【A Simple Math Problem】【HDU - 1757 】（矩阵快速幂）
题目：
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
解题报告：求解一个奇妙的数学问题，特殊的函数，当参数是小于10的时候返回自己，然后大于10的话，满足等式 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10)。利用矩阵快速幂就可以求解，难点就是想到利用矩阵快速幂，中间的取模是关键，要不然会wa。
ac代码：
#include
#include
#include
#include
#define maxn 10
using namespace std;
typedef long long ll;
int k;
int m;
struct Mat{
ll v[maxn][maxn];
Mat ()
{
memset(v,0,sizeof(v));
for(int i=0;i<10;i++)
v[i][i]=1;
}
};
Mat f;
Mat temp;
Mat mul(Mat a,Mat b)
{
Mat res;
for(int i=0;i>=1;
}
res=mul(res,a);
return res.v[0][0];
}
int main()
{
int i;
for(i=0;i<10;i++)
{
f.v[i][0]=9-i;
for(int j=1;j<10;j++)
f.v[i][j]=0;
}
while(scanf("%d%d",&k,&m)!=EOF)
{
for(int i=0;i