233 Matrix HDU

f[n][m]=f[n-1][m]+f[n][m-1]的形式，可以通过下三角矩阵的线性组合体现出来

//#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>  
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip>  
using namespace std;  
  
const double pi=acos(-1.0);  
#define ll long long  
#define pb push_back

#define sqr(a) ((a)*(a))
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))

const double eps=1e-10;
const int maxn=5e4+56;
const int inf=0x3f3f3f3f;
const ll mod=1e7+7;

ll arr[maxn];

struct mat{  
    ll a[15][15];  
    ll n,m;  
    mat(){memset(a,0,sizeof a);n=0;m=0;}  
    mat(ll x,ll y){memset(a,0,sizeof a);n=x;m=y;}  
    mat operator* (const mat &rhs)const{  
        mat ans;  
        ans.n=n;ans.m=rhs.m;  
        for(int i=1;i<=n;i++){  
            for(int j=1;j<=rhs.m;j++){  
                for(int k=1;k<=m;k++){  
                    ans.a[i][j]=(ans.a[i][j]+a[i][k]*rhs.a[k][j]+mod)%mod;  
                }  
            }  
        }  
        return ans;  
    }  
    mat operator^ (ll rhs)const{  
        mat ans(n,n),b=*this;  
        for(int i=1;i<=n;i++)ans.a[i][i]=1;  
        for(;rhs;rhs>>=1,b=b*b)  
            if(rhs&1)ans=ans*b;  
        return ans;  
    }   
};

int main(){
	ll n,m;
	while(~scanf("%lld%lld",&n,&m)){
		for(int i=2;i<=n+1;i++){scanf("%lld",&arr[i]);}
		arr[1]=1ll*23;arr[n+2]=1ll*3;
		mat mult(n+2,n+2);
		mult.a[n+2][n+2]=1;
		for(int i=1;i<=n+1;i++)for(int j=1;j<=n+2;j++){
			if(j==1){mult.a[i][j]=10;}
			else if(i>=j){mult.a[i][j]=1;}
			else if(j==n+2){mult.a[i][j]=1;}
		}
		mat fin=mult^(m);
		ll ans=0;
		for(int j=1;j<=n+2;j++){
			ans=(ans+fin.a[n+1][j]*arr[j]%mod)%mod;
		}
		printf("%lld
",ans);
	}
}