hdu 5015 矩阵快速幂

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2722    Accepted Submission(s): 1572

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

1 112 20 03 723 47 16

 

Sample Output

234279972937
Hint
题意：构建一个矩阵 啊a(0,0) = 0; a(n,0)由键盘输入 a(0,1)=233 a(0,2)=2333 a(0,3)=23333......; a(i,j)=a(i-1.j)+a(i,j-1)
求a(n,m)
很典型的一个矩阵快速幂的题
最重要的是推导出快速幂的矩阵
 
思路：
第一列元素为：
0
a1
a2
a3
a4
转化为：
23
a1
a2
a3
a4
3
则第二列为：
23*10+3
23*10+3+a1
23*10+3+a1+a2
23*10+3+a1+a2+a3
23*10+3+a1+a2+a3+a4
3
根据前后两列的递推关系，有等式可得矩阵A的元素为：
其中a(0,0)=23
ac代码
#include<iostream>#include<cstdio>#include<cstring>#define MOD 10000007using namespace std;typedef long long LL;int n,m;//矩阵结构体struct Mat{    LL mat[15][15];};//乘法模板Mat operator*(Mat a,Mat b){        Mat c;        memset(c.mat,0,sizeof(c.mat));        for(int k = 0; k < n+2;k++)            for(int i = 0; i < n+2; i++)            for(int j = 0; j < n+2; j++)            c.mat[i][j] = (c.mat[i][j] + a.mat[i][k]*b.mat[k][j])%MOD;        return c;}int main(){    while(~scanf("%d %d",&n,&m))    {        Mat sum,temp;        memset(sum.mat,0,sizeof(sum.mat));        memset(temp.mat,0,sizeof(temp.mat));        sum.mat[0][0] = 23;        sum.mat[n+1][0] = 3;        for(int i = 1; i < n+1; i++)            scanf("%lld",&sum.mat[i][0]);        if(m == 0)        {            if(n == 0)            printf("0\n");            else                printf("%lld\n",sum.mat[n][0]);        }        else        {            //构建快速幂矩阵            for(int i = 1; i <= n; i++)            {                for(int j = 1; j <= i; j++)                {                    temp.mat[i][j] = 1;                }            }            for(int i = 0; i < n+2; i++)                temp.mat[i][0] = 10;            for(int i = 0; i < n+2; i++)                temp.mat[n+1][i] = 0;              for(int i = 0; i < n+2; i++)                temp.mat[i][n+1] = 1;            Mat res;            memset(res.mat,0,sizeof(res.mat));            for(int i = 0; i < n+2; i++)                res.mat[i][i] = 1;                //m次方            while(m>0)            {                if(m&1)                    res = res*temp;                temp = temp*temp;                m/=2;            }        sum = res*sum;//特别注意这里  是和res相乘            printf("%lld\n",sum.mat[n][0]);        }    }    return 0;}