UVa 10870 Recurrences / 矩阵快速幂

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给你一个数列的前d项第n项(n > d) f(n) = a₁ f(n - 1) + a₂ f(n - 2) + a₃ f(n - 3) + ... + a_d f(n - d), for n > d.

n很大可以构造一个矩阵

f(n) = A*f(n-1)

例如n=5

0   1   0   0   0             f[1]                           f[2]0   0   1   0   0             f[2]                           f[3]0   0   0   1   0      *      f[3]            =              f[4]                   0   0   0   0   1             f[4]                           f[5]a5  a4  a3  a2  a1            f[5]                           f[6]

f[n] = A^(n-d)*f[d];f[n] = A^(n-d)*f[d];

所以可以快速幂出A矩阵的n-d次在乘以f[d]

#include <cstdio>#include <cstring>const int maxn = 20;struct Matrix{long long a[maxn][maxn];};Matrix a, c;long long b[maxn];long long n, m;int d;Matrix matrix(Matrix x, Matrix y){Matrix z;memset(z.a, 0, sizeof(z.a));for(int i = 1; i <= d; i++){for(int j = 1; j <= d; j++){for(int k = 1; k <= d; k++){z.a[i][j] += x.a[i][k] * y.a[k][j];z.a[i][j] %= m;}}}return z;}void matrix_pow(long long n){while(n){if(n&1)c = matrix(c, a);a = matrix(a, a);n >>= 1;}}int main(){while(scanf("%d %d %d", &d, &n, &m), d || n || m){memset(a.a, 0, sizeof(a.a));memset(c.a, 0, sizeof(c.a));for(int i = d; i >= 1; i--)scanf("%d", &a.a[d][i]);for(int i = 1; i <= d; i++)scanf("%d", &b[i]);for(int i = 1; i < d; i++)a.a[i][i+1] = 1;for(int i = 1; i <= d; i++)c.a[i][i] = 1;long long ans = 0;if(d < n){matrix_pow(n-d);//n-d个矩阵a相乘存在c c初始化为单位矩阵 for(int i = 1;i <= d; i++){ans += c.a[d][i]*b[i];ans %= m;}}elseans = b[n]%m;printf("%d\n", ans);} return 0;}/*1 1 10021 2 10 1001 11 1 3 2147483647 1234512345678 0 123451 2 30 0 0*/

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