题目链接：​​点我​​

Problem 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<2*10^9 , m < 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

矩阵快速幂求线性递推式，主要是把转换矩阵求出来，然后直接把矩阵套进快速幂

详细见代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<iomanip>
#include<cctype>
using namespace std;
#define ll long long
#define edl putchar('\n')
#define sscc ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define FORLL(i,a,b) for(ll i=a;i<=b;i++)
#define ROFLL(i,a,b) for(ll i=a;i>=b;i--)
#define mst(a) memset(a,0,ssizeof(a))
#define mstn(a,n) memset(a,n,ssizeof(a))
#define zero(x)(((x)>0?(x):-(x))<eps)
const int ssize=15;
int i,j;
int n,m,f[ssize],a[ssize];
struct Matrix {
    ll a[ssize][ssize];
    Matrix() {
        memset(a,0,sizeof(a));
    }
    void init() {
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                a[i][j]=(i==j);
    }
    Matrix operator + (const Matrix &B)const {
        Matrix C;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                C.a[i][j]=(a[i][j]+B.a[i][j])%m;
        return C;
    }
    Matrix operator * (const Matrix &B)const {
        Matrix C;
        for(int i=0; i<n; i++)
            for(int k=0; k<n; k++)
                for(int j=0; j<n; j++)
                    C.a[i][j]=(C.a[i][j]%m+1LL*a[i][k]*B.a[k][j]%m+m)%m;
        return C;
    }
    Matrix operator ^ (const ll &t)const {
        Matrix A=(*this),res;
        res.init();
        ll p=t;
        while(p) {
            if(p&1)res=res*A;
            A=A*A;
            p>>=1;
        }
        return res;
    }
    
};

int main() {
    int k,i;
    while(scanf("%d%d",&k,&m)!=EOF)
    {
        n=10;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        f[n]=0;
        for(i=0;i<n;i++)   //求前10项
            f[i]=i;
        for(i=0;i<n;i++)
            f[n]+=a[i]*f[n-1-i];
        if(k<=n)
            printf("%d\n",f[k]);
        else
        {
            Matrix A;
            for(i=1;i<n;i++)    //构造矩阵A
                A.a[i][i-1]=1;
            for(i=0;i<n;i++)
                A.a[i][n-1]=a[n-1-i];
              //A^(k-10)
            
            Matrix res=A^(k-10);
            Matrix origin;
            for(i=0;i<n;i++)   //前10项的矩阵[f(1),f(2),f(3)....f(10)]
                origin.a[0][i]=f[i+1];
                //[f(1),f(2),f(3)....f(10)] * A^(n-10)
            Matrix ans;
            ans=origin*res;
            printf("%d\n",ans.a[0][n-1]%m);
        }
    }
}