UVA - 10870 Recurrences (矩阵快速幂)
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 twosets of d non-negative integers. The first set contains coefficients: a1,a2, ..., ad. 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) (mod m)on a separate line. It must be a non-negative integer, less than m.
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.
题意：考虑一个递推关系 f[n] = a1*f[n-1] + a2*f[n-2] + .... ad*f[n-d],求f[n] % m
思路：和经典的Fibonacci的递推矩阵构造类似
#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;
}