CCPC-Wannafly & Comet OJ joy summer season (2019) E

Face questions

 

The problem of violence Well, consider a lot of things.

First set f (x) is the expected number of steps x step from the finish to go, we can find:

    1.x> time = k, x can be transferred to the index points are <x.

    2.x <When k, x, or it may go back to the larger point index.

 

Because k very small, so we can put f (0) (apparently 0), f (1), f (2), ....., f (k-1) Gaussian elimination of violence out (this is not you We will try to be written for each equation x 0 <x <k out, then f (x) to the left the whole item, other items whole to the right, an equation can be obtained; this column k -1 equations, unknown exactly k-1, Gaussian elimination template).

2. So that we solved the case. 1. So for a lot of cases, we can use from pushing subscript 2 smaller f () to do matrix multiplication.

 

So that's two questions: first, then build a matrix Gaussian elimination, and then build a fast power matrix, the matrix vector multiply the last to do it again you can get the answer.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=23,ha=1000000007;

inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;}

inline int ksm(int x,int y){
	int an=1;
	for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha;
	return an;
}

int k,inv,Inv,ans;
ll n;

struct node{
	int a[N][N];
	
	inline void clear(){ memset(a,0,sizeof(a));}
	inline void Base(){
		clear();
		for(int i=0;i<=k;i++) a[i][i]=1;
	}
	
    node operator *(const node &u)const{
    	node r; r.clear();
    	for(int l=0;l<=k;l++)
    	    for(int i=0;i<=k;i++)
    	        for(int j=0;j<=k;j++) ADD(r.a[i][j],a[i][l]*(ll)u.a[l][j]%ha);
    	return r;
	}
	
	inline void build(){
		clear(),a[k][k]=1;
		
		for(int i=1;i<k;i++){
			a[i][k]=(i*2>k)?1:k*(ll)Inv%ha,a[i][i]=1;
			
			for(int j=1,to;j<=k;j++){
				to=i-j;
				if(to<0) to=-to;
				if(to&&to!=i) ADD(a[i][to],ha-((i*2>k)?inv:Inv));
			}
		}
	}
	
	inline void solve(){
		for(int i=1,ne;i<k;i++){
			for(int j=i;j<k;j++) if(a[j][i]){ ne=j; break;}
			if(ne!=i) for(int l=i;l<=k;l++) swap(a[ne][l],a[i][l]);
			
			int ni=ksm(a[i][i],ha-2),now;
			for(int j=i+1;j<k;j++){
				now=ni*(ll)a[j][i]%ha;
				for(int l=i;l<=k;l++) ADD(a[j][l],ha-a[i][l]*(ll)now%ha);
			}
		}
		
		for(int i=k-1;i;i--){
			for(int j=i+1;j<k;j++) ADD(a[i][k],ha-a[i][j]*(ll)a[j][k]%ha);
			a[i][k]=a[i][k]*(ll)ksm(a[i][i],ha-2)%ha;
//			printf("%d %d\n",i,a[i][k]);
		}
	}
	
	inline void init(){
        clear();
        for(int i=k-2;i>=0;i--) a[i+1][i]=1;
        for(int i=0;i<k;i++) a[i][k-1]=inv;
        a[k][k-1]=1,a[k][k]=1;
	}
}A,X,ANS;

int main(){
	scanf("%lld%d",&n,&k),inv=ksm(k,ha-2),Inv=ksm(k-1,ha-2);
	if(k==1){ printf("%d\n",n%ha); return 0;}
	
	A.build(),A.solve(),X.init();
	n-=k-1,ANS.Base ();
	for(;n;n>>=1,X=X*X) if(n&1) ANS=ANS*X;
	
	for(int i=0;i<=k;i++) ADD(ans,A.a[i][k]*(ll)ANS.a[i][k-1]%ha);
	printf("%d\n",ans);
	return 0;
}

  

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Origin www.cnblogs.com/JYYHH/p/11265471.html