nefu 459 矩阵连乘

http://acm.nefu.edu.cn/JudgeOnline/problemshow.php?problem_id=459

description

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.

input

The problem contains mutiple test cases.Please process to the end of file. In each case, there will be two lines. In the first line , there are two positive integers k and m. ( k&lt;2*10^9 , m &lt; 10^5 ) In the second line , there are ten integers represent a0 ~ a9.

output

For each case, output f(k) % m in one line.

sample_input

10 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0

sample_output

45 104

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
typedef long long LL;
const int N=10;
struct Matrix
{
    LL m[N][N];
};
LL mod;
Matrix I=
{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,
 0,0,0,0,0,0,0,0,0,1};
Matrix multi(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++)
        {
            c.m[i][j]=0;
            for(int k=0;k<N;k++)
            {
                c.m[i][j]+=a.m[i][k]*b.m[k][j]%mod;
            }
            c.m[i][j]%=mod;
        }
    return c;
}
Matrix quick_mod(Matrix a,LL k)
{
    Matrix ans=I;
    while(k!=0)
    {
        if(k&1)
        {
            ans=multi(ans,a);
        }
        k>>=1;
        a=multi(a,a);
    }
    return ans;
}
int main()
{
    LL k;
    int a0,a1,a2,a3,a4,a5,a6,a7,a8,a9;
    while(~scanf("%lld%lld",&k,&mod))
    {

        scanf("%d%d%d%d%d%d%d%d%d%d",&a0,&a1,&a2,&a3,&a4,&a5,&a6,&a7,&a8,&a9);
        if(k<10)
        {
            printf("%d\n",k%mod);
            continue;
        }
        Matrix A={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};
        Matrix ans=quick_mod(A,k-9);
        LL t=((ans.m[0][0]*9)%mod+(ans.m[0][1]*8)%mod+(ans.m[0][2]*7)%mod+(ans.m[0][3]*6)%mod+(ans.m[0][4]*5)%mod+(ans.m[0][5]*4)%mod+(ans.m[0][6]*3)%mod+(ans.m[0][7]*2)%mod+(ans.m[0][8]*1)%mod+(ans.m[0][9]*0)%mod)%mod;
        printf("%lld\n",t);
    }
    return 0;
}