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
题意：按f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10) (x>=10) ； f(x) = x（x<10）来计算f(x)%m的值。
分析：这题要用递推，并且k值很大，所以需要用矩阵快速幂。
构造的矩阵是：
|
* |
|
= |
|
写个结构类型代表矩阵，以及矩阵的相乘的函数和矩阵快速幂的函数，注意一下初始化。
#include<stdio.h>
#include<string.h>
int n,k,m;
struct matrix
{
int a[15][15];
int row,col;
void init(int r,int c){
memset(a,0,sizeof(a));
row=r;col=c;
}
} big,f,u;
matrix mul(matrix a,matrix b)
{
matrix c;
c.init(a.row,b.col);
for(int i=0; i<a.row; i++)
for(int j=0; j<b.col; j++)
{
for(int k=0; k<a.col; k++)
c.a[i][j]+=(a.a[i][k]*b.a[k][j])%m;
c.a[i][j]%=m;
}
return c;
}
void init()
{
big.init(10,10);
f.init(10,1);
for(int i=1; i<10; i++)
big.a[i][i-1]=1;
for(int i=0; i<10; i++)
f.a[i][0]=9-i;
}
matrix qpow(matrix a,int k)
{
matrix ans;
ans.init(a.row,a.col);
for(int i=0;i<a.row;i++)
ans.a[i][i]=1;
while(k)
{
if(k&1)ans=mul(ans,a);
a=mul(a,a);
k>>=1;
}
return ans;
}
int main()
{
init();
while(~scanf("%d%d",&k,&m))
{
for(int i=0; i<10; i++)
scanf("%d",&big.a[0][i]);
u=mul(qpow(big,k-9),f);
printf("%d\n",u.a[0][0]%m);
}
return 0;
}