8.3.2 A Simple Math Problem

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 44 Accepted Submission(s): 42

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

思路：递推数列都可以用矩阵解的～例如一个递推数列Fn=Fn-1+Fn-2+..+Fn-k

构造Fk,Fk+1,Fk+2..Fn-1,Fn

 --->Fk+1,Fk+2,....Fn,Fn+1即可……

#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

typedef long long ll;
const int maxn=10;
struct matrix
{
    int m[maxn][maxn];
};
matrix c,base,ans;
int n,k,mod,sum,x;
int a[maxn];


void close()
{
exit(0);
}

matrix mul(matrix a,matrix b)
{
    memset(c.m,0,sizeof(c.m));
    for (int i=0;i<=9;i++)
        for (int j=0;j<=9;j++)
            for (int k=0;k<=9;k++)
                c.m[i][j]=(c.m[i][j]+(ll)a.m[i][k]*b.m[k][j]) % mod;
    return c;
}

void print(matrix a)
{
    for (int i=0;i<=9;i++)
    {
        for (int j=0;j<=9;j++)
            printf("%d ",a.m[i][j]);
        printf("\n");
    }
}

void work()
{
    k-=9;
    memset(ans.m,0,sizeof(ans.m));
    for (int i=0;i<=9;i++)
        ans.m[i][i]=1;
    while (k!=0)
    {
        if (k & 1)
            ans=mul(base,ans);
        k/=2;
        base=mul(base,base);
    }
    sum=0;
    for (int i=0;i<=9;i++)
        sum=(sum + (ll)(i)*ans.m[9][i]) % mod;
    printf("%d\n",sum);
}


void init()
{
    while(scanf("%d %d",&k,&mod)!=EOF)
    {
        if (k<=9)
        {
            printf("%d\n",k);
            continue;
        }
        for (int i=0;i<=9;i++)
            scanf("%d",&a[9-i]);
        memset(base.m,0,sizeof(base.m));
        for (int i=0;i<=8;i++)
            base.m[i][i+1]=1;
        for (int i=0;i<=9;i++)
            base.m[9][i]=a[i];
        work();
    }
}

int main ()
{
    init();
    close();
    return 0;
}