构造矩阵+快速幂

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means
a_0,1 = 233,a_0,2 = 2333,a_0,3 = 23333...) Besides, in 233 matrix, we got a_i,j = a_i-1,j +a_i,j-1( i,j ≠ 0). Now you have known a_1,0,a_2,0,...,a_n,0, could you tell
me a_n,m in the 233 matrix?

Input

There are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 10⁹). The second line contains n integers, a_1,0,a_2,0,...,a_n,0(0 ≤ a_i,0 < 2³¹).

Output

For each case, output a_n,m mod 10000007.

Sample Input

1 1
1
2 2
0 0
3 7
23 47 16

Sample Output

234
2799
72937

Hint

Source

2014 ACM/ICPC Asia Regional Xi‘an Online

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

const LL MOD=10000007LL;

int n,m;
LL a[20];

struct Matrix
{
	int x,y;
	LL matrix[12][12];
	Matrix()
	{
		x=y=0;
		memset(matrix,0,sizeof(matrix));
	}
};

void init(Matrix& m)
{
	m.x=m.y=n+2;
	for(int i=0;i<m.x-2;i++)
		for(int j=i;j<m.y-1;j++)
			m.matrix[i][j]=1;
	m.matrix[m.x-2][m.y-2]=10;
	m.matrix[m.x-1][m.y-1]=m.matrix[m.x-2][m.y-1]=1;
}

Matrix CHENGFA(Matrix& a,Matrix& b)
{
	int n=a.x;
	Matrix ret;
	ret.x=ret.y=n;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			LL temp=0;
			for(int k=0;k<n;k++)
			{
				temp=(temp+(a.matrix[i][k]*b.matrix[k][j])%MOD)%MOD;
			}
			ret.matrix[i][j]=temp%MOD;
		}
	}
	return ret;
}

Matrix MatrixQuickPow(Matrix m,int k)
{
	Matrix e;
	int n=m.x;
	e.x=e.y=n;
	for(int i=0;i<n;i++) e.matrix[i][i]=1;
	while(k)
	{
		if(k%2)
			e=CHENGFA(e,m);
		m=CHENGFA(m,m);
		k/=2;
	}
	return e;
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=n-1;i>=0;i--)
			scanf("%lld",a+i);
		a[n]=233; a[n+1]=3;
		Matrix M,ED;
		init(M);
		ED=MatrixQuickPow(M,m-1);
		LL ans=0;
		for(int i=0;i<n+1;i++)
		{
			LL temp=0;
			for(int j=0;j<n+2;j++)
			{
				temp=(temp+(ED.matrix[i][j]*a[j])%MOD)%MOD;
			}
			ans=(ans+temp)%MOD;
		}
		printf("%I64d\n",ans);
	}
	return 0;
}