qwq\n";
a=a*a;b>>=1;
}
return c;
}
inline int get(int x){
int res=0;
while(x){
if(x&1)res++;
x>>=1;
}
return res;
}
int cnt[500],tot=-1,mp[500];
int use[300];
inline bool check(int x,int y){
// memset(use,0,sizeof(use));
int L=0;
for(int i=m-1;i>=0;--i)use[++L]=x>>i&1;
for(int i=m-1;i>=0;--i)use[++L]=y>>i&1;
for(int i=1;i<=L-m+1;++i){
int r=0;
for(int j=0;j<m;++j)
r+=use[i+j];
if(r>cc)return false;
}
return true;
}
int main(){
scanf("%lld%d",&n,&m);
//memset(cnt,-1,sizeof(cnt));
// memset(mp,-1,sizeof(mp));
cnt[0]=0;
cc=m/2;
t=(1<<m);
for(int i=1;i<(1<<m);++i)cnt[i]=cnt[i-(i&-i)]+1;
int H=(1<<m)-1;
//cout<<cnt[8]<<endl;
for(int i=0;i<(1<<m);++i){
// int nt=cnt[i]>>1;
//if(cnt[nt]>cc)continue;
if(cnt[i]>cc)continue;
int nt=(i<<1)&H;
if(cnt[nt]>cc)continue;
if(cnt[nt]<=cc)w.A[i][nt]=1;
nt|=1;
if(cnt[nt]<=cc)w.A[i][nt]=1;
//状态i所累加的值在mpnt列
}
//for(int i=0;i<tot;++i)d.A[i][0]=1;
/*for(int i=0;i<tot;++i){
memset(d.A,0,sizeof(d.A));
d.A[i][0]=1;
Mat AA=qpow(w,n-m);
//d.A[0][j]=1;
//Mat e=AA*d;
for(int j=0;j<tot;++j)
if(check(cnt[j],cnt[i]))
ans+=AA.A[j][0],ans%=mod;
}*/
for(int i=0;i<(1<<m);++i){
if(cnt[i]>cc)continue;
memset(d.A,0,sizeof(d.A));
d.A[0][i]=1;
//cout<<"qwq\n";
Mat AA=qpow(w,n-m);
// cout<<"qwq\n";
for(int j=0;j<(1<<m);++j){
if(cnt[j]>cc)continue;
if(check(j,i))ans+=AA.A[0][j],ans%=mod;
}
}
cout<<ans<<endl;
return 0;
}