HDOJ 1757 A Simple Math Problem

矩阵连乘+递归形式的快速幂

盗用一张图：

把问题转化为求矩阵的n-9次幂就行了；

A Simple Math Problem

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 4   Accepted Submission(s) : 2

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

 

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

 

Output

For each case, output f(k) % m in one line.

 

Sample Input

10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0

 

Sample Output

45
104

 

Author

linle

 

Source

2007省赛集训队练习赛（6）_linle专场

 

#include<cstdio>
#include<cstring>
const int N=10;
using namespace std;
int k,m;

struct Matrix
{
    int map[N][N];
};

Matrix matrix;

void init()
{
     for(int i=0;i<N;i++)
     {
         scanf("%d",&matrix.map[0][i]);
     }
     for(int i=1;i<N;i++)
     {
         for(int j=0;j<N;j++)
         {
             if(i==(j+1))matrix.map[i][j]=1;
             else matrix.map[i][j]=0;
         }
     }
}

Matrix mul(Matrix a,Matrix b)
{
     Matrix c;
     for(int i=0;i<N;i++)
    {
         for(int j=0;j<N;j++)
        {
             c.map[i][j]=0;
             for(int k=0;k<N;k++)
             {
                 c.map[i][j]+=a.map[i][k]*b.map[k][j];
             }
             c.map[i][j]%=m;
         }
     }
     return c;
 }

 Matrix quickpow(int k)
 {
     if(k==1)  return matrix;
     if(k&1) return mul(matrix,quickpow(k-1));
     else
     {
         Matrix temp=quickpow(k>>1);
         return mul(temp,temp);
     }
 }

 int main()
 {
     while(scanf("%d%d",&k,&m)!=EOF)
     {
         init();
         if(k<10)
         {
             printf("%d\n",k%m);
             continue;
         }
         Matrix temp=quickpow(k-9);
         int ans=0;
         for(int i=0;i<N;i++)
         {
             ans+=temp.map[0][i]*(N-i-1);
             ans%=m;
         }
         printf("%d\n",ans);
     }
     return 0;
 }