HDU5015 233 Matrix

题意

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4039    Accepted Submission(s): 2287

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

 

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分析

参照whatbeg的题解。

题意：给出矩阵的第0行（233,2333,23333,...）和第0列a1,a2,...an（n<=10，m<=10^9），给出式子： A[i][j] = A[i-1][j] + A[i][j-1]，要求A[n][m]。

解法：看到n<=10和m<=10^9 应该对矩阵有些想法，现在我们假设要求A[a][b],则A[a][b] = A[a][b-1] + A[a-1][b] = A[a][b-1] + A[a-1][b-1] + A[a-2][b] = ...

这样相当于右图：，红色部分为绿色部分之和，而顶上的绿色部分很好求，左边的绿色部分(最多10个)其实就是：A[1][m-1],A[2][m-1]..A[n][m-1],即对每个1<=i<=n, A[i][m]都可由A[1][m-1],A[2][m-1]..A[n][m-1]，于是建立12*12的矩阵：，将中间矩阵求m-1次幂，与右边[A[0][1],A[1][1]..A[n][1],3]^T相乘，结果就可以得出了。

时间复杂度\(O(n^3 \log m)\)

代码

#include<iostream>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
co int N=12,mod=10000007;
int n,m,t,A[N][N],ANS[N][N];
void mul(int a[N][N],int b[N][N]){
    static int c[N][N];
    for(int k=0;k<t;++k)
        for(int j=0;j<t;++j)if(b[k][j])
            for(int i=0;i<t;++i)if(a[i][k])
                (c[i][j]+=(ll)a[i][k]*b[k][j]%mod)%=mod;
    for(int i=0;i<t;++i)for(int j=0;j<t;++j)
        b[i][j]=c[i][j],c[i][j]=0;
}
int main(){
//  freopen(".in","r",stdin),freopen(".out","w",stdout);
    while(~scanf("%d%d",&n,&m)){
        memset(ANS,0,sizeof ANS),memset(A,0,sizeof A);
        ANS[0][0]=23,ANS[n+1][0]=3;
        for(int i=1;i<=n;++i) ANS[i][0]=read<int>()%mod;
        for(int i=0;i<=n;++i) A[i][0]=10,A[i][n+1]=1;
        for(int i=1;i<=n;++i)for(int j=i;j<=n;++j) A[j][i]=1;
        A[n+1][n+1]=1;
        for(t=n+2;m;m>>=1,mul(A,A))
            if(m&1) mul(A,ANS);
        printf("%d\n",ANS[n][0]);
    }
    return 0;
}