A Simple Math Problem（矩阵快速幂）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2547    Accepted Submission(s): 1485

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

      

      题意：

      给出 k（ < 2 X 10 ^ 9） 和 m（ < 10 ^ 5），数列的规则：

      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);
      求出 f(k)%m 。

 

      思路：

      矩阵快速幂。根据式子可以得出：

 

化简后可以得出 ：

 

 

 

      AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

typedef vector<int> vec;
typedef vector<vec> met;

met mul (met a, met b, int mod) {
    met c(a.size(), vec(b[0].size()));

    for (int i = 0; i < a.size(); ++i) {
        for (int j = 0; j < b[0].size(); ++j) {
            for (int k = 0; k < b.size(); ++k) {
                c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod;
            }
        }
    }

    return c;
}

met pow (met a, int n, int mod) {

    met b(a.size(), vec(a[0].size()));
    for (int i = 0; i < a.size(); ++i) {
        b[i][i] = 1;
    }

    while (n > 0) {
        if (n & 1) b = mul(b, a, mod);
        a = mul(a, a, mod);
        n >>= 1;
    }

    return b;
}

int main () {
    int k, mod;

    while (~scanf("%d%d", &k, &mod)) {
        met a(10, vec(10));

        for (int i = 0; i < a[0].size(); ++i)
            scanf("%d", &a[0][i]);

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

        if (k >= 10) {
                int sum = 0;
                a = pow(a, k - 9, mod);

                for (int i = 0; i < 10; ++i) {
                    sum = (sum + a[0][i] * (9 - i)) % mod;
                }

                printf("%d\n", sum % mod);
        } else printf("%d\n", k % mod);

    }

    return 0;
}