hdu 5015 233 Matrix 矩阵快速幂

233 Matrix Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 98    Accepted Submission(s): 47

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 
     
 
      
    
   
   
   
   
   
 
   
 
   
    
    
    
     
  
  
  
  
  

  

  
   
   
   
   

构造矩阵就行了。具体看代码把，可以用print 把矩阵打出来看；

不知道为什么G++能过，C++数组越界，路过大神求指点。

#include<stdio.h>
#include<string.h>
#define Matr 15 //矩阵大小,注意能小就小   矩阵从1开始   所以Matr 要+1
#define ll __int64
struct mat//矩阵结构体，a表示内容，size大小 矩阵从1开始   但size不用加一
{
    ll a[Matr][Matr];
    mat()//构造函数
    {
        memset(a,0,sizeof(a));
    }
};
int Size;
ll mod= 10000007;

mat multi(mat m1,mat m2)//两个相等矩阵的乘法，对于稀疏矩阵，有0处不用运算的优化
{
    mat ans=mat();
    for(int i=1;i<=Size;i++)
        for(int j=1;j<=Size;j++)
            if(m1.a[i][j])//稀疏矩阵优化
                for(int k=1;k<=Size;k++)
                    ans.a[i][k]=(ans.a[i][k]+m1.a[i][j]*m2.a[j][k])%mod; //i行k列第j项
    return ans;
}

mat quickmulti(mat m,ll n)//二分快速幂
{
    mat ans=mat();
    int i;
    for(i=1;i<=Size;i++)ans.a[i][i]=1;
    while(n)
    {
        if(n&1)ans=multi(m,ans);
        m=multi(m,m);
        n>>=1;
    }
    return ans;
}

void print(mat m)//输出矩阵信息，debug用
{
    int i,j;
    printf("%d\n",Size);
    for(i=1;i<=Size;i++)
    {
        for(j=1;j<=Size;j++)
<span style="white-space:pre">            </span>printf("%I64d ",m.a[i][j]);
        printf("\n");
    }
}
int main()
{
    int i;
    mat gou,chu=mat();//构造矩阵  初始矩阵
    int n,m;
    __int64 ans;
    while(scanf("%I64d%I64d",&n,&m)!=EOF)//n+2*n+2
    {
        Size=n+2;
        chu=mat();
        chu.a[1][n+1]=23;
        for(i=1;i<=n;i++)
        {
            scanf("%I64d",&chu.a[1][i]);
            chu.a[1][i]%=mod;
        }
        chu.a[1][n+2]=3;
        gou=mat();
        for(int j=1;j<=n;j++)
        {
            if(j==1)
            {
                gou.a[n+1][j]=10;
                gou.a[n+2][j]=1;
            }
            gou.a[j][j]=1;
            for(int i=1;i<=n+2;i++)
            {
                gou.a[i][j]+=gou.a[i][j-1];
            }
        }
        gou.a[n+1][n+1]=10;
        gou.a[n+2][n+2]=1;
        gou.a[n+2][n+1]=1;
        ans=multi(chu,quickmulti(gou,m)).a[1][n];
        printf("%I64d\n",ans%mod);
    }
    return 0;
}