233 Matrix HDU

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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\n",ans);}}

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