HDU 1757 A Simple Math Problem (矩阵快速幂模板)

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5126    Accepted Submission(s): 3104

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

 

给出 f(n) 的关系:  

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);

构造矩阵  

 

然后 利用 矩阵快速幂:

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cstring>

typedef long long ll;

using namespace std;
const int MAXN=10;
const int N=20;
int num[10];
int MOD;
struct Matrix{
	ll arr[N][N];
	void init()
	{
		memset(arr,0,sizeof(arr));
		for(int i=1;i<MAXN;i++)
			arr[i][i-1]=1;
        for(int i=0;i<MAXN;i++)
            arr[0][i]=num[i];
	}
	void iinit()//单位矩阵
	{
	    memset(arr,0,sizeof(arr));
	    for(int i=0;i<MAXN;i++)
            arr[i][i]=1;
	}
}A;
Matrix mul(Matrix X,Matrix Y)//
{
	Matrix ans;
	for(int i=0;i<MAXN;i++)
		for(int j=0;j<MAXN;j++){
			ans.arr[i][j]=0;
			for(int k=0;k<MAXN;k++)
				ans.arr[i][j]+=X.arr[i][k]*Y.arr[k][j];
			ans.arr[i][j]%=MOD;
		}
	return ans;
}
Matrix Q_pow(Matrix B,int n)// ¾ØÕó¿ìËÙÃÝ
{
	Matrix ans;
	ans.iinit();
	while(n)
	{
		if(n&1)
			ans=mul(ans,B);
		n>>=1;
		B=mul(B,B);
	}
	return ans;
}

int main()
{
    Matrix ans;
    int k;
    while(~scanf("%d %d",&k,&MOD))
    {
        memset(num,0,sizeof(num));
        for(int i=0;i<=9;i++)
            scanf("%d",&num[i]);
        ans.init();//初始化
        int res=0;
        if(k<10){
            printf("%d\n",k);
            continue;
        }
        ans=Q_pow(ans,k-9);
        for(int i=0;i<=9;i++)
            res+=(ans.arr[0][i]*(9-i));
        printf("%d\n",res%MOD);
    }
    return 0;
}