hdu1757 构造矩阵

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

构造一个矩阵

{

a0 a1 a2 a3 a4 a5 a6 a7 a8 a9

1  0   0   0   0  0   0  0   0   0

0  1   0   0   0  0   0  0   0   0

0  0   1   0   0  0   0  0   0   0

0  0   0   1   0  0   0  0   0   0

0  0   0   0   1  0   0  0   0   0

0  0   0   0   0  1   0  0   0   0

0  0   0   0   0  0   1  0   0   0

0  0   0   0   0  0   0  1   0   0

0  0   0   0   0  0   0  0   1   0

}得解

#include <iostream>
#include <algorithm>
#include <string.h>
#include <cstdio>
#include <math.h>
using namespace std;
typedef long long LL;
const int maxn=;
int MOD;
struct Matrix
{
    LL m[maxn][maxn];
    Matrix mult(Matrix &rhs)
    {
        Matrix ans;
        for(int i=; i<; i++)
            for(int j=; j<; j++)
             {
                ans.m[i][j]=;
                for(int k=; k<; k++)
                ans.m[i][j]=(ans.m[i][j]+(m[i][k]*rhs.m[k][j])%MOD )%MOD;

             }
       return ans;
    }
};
int a[];
void powk(int n)
{
    Matrix A,ans;
    memset(ans.m,,sizeof(ans.m));
    memset(A.m,,sizeof(A.m));
    for(int i=; i<; i++){
        ans.m[i][i]=;
        A.m[][i]=a[i];
        A.m[i+][i]=;
    }
    while(n)
        {
             if(n&)ans=ans.mult(A);
              n>>=;
              A=A.mult(A);
        }
     LL AS=;
     for(LL i=; i<; i++){
         AS=(AS+(ans.m[][i]*(-i+))%MOD )%MOD;
     }
     printf("%I64d\n",AS);
}
int main()
{
    int k;
    while(scanf("%d%d",&k,&MOD)==)
    {
          for(int i=; i<; i++)scanf("%d",&a[i]);
          if(k<){
            printf("%d\n",k%MOD); continue;
          }
          powk(k-);
    }
    return ;
}