*HDU 1757 矩阵乘法

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4307    Accepted Submission(s): 2586

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

 

题意：

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 .

上面的递推式计算f(k)。计算结果%m;

代码：

 //普通递推会超时，用矩阵乘法。构造矩阵时右上角的(n-1)*(n-1)矩阵置为单位矩阵，第n行从 右 到 左 填入系数，快速幂之后第n行作为行矩阵乘递推式从 小 到 大 排列的列矩阵得到结果
 //或者构造左下角的单位矩阵，第一行从 左 到 右 填入系数，快速幂之后第1行作为行矩阵乘递推式从 大 到 小 排列的列矩阵得到结果
 #include<bits\stdc++.h>
 using namespace std;
 int a[],f[]={,,,,,,,,,};
 int k,m;
 struct Lu
 {
     int mp[][];
 }L1;
 void init()
 {
     memset(L1.mp,,sizeof(L1.mp));
     for(int i=;i<=;i++)
     {
         L1.mp[i][i+]=;
     }
     for(int i=;i<=;i++)
     {
         L1.mp[][i]=a[-i];
     }
 }
 Lu mult(Lu a,Lu b)
 {
     Lu c;
     for(int i=;i<=;i++)
     for(int j=;j<=;j++)
     {
         c.mp[i][j]=;
         for(int k=;k<=;k++)
         c.mp[i][j]+=(a.mp[i][k]*b.mp[k][j])%m;
         c.mp[i][j]%=m;
     }
     return c;
 }
 Lu solve(int x)
 {
     if(x==)
     return L1;
     if(x&)
     {
         Lu q=solve(x-);
         return mult(q,L1);
     }
     else
     {
         Lu q=solve(x/);
         return mult(q,q);
     }
 }
 int main()
 {
     while(scanf("%d%d",&k,&m)!=EOF)
     {
         for(int i=;i<=;i++)
         scanf("%d",&a[i]);
         if(k<)
         {
             printf("%d\n",k%m);
             continue;
         }
         init();
         Lu tem=solve(k-);
         int ans=;
         for(int i=;i<=;i++)
         ans+=(tem.mp[][i]*f[i])%m;
         printf("%d\n",ans%m);
     }
     return ;
 }

 //普通递推会超时，用矩阵乘法。
 #include<bits\stdc++.h>
 using namespace std;
 int a[],f[]={,,,,,,,,,};
 int k,m;
 struct Lu
 {
     int mp[][];
 }L1;
 void init()
 {
     memset(L1.mp,,sizeof(L1.mp));
     for(int i=;i<=;i++)
     {
         L1.mp[i][i-]=;
     }
     for(int i=;i<=;i++)
     {
         L1.mp[][i]=a[i];
     }
 }
 Lu mult(Lu a,Lu b)
 {
     Lu c;
     for(int i=;i<=;i++)
     for(int j=;j<=;j++)
     {
         c.mp[i][j]=;
         for(int k=;k<=;k++)
         c.mp[i][j]+=(a.mp[i][k]*b.mp[k][j])%m;
         c.mp[i][j]%=m;
     }
     return c;
 }
 Lu solve(int x)
 {
     if(x==)
     return L1;
     if(x&)
     {
         Lu q=solve(x-);
         return mult(q,L1);
     }
     else
     {
         Lu q=solve(x/);
         return mult(q,q);
     }
 }
 int main()
 {
     while(scanf("%d%d",&k,&m)!=EOF)
     {
         for(int i=;i<=;i++)
         scanf("%d",&a[i]);
         if(k<)
         {
             printf("%d\n",k%m);
             continue;
         }
         init();
         Lu tem=solve(k-);
         int ans=;
         for(int i=;i<=;i++)
         ans+=(tem.mp[][i]*f[i])%m;
         printf("%d\n",ans%m);
     }
     return ;
 }