【bzoj 2326】【HNOI 2011】数学作业

题解：

　　矩阵裸体。

　　

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 typedef long long ll;
 ll n,mod;
 struct Matrix{
     ll a[][];
     Matrix (){
         memset(a,,sizeof(a));
     }
     inline void e(){
         for(int i=;i<=;i++)
             a[i][i]=;
     }
     friend Matrix operator *(Matrix x,Matrix y){
         Matrix c;
         for(int k=;k<=;k++)
             for(int i=;i<=;i++)
                 for(int j=;j<=;j++){
                     c.a[i][j]+=x.a[i][k]*y.a[k][j];
                     if(c.a[i][j]>mod)
                         c.a[i][j]%=mod;
                 }
         return c;
     }
     friend Matrix operator ^(Matrix a,ll b){
         Matrix ans;ans.e();
         while(b){
             if(b&) ans=ans*a;
             b>>=;
             a=a*a;
         }return ans;
     }
     inline void kmod(ll k){
         for(int i=;i<=;i++)
             for(int j=i;j<=;j++)
                 a[i][j]=;
         a[][]=k%mod;
     }
 };
 int main(){
     scanf("%lld%lld",&n,&mod);
     Matrix ans;
     ans.a[][]=;
     ans.a[][]=;
     ans.a[][]=;
     int i;ll j;
     for(i=,j=;j*<=n;j*=,i++){
         Matrix b;b.kmod(j*);
         ans=(b^(j*-j))*ans;
     }//printf("%lld\n",ans.a[1][1]);
     Matrix b;b.kmod(j*);
     if(n-j+>)
         ans=(b^(n-j+))*ans;
     printf("%lld\n",ans.a[][]);
 }