HDU A Simple Math Problem （矩阵快速幂）

题面

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

思路

直接看就是快速幂，我们关注一下这个2*1e9的数据量，这个数据量的话，复杂度肯定要降到log级别的，所以快速幂并木有错误。那么重点还是构造矩阵，我们让第一列是题目给的递推式子，其余为单位偏移作用。做n-9次的快速幂后和f1-f9去相乘就好啦。那么有几个细节注意，矩阵乘法的时候要取两遍的mod，然后最后的矩阵乘法不要写反，矩阵的相乘并不满足交换律，还有就是数组不要开小，会re。这些矩阵快速幂的题目都一个套路，我们真正要做的就是去把题目中隐含的或者直接给的递推式求出来，然后构造相应的辅助矩阵，最后完成快速幂就好。

代码实现

#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;
const int inf =0x3f3f3f3f;
const int maxn=11;
struct Matrix {
    int maze[maxn][maxn];
    Matrix () {
        memset (maze,0,sizeof (maze));
    }
};
int n,mod;
Matrix mul (Matrix a,Matrix b) {
    Matrix ans;
    for (int i=1;i<=10;i++) 
     for (int j=1;j<=10;j++){
       for (int k=1;k<=10;k++) 
           ans.maze[i][j]=(ans.maze[i][j]+(a.maze[i][k]*b.maze[k][j])%mod)%mod;
       }
       return ans;
}
Matrix fats_pow (Matrix a,int p) {
    Matrix ans;
    for (int i=1;i<=10;i++) ans.maze[i][i]=1;
    while (p) {
        if (p&1) ans=mul (ans,a);
        a=mul(a,a);
        p>>=1;
    }
    return ans;
}
int main () { 
    while (cin>>n>>mod) {
        Matrix fz;
        for (int i=1;i<=10;i++) cin>>fz.maze[i][1];
        for (int i=1;i<=9;i++) fz.maze[i][i+1]=1;
        if (n<10) {
            cout<<n%mod<<endl;
            continue;
        } 
        else {
            Matrix k;
            for (int i=1;i<=10;i++) k.maze[1][i]=10-i;
            fz=fats_pow (fz,n-9);
            k=mul (k,fz);
            cout<<k.maze[1][1]%mod<<endl;
        }
    }   
    return 0;
}