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hdu 5015 233 Matrix （矩阵高速幂）

2024-10-22 00:03:41   211人阅读

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 749    Accepted Submission(s): 453


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 a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell me an,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, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).
 

Output
For each case, output an,m mod 10000007.
 

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

Sample Output
234
2799
72937


Hint


技术分享

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

依据前后两列的递推关系，有等式可得矩阵A的元素为：技术分享   

#include"iostream"
#include"stdio.h"
#include"string.h"
#include"algorithm"
#include"queue"
#include"vector"
using namespace std;
#define N 15
#define LL __int64
const int mod=10000007;
int n;
int b[N];
struct Mat
{
    LL mat[N][N];
}a,ans;
Mat operator*(Mat a,Mat b)
{
    int i,j,k;
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(i=0; i<=n+1; i++)
    {
        for(j=0; j<=n+1; j++)
        {
            c.mat[i][j]=0;
            for(k=0; k<=n+1; k++)
            {
                if(a.mat[i][k]&&b.mat[k][j])
                {
                    c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                    c.mat[i][j]%=mod;
                }
            }
        }
    }
    return c;
}
void mult(int k)
{
    int i;
    memset(ans.mat,0,sizeof(ans.mat));
    for(i=0;i<=n+1;i++)
        ans.mat[i][i]=1;
    while(k)
    {
        if(k&1)
            ans=ans*a;
        k>>=1;
        a=a*a;
    }
}
void inti()
{
    int i,j;
    b[0]=23;
    b[n+1]=3;
    for(i=1; i<=n; i++)
        scanf("%d",&b[i]);
    memset(a.mat,0,sizeof(a.mat));
    for(i=0; i<=n; i++)
    {
        a.mat[i][0]=10;
        a.mat[i][n+1]=1;
    }
    a.mat[n+1][n+1]=1;
    for(i=1; i<n+1; i++)
    {
        for(j=1; j<=i; j++)
        {
            a.mat[i][j]=1;
        }
    }
}
int main()
{
    int i,m;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        inti();
        mult(m);
        LL s=0;
        for(i=0;i<=n+1;i++)
            s=(s+(ans.mat[n][i]*b[i])%mod)%mod;
        printf("%I64d\n",s);
    }
    return 0;
}




hdu 5015 233 Matrix （矩阵高速幂）


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