hdu 1757 A Simple Math Problem（矩阵优化DP）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3984    Accepted Submission(s): 2410

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
  
  

 

Author

linle

 

Source

2007省赛集训队练习赛（6）_linle专场

 

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题解：矩阵优化dp

优化矩阵   ：

0 0 0 0 0 0 0 0 0 a9
1 0 0 0 0 0 0 0 0 a8
0 1 0 0 0 0 0 0 0 a7 
0 0 1 0 0 0 0 0 0 a6
0 0 0 1 0 0 0 0 0 a5 
0 0 0 0 1 0 0 0 0 a4 
0 0 0 0 0 1 0 0 0 a3
0 0 0 0 0 0 1 0 0 a2
0 0 0 0 0 0 0 1 0 a1
0 0 0 0 0 0 0 0 1 a0

因为我们初始的矩阵是

{ f(x-9),f(x-8),f(x-7),....,f(x)}  我们要推到下一个状态即{f(x-8),f(x-7),....,f(x),f(x+1)}  ，注意乘的时候是用初始矩阵的行×优化矩阵的列，这样我们求出优化矩阵A^(n-9) 乘初始矩阵，得到的初始矩阵的最后一个数就是答案。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define N 10
using namespace std;
int n,m,p,f[20];
struct data
{
	int a[20][20];
}a;
void clear(data &x)
{
    for (int i=1;i<=N;i++)
     for (int j=1;j<=N;j++)
      x.a[i][j]=0;
}
void change(data &a1,data b)
{
	for (int i=1;i<=N;i++)
	 for (int j=1;j<=N;j++)
	  a1.a[i][j]=b.a[i][j];
}
data mul(data a1,data b)
{
	data c;
	for (int i=1;i<=N;i++)
	 for (int j=1;j<=N;j++)
	  {
	  	c.a[i][j]=0;
	  	for (int k=1;k<=N;k++)
	  	 c.a[i][j]=(c.a[i][j]+a1.a[i][k]*b.a[k][j]%p)%p;
	  }
	return c;
}
data pow(data num,int x)
{
	data ans; clear(ans);
	for (int i=1;i<=N;i++)  ans.a[i][i]=1;
	data base; clear(base);
	change(base,num);
	while (x)
	{
		if (x&1)  ans=mul(ans,base);
		x>>=1;
		base=mul(base,base);
	}
	return ans;
}
int main()
{
	clear(a);
	for (int i=2;i<=10;i++)
	 a.a[i][i-1]=1;
	for (int i=1;i<=10;i++)
	 f[i]=i-1;
	while (scanf("%d%d",&n,&p)!=EOF)
	{
		for (int i=1;i<=10;i++)
		 scanf("%d",&a.a[N-i+1][10]);
		n-=9;
		data ans=pow(a,n);
		int t[20]; memset(t,0,sizeof(t));
		for (int j=1;j<=N;j++)
		 {
		 	t[j]=0;
		 	for (int k=1;k<=N;k++)
		 	 t[j]=(t[j]+f[k]*ans.a[k][j]%p)%p;
		 }
		printf("%d\n",t[10]);
	}
}