A Simple Math Problem 矩阵打水题

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　A Simple Math Problem

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

 

 

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long LL;
const int INF = 0x4fffffff;
const double EXP = 1e-5;
const int MS = 1005;
const int SIZE = 10;

typedef vector<vector<int> > mat;

mat matrix(SIZE,vector<int>(SIZE));

int n,mod;

mat mul(mat &m1,mat &m2)
{
    mat m(SIZE,vector<int>(SIZE));
    for(int i=0;i<SIZE;i++)
        for(int j=0;j<SIZE;j++)
            for(int k=0;k<SIZE;k++)
        {
            m[i][j]=(m1[i][k]*m2[k][j]+m[i][j])%mod;
        }
    return m;
}

mat pow(int n)
{
    mat m(SIZE,vector<int>(SIZE));
    for(int i=0;i<SIZE;i++)
        m[i][i]=1;
    while(n)
    {
        if(n&1)
            m=mul(m,matrix);
        matrix=mul(matrix,matrix);
        n>>=1;
    }
    return m;
}

int main()
{
    while(scanf("%d%d",&n,&mod)!=EOF)
    {
        for(int i=0;i<SIZE;i++)
            for(int j=0;j<SIZE;j++)
                matrix[i][j]=(i-1)==j;
        for(int i=0;i<SIZE;i++)
            scanf("%d",&matrix[0][i]);

        if(n<10)
        {
            printf("%d
",n%mod);
            continue;
        }
        matrix=pow(n-9);
        int ans=0;
        for(int i=0;i<SIZE;i++)
            ans+=matrix[0][i]*(9-i)%mod;
        printf("%d
",ans%mod);
    }
}