HDU - 1757 A Simple Math Problem （构造矩阵）

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 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0

 

Sample Output

45104

 题意：求f(k)%m
思路：还是构造矩阵跟HDU-2604类似

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;const int maxn = 10;int n, m;struct Matrix {int v[maxn][maxn];Matrix() {}Matrix(int x) {init();for (int i = 0; i < maxn; i++) v[i][i] = x;}void init() {memset(v, 0, sizeof(v));}Matrix operator *(Matrix const &b) const {Matrix c;c.init();for (int i = 0; i < maxn; i++)for (int j = 0; j < maxn; j++)for (int k = 0; k < maxn; k++)c.v[i][j] = (c.v[i][j] + (v[i][k]*b.v[k][j])%m) % m;return c;}Matrix operator ^(int b) {Matrix a = *this, res(1);while (b) {if (b & 1)res = res * a;a = a * a;b >>= 1;}return res;}} a, b, tmp;int main() {while (scanf("%d%d", &n, &m) != EOF) {a.init();for (int i = 1; i < 10; i++)a.v[i][i-1] = 1;for (int i = 0; i < 10; i++)scanf("%d", &a.v[0][i]);if (n < 10) {printf("%d\n", n%m);continue;}tmp = a^(n-9);int ans = 0;for (int i = 0; i < 10; i++)ans = (ans + tmp.v[0][i]*(9-i)) % m;printf("%d\n", ans);}return 0;}