233 Matrix（hdu5015 矩阵）

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1190    Accepted Submission(s): 700

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

我们这样看：已知a₁₁ ,a₂₁ ,a₃₁ ,a₄₁  。。。求后面的

a₁₂ = a₁₁ +233;

a₂₂ = a₁₁ + a₂₁ +233;

a₃₂ = a₁₁ + a₂₁ +a₃₁ +233;

a₄₂ = a₁₁ + a₂₁ +a₃₁ +a₄₁ +233;

.........

同理：后面的列也一样：

a₁₃ = a₁₂ +233;

a₂₃ = a₁₂ + a₂₂ +233;

a₃₃ = a₁₂ + a₂₂ +a₃₂ +233;

a₄₃ = a₁₂ + a₂₂ +a₃₂ +a₄₂ +233;

...........

ss所以有矩阵：

233

a11

a21

a31

a41

...

3

*

10	1	1	1	1	...	0
0	1	1	1	1	...	0
0	0	1	1	1	...	0
0	0	0	1	1	...	0
0	0	0	0	1	...	0
...	...	...	...	...	...	...
1	0	0	0	0	...	1

=

......................................................................................................................................................

z转载请注明出处：寻找&星空の孩子

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5015

#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL __int64
#define mod 10000007

LL N,M;

struct matrix
{
    LL m[][];
};
LL a[];

matrix multiply(matrix x,matrix y)
{
    matrix temp;
    memset(temp.m,,sizeof(temp.m));
    for(int i=; i<N+; i++)
    {
        for(int j=; j<N+; j++)
        {
            if(x.m[i][j]==) continue;
            for(int k=; k<N+; k++)
            {
                if(y.m[j][k]==) continue;
                temp.m[i][k]+=x.m[i][j]*y.m[j][k]%mod;
                temp.m[i][k]%=mod;
            }
        }
    }
    return temp;
}

matrix quickmod(matrix a,LL n)
{
    matrix res;
    memset(res.m,,sizeof(res.m));
    for(int i=;i<N+;i++) res.m[i][i]=;
    while(n)
    {
        if(n&)
            res=multiply(res,a);
        n>>=;
        a=multiply(a,a);
    }
    return res;
}
int main()
{
    int n,k;
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        a[]=;
        a[N+]=;
        for(int i=;i<=N;i++)
        {
            scanf("%d",&a[i]);
        }

        matrix ans;
        memset(ans.m,,sizeof(ans.m));
        ans.m[][]=;
        ans.m[N+][]=;
        ans.m[N+][N+]=;
        for(int j=;j<=N;j++)
        {
            for(int i=;i<=j;i++)
            {
                ans.m[i][j]=;
            }
        }

        ans=quickmod(ans,M);//M次幂定位到纵坐标。

        LL ant=;
        for(int i=;i<N+;i++)//横坐标是N，即，乘以矩阵的N列。
        {
            ant=(ant+a[i]*ans.m[i][N])%mod;
        }
        printf("%I64d\n",ant);
    }
    return ;
}

本来要做新题的，可是遇到不会的了。。。hdu4767 Bell 现在卡在  中国剩余定理，还要好好梳理梳理！

加油！少年！！！