HDU - 1757 A Simple Math Problem 矩阵快速幂

解题思路：借用过了别人写的。。
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);

|f(10) | |a0 a1 a2 …a8 a9| |f(9)|
| f(9) | | 1 0 0 … 0 0| |f(8)|
| …..| = |.. … … … ..| | .. |
| f(2) | | 0 0 0 … 0 0| |f(1)|
| f(1) | | 0 0 0 … 1 0| |f(0)|

另A举证为10*10的举证，如上图。
可以推出：
(f(n),f(n-1),…,f(n-9))^(-1) = A^(n-9)*(f(9),f(8),…,f(0))^(-1)

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
#define N 10
struct Matrix{
    ll mar[N][N];
}a, b, tmp;
ll n, mod;

Matrix matrixMul(Matrix aa, Matrix bb) {
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++) {
            tmp.mar[i][j] = 0;
            for(int k = 0; k < N; k++)
                tmp.mar[i][j] += (aa.mar[i][k] * bb.mar[k][j]) % mod;
        }
    return tmp;
}

void solve() {
    while(n) {
        if(n & 1)
            b = matrixMul(b,a);
        a = matrixMul(a,a);
        n >>= 1;
    }
}

int main() {
    while(scanf("%I64d%I64d", &n, &mod) != EOF) {
        for(int i = 0; i < N; i++)
            for(int j = 0; j < N; j++)
                a.mar[i][j] = b.mar[i][j] = 0;

        for(int i = 0; i < N; i++) 
            b.mar[i][i] = 1;

        for(int i = 1; i < N; i++)
            a.mar[i][i-1] = 1;

        for(int i = 0; i < N; i++)
            scanf("%I64d", &a.mar[0][i]);
        if(n <= 9) {
            printf("%I64d\n", n);
            continue;
        }
        n -= 9;
        solve();
        long long ans = 0;
        for(int i = 0; i < N; i++) 
            ans += (b.mar[0][i] * (9 - i)) % mod; 
        printf("%I64d\n", ans % mod);
    }
    return 0;
}