hdu 1757(快速矩阵幂)

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4597    Accepted Submission(s): 2777

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：快速矩阵幂，递归会超时
#include <bits/stdc++.h>
using namespace std;
const int N=10;
long long int mod;
struct M{
	long long int m[N][N];
};
M 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,};  
M A={0,0,0,0,0,0,0,0,0,0,                //所求矩阵 
    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};
M fun(M a,M b)            //a*b矩阵 
{
	M 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;
}
M power( int k)                   //快速矩阵幂 
{
	M ans=I,p=A;
	while(k)
	{
		if(k&1)
		{
			ans=fun(ans,p);
			k--;
		}
		k>>=1;
		p=fun(p,p);
	}
	return ans;
}
int main()
{
	long long int n;
	                                  
	while(~scanf("%lld %lld",&n,&mod))
	{
		M ans;
		long long int ret=0;
		for(int i=0;i<10;i++)
		scanf("%d",&A.m[0][i]);
		
		if(n<10)                         //特殊情况 
		{
			cout<<n%mod<<"\n";
			continue;
		}
		ans=power(n-9);             //注意 
		for(int i=0;i<10;i++)
		{
			ret+=(ans.m[0][i]*(9-i))%mod; 
		}
		cout<<ret%mod<<"\n";          
	}
	return 0;
 }