【HDU-oj】-1757-A Simple Math Problem（矩陣，十一階）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4503    Accepted Submission(s): 2714

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

 

題解：這裏構造的矩陣是一個十一階的，方法和之前的矩陣乘法一樣。

A： Fn      Fn-1......Fn-10

B: Fn-1   Fn-2......Fn-11

要由下面的矩陣的到上面的就要乘以一個十一階的已知矩陣，根據矩陣乘法可以求出就這個矩陣爲：

暫時記爲c;若求n=11，則輸出最上面矩陣的A[1][1]；

用B*c^1就好。這樣我們就需要知道B矩陣的值即：F10  F9  F8 ......F1  F0，可以根據題意求出

代碼：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
int MOD;
struct Matrix 
{
	int h,w;
	int m[15][15];
};
Matrix Matrix_multiply(Matrix a,Matrix b)
{
	Matrix c;
	c.h=a.h;
	c.w=b.w;
	CLR(c.m,0);
	for(int i=1;i<=a.h;i++)
	{
		for(int j=1;j<=a.w;j++)
		{
			if(a.m[i][j]==0)
				continue;
			for(int k=1;k<=b.w;k++)
				c.m[i][k]=(c.m[i][k]+a.m[i][j]*b.m[j][k]%MOD)%MOD;
		}
	}
	return c;
}
Matrix Quick(Matrix a,int n)
{
	Matrix ans;
	ans.h=ans.w=a.h;
	CLR(ans.m,0);
	for(int i=1;i<=a.h;i++)	
		ans.m[i][i]=1;
	while(n)
	{
		if(n&1)
			ans=Matrix_multiply(ans,a);
		n>>=1;
		a=Matrix_multiply(a,a);
	}
	return ans;
}
int main()
{
	int k;
	int a[11];
	while(~scanf("%d%d",&k,&MOD))
	{
		if(k<10)
			printf("%d\n",k%MOD);
		else
		{	
			Matrix pr,now;
			pr.h=pr.w=11;
			CLR(pr.m,0);
			for(int i=1;i<=10;i++) 
			{
				scanf("%d",&a[i]);
				pr.m[i][1]=a[i];
			}
			for(int i=2;i<=11;i++)		
				pr.m[i-1][i]=1;
			now.h=1,now.w=11;
			CLR(now.m,0);
			now.m[1][1]=(9*a[1]+8*a[2]+7*a[3]+6*a[4]+5*a[5]+4*a[6]+3*a[7]+2*a[8]+a[9]);
			for(int i=2;i<=10;i++)
				now.m[1][i]=11-i;
			Matrix ans;
			ans=Matrix_multiply(now,Quick(pr,k-10));
			printf("%d\n",ans.m[1][1]);
		}
	}
	return 0;
}