HDU 1757 A Simple Math Problem

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

 

矩阵快速幂，但是我没有自己推出来，感觉都好神（是你太弱了吧。。）。别忘了取模，各种地方都要取！

CODE:

#include <iostream>
#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 10 + 5

using namespace std;

int k, m;
struct node{
    int Mtx[MAX_N][MAX_N];
}med;

void Pre_Set(){
    REP(i, 1, 10) scanf("%d", &med.Mtx[1][i]);
    REP(i, 2, 10) REP(j, 1, 10){
        if(i - j == 1) med.Mtx[i][j] = 1;
        else med.Mtx[i][j] = 0;
    } 
}

node operator*(node a,node b){
    node c;
    REP(i, 1, 10) REP(j, 1, 10){
        c.Mtx[i][j] = 0;
        REP(k, 1, 10) c.Mtx[i][j] += a.Mtx[i][k] * b.Mtx[k][j]; 
        c.Mtx[i][j] %= m;
    }
    return c;
}

node operator^(node a,int k){
    if(k == 0){
        memset(a.Mtx, 0, sizeof(a.Mtx));
        REP(i, 1, 10) a.Mtx[i][i] = 1;
        return a;
    }
    if(k == 1) return a;
    node c = a ^ (k >> 1);
    if(k & 1) return c * c *a;
    return c * c;
}

int main(){
    while(scanf("%d%d", &k, &m) != EOF){
        Pre_Set();
        if(k < 10) { cout << k % m << endl; }
        else {
            med = med ^ (k - 9);
            int ans = 0;
            REP(i, 1, 10){
                ans = (ans + med.Mtx[1][i] * (10 - i)) % m;
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}