UVA 10870 recurrences (math: Matrix fast Power)

Give a recursive relationship that allows you to find the nth item in the relationship

It is common to think that recursion with large data can consider using matrix fast power

Here the matrix is no longer described, there is a petition

The code is as follows:

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAXN usi

NG namespace Std;
int X[MAXN];
int MOD, n, D; struct Matrix {int M[MAXN][MAXN];}

A, F;
    Matrix Mul (Matrix A, matrix B) {matrix res;
            for (int i=1, i<=d; ++i) {for (int j=1; j<=d; ++j) {res.m[i][j] = 0;
            for (int k=1; k<=d; ++k) {res.m[i][j] = (res.m[i][j]+a.m[i][k]*b.m[k][j]%mod+mod)%mod;
}}} return res;
    } Matrix Pow (matrix A, int b) {Matrix res;
    memset (res.m, 0, sizeof (RES.M));
    for (int i=1; i<=d; ++i) res.m[i][i] = 1;
        while (b) {if (b & 1) res = Mul (res, a);
        A = Mul (A, a);
    b >>= 1;
} return res; } int main (void) {while (~scanf ("%d%d%d", &d, &n, &mod) && (d| | n| |
        MOD) {memset (a.m., 0, sizeof (a.m.));

        memset (f.m, 0, sizeof (F.M)); for (int i=1; i&Lt;=d;
        ++i) scanf ("%d", &x[i]);
        for (int i=1; i<d; ++i) a.m[i][i+1] = 1;
        for (int i=1; i<=d; ++i) {a.m[d][i] = X[d-i+1]%mod;
            } for (int i=1; i<=d; ++i) {scanf ("%d", &f.m[i][1]);
        F.M[I][1]%= MOD;
        } Matrix tmp = Pow (A, n-d);
        Matrix ans = Mul (tmp, F);
    printf ("%d\n", (ans.m[d][1]+mod)%mod);
} return 0; }