UVa 10870 (矩阵快速幂) Recurrences

给出一个d阶线性递推关系，求f(n) mod m的值。

,

 求出A^n-dv₀，该向量的最后一个元素就是所求。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 20;

typedef long long Matrix[maxn][maxn];
typedef long long Vector[maxn];

int d, n, m;

void matrix_mul(Matrix A, Matrix B, Matrix res)
{
    Matrix C;
    memset(C, 0, sizeof(C));
    for(int i = 0; i < d; i++)
        for(int j = 0; j < d; j++)
            for(int k = 0; k < d; k++)
                C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % m;
    memcpy(res, C, sizeof(C));
}

void matrix_pow(Matrix A, int n, Matrix res)
{
    Matrix a, r;
    memcpy(a, A, sizeof(a));
    memset(r, 0, sizeof(r));
    for(int i = 0; i < d; i++) r[i][i] = 1;
    while(n)
    {
        if(n&1) matrix_mul(r, a, r);
        n >>= 1;
        matrix_mul(a, a, a);
    }
    memcpy(res, r, sizeof(r));
}

int main()
{
    //freopen("in.txt", "r", stdin);

    while(cin >> d >> n >> m && d)
    {
        Matrix A;
        memset(A, 0, sizeof(A));
        Vector a, f;
        for(int i = 0; i < d; i++) { cin >> a[i]; a[i] %= m; }
        for(int i = 0; i < d; i++) { cin >> f[i]; f[i] %= m; }
        if(n <= d) { cout << f[n-1] << "
"; continue; }
        for(int i = 0; i < d-1; i++) A[i][i+1] = 1;
        for(int i = 0; i < d; i++) A[d-1][i] = a[d-i-1];
        matrix_pow(A, n-d, A);
        long long ans = 0;
        for(int i = 0; i < d; i++) ans = (ans + A[d-1][i]*f[i]) % m;
        cout << ans << "
";
    }

    return 0;
}