HDU 5015 233 Matrix --矩阵快速幂

最新推荐文章于 2020-11-11 23:26:07 发布

原创最新推荐文章于 2020-11-11 23:26:07 发布 · 181 阅读

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本文介绍了一种利用矩阵快速幂解决特定类型递推问题的方法。通过构造特定的矩阵A和B，并采用快速幂运算，可以在O(log m)的时间复杂度内求解an,m的值。该方法适用于形如ai,j=ai-1,j+ai,j-1的递推问题，其中a0,i由特定规律定义。

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题意：

ai,j = a i-1,j +a i,j-1( i,j > 0) , a0,i=23333...., i+1个3，ai，0为输入的数值

给出n，m，给出（a0，1 ），（ a0，n）

求an，m

思路：

我们知道这是NXM的矩阵

试着画出该矩阵，想推出ai，j发现推不出来，但是可以推出下一列

应该一目了然了吧

#include<bits/stdc++.h>
#define N  10000007
using namespace std;
struct node
{
	long long a[12][12];
};
const node A={
	          { 
			    {1,1 ,0,0,0,0,0,0,0,0,0,0},
	            {0,10,1,1,1,1,1,1,1,1,1,1},
				{0,0 ,1,1,1,1,1,1,1,1,1,1},
				{0,0 ,0,1,1,1,1,1,1,1,1,1},
				{0,0 ,0,0,1,1,1,1,1,1,1,1},
				{0,0 ,0,0,0,1,1,1,1,1,1,1},
				{0,0 ,0,0,0,0,1,1,1,1,1,1},
				{0,0 ,0,0,0,0,0,1,1,1,1,1},
				{0,0 ,0,0,0,0,0,0,1,1,1,1},
				{0,0 ,0,0,0,0,0,0,0,1,1,1},
				{0,0 ,0,0,0,0,0,0,0,0,1,1},
				{0,0 ,0,0,0,0,0,0,0,0,0,1}  
			  }
             };
const node B={
	          {
	          	{3,233,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0},
				{0,0,0,0,0,0,0,0,0,0,0,0}
	          }
             };
node cheng(node a,node b,bool f)
{
	node c;
	memset(c.a,0,sizeof(c.a));
	for(int i=0;i<12;i++){
	if(f&&i==1) break;
	for(int j=0;j<12;j++)
	for(int k=0;k<12;k++)
	c.a[i][j]=(c.a[i][j]+(a.a[i][k]*b.a[k][j])%N)%N;
	}
	return c;
}

int main()
{
	int t,n,m,sum;
	node a,b;
	while(~scanf("%d%d",&n,&m))
    {
    	if(n==m&&n==0)
    	{
    		printf("0\n");
    		continue;
    	}
    	a=A;
    	b=B;
    	for(int i=0;i<n;i++)
    	scanf("%lld",&(b.a[0][2+i]));
    
    	while(m)
    	{
    		if(m&1) b=cheng(b,a,true);
    		m>>=1;
    		a=cheng(a,a,false);
    	}
    	printf("%lld\n",(b.a[0][n+1]%N+N)%N);
    }	
	return 0;
}