233 Matrix题解

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1355 Accepted Submission(s): 806

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

    1 112 20 03 723 47 16 
  

Sample Output

    234279972937 
    
     Hint

Source

2014 ACM/ICPC Asia Regional Xi'an Online

思路：

将第一列改写成如下：

3(这个3是为了计算下一列而存在的)

然后计算出第二列:

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（为了继续计算下一列保留这个3）

然后计算第三列……

受以上的启发推导出公式：

最后，目测测试数据很大，应该是用快速幂矩阵。

AC C++：

#include <cstdio>      
#include <algorithm> 
#define MOD 10000007
#define INF 0x3f3f3f3f  
#define N 10005  
//#define DEBUG
typedef long long LL; 
  
struct Matrix  
{  
    LL m[15][15];  
};  

int n; 
LL m;
LL a[20]; 
Matrix I; 

  
Matrix mult(Matrix a,Matrix b)  
{  
    Matrix c;  
    for(int i=0;i<n+2;i++)  
        for(int j=0;j<n+2;j++)  
        {  
             c.m[i][j]=0;  
             for(int k=0;k<n+2;k++)  
                 c.m[i][j]+=(a.m[i][k]*b.m[k][j]);  
             c.m[i][j]%=MOD;  
        }  
    return c;  
}  

Matrix quick_mod(Matrix a,LL n)  
{  
    Matrix c=I;  
    while(n)  
    {  
        if(n&1)  c=mult(c,a);  
        n>>=1;  
        a=mult(a,a);  
    }  
    return c;  
}  
  
int main()  
{  
    while(~scanf("%d%lld",&n,&m))  
    {  
        a[0]=3;  
        a[1]=23;  
        for(int i=2;i<n+2;i++)  
            scanf("%lld",&a[i]); 
			 
        for(int i=0;i<n+2;i++)  
            for(int j=0;j<n+2;j++)  
                if(i==j)  
                  I.m[i][j]=1;  
                else  
                  I.m[i][j]=0;
				    
#ifdef DEBUG
        for(int i=0;i<n+2;i++) 
        { 
            for(int j=0;j<n+2;j++) 
               printf("%lld ",I.m[i][j]); 
            printf("\n"); 
        }
#endif
		  
        Matrix A;  
        for(int i=0;i<n+2;i++)  
            for(int j=0;j<n+2;j++)  
                 if(i>=j)  
                    A.m[i][j]=1;  
                 else  
                    A.m[i][j]=0;  
        for(int i=1;i<n+2;i++)  
            A.m[i][1]=10;  
        LL ans=0;  
        Matrix B=quick_mod(A,m);
		  
#ifdef DEBUG
       for(int i=0;i<n+2;i++) 
        { 
            for(int j=0;j<n+2;j++) 
               printf("%lld ",B.m[i][j]); 
            printf("\n"); 
        }
#endif
		  
        for(int i=0;i<n+2;i++)  
            ans=(ans+B.m[n+1][i]*a[i]%MOD)%MOD;  
        printf("%lld\n",ans);  
    }  
    return 0;  
}

引用出处：

http://blog.csdn.net/ydd97/article/details/47684463

http://blog.csdn.net/lvshubao1314/article/details/39288393

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