UV-10870 recurrences (matrix power)

Consider recurrent functions ofthe following form:

F (n) = A1 F (n-1) + A2 F (n-2) + A3 F (n-3) +... + Ad f (n-d), for N> D.
A1, A2,..., ad-Arbitrary Constants.

A famous example is the Fibonacci sequence, defined as: F (1) = 1, F (2) = 1, F (n) = f (n-1) + f (n-2 ). here d = 2, a1 = 1, a2 = 1.

Every such function is completely described byspecifying D (which is called the order of recurrence), values of dcoefficients: A1, A2 ,..., AD, and values off (1), F (2 ),..., F (d ). you'll be given these numbers, and two integers n and M. your program's job is to compute F (n) Modulo M.

Input

Input file contains several test cases. Each testcase begins with three integers:D,N,M, Followed by twosetsDNon-negative integers. The first set contains coefficients: A1, A2,...,D. The second set gives values of F (1), F (2),..., F (D).

You can assume that: 1 <=D<= 15, 1 <=N<= 231-1, 1 <=M<= 46340. allnumbers in the input will fit in signed 32-bit integer.

Input is terminated by line containing threezeroes instead of d, n, m. Two consecutive test cases are separated by a blankline.

Output

For each test case, print the value of F (N) (ModM) On a separate line. It must be a non-negative integer, lessM.

 

Sampleinput output for sample input

1 1 100 2 1   2 10 100 1 1 1 1   3 2147483647 12345 12345678 0 12345 1 2 3   0 0 0

1 55 423

Problem setter: maxfurlong

Special thanks: derekkisman, EPS.

Consider a recursive relationship f [N] = A1 * f [n-1] + A2 * f [N-2] + .... ad * f [n-d], evaluate f [N] % m

Idea: similar to the recursive matrix structure of the classic Fibonacci

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>typedef long long ll;using namespace std;const int maxn = 30;int n, m, d;struct Matrix {ll 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 < d; i++)for (int j = 0; j < d; j++)for (int k = 0; k < d; 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 tmp = *this, res(1);while (b) {if (b & 1)res = res * tmp;tmp = tmp * tmp;b >>= 1;}return res;}} a, b, res;ll f[maxn];int main() {while (scanf("%d%d%d", &d, &n, &m) != EOF && n+m+d) {a.init();b.init();for (int i = 0; i < d; i++)scanf("%lld", &a.v[0][i]);for (int i = 0; i < d; i++)scanf("%lld", &f[i]);for (int i = 1; i < d; i++)a.v[i][i-1] = 1;if (n == 1) {printf("%lld\n", f[n-1]);continue;}b = a ^ (n-d);ll ans = 0;for (int i = 0; i < d; i++)ans = (ans + (f[d - 1 - i] * b.v[0][i]) % m) % m;printf("%lld\n", ans);}return 0;}

UV-10870 recurrences (matrix power)