【矩阵快速幂】 hdu5015 233Matrix

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2178    Accepted Submission(s): 1273


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 ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).
 

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

转载:

很显然矩阵的第一列为:

0

a[1]

a[2]

a[3]

a[4]

我们转化一下,转化为

23

a[1]

a[2]

a[3]

a[4]

3

那么由第一列转移到第二列则为

23*10+3

a[1]+23*10+3

a[2]+a[1]+23*10+3

a[3]+a[2]+a[1]+23*10+3

a[4]+a[3]+a[2]+a[1]+23*10+3

3

很显然转移矩阵A就出来了:

10 0 0 0 0 1

10 1 0 0 0 1

10 1 1 0 0 1

10 1 1 1 0 1

10 1 1 1 1 1

 0  0 0 0 0 1

那么最后一列就是A的m次方*第一列。


我不会构造这题矩阵,要多练多想啊!!

#include
#include
#include
#include
#include
using namespace std;
typedef pair P;
typedef long long LL;
const int maxn=100;
#define N 15
#define mod 10000007

int n,m;
struct Matrix
{
    LL mat[N][N];
    Matrix()
    {
        memset(mat,0,sizeof(mat));
    }
};

Matrix mul(Matrix a,Matrix b)
{
    Matrix res;

    for(int i=0; i>=1;
    }
    return res;
}

void print()
{
    cout<<"The matrix -->"<

 手残,mod打错了,耽误了一小时 ;

手残,mod少打了一个零,耽误一小时;

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