UVA - 12470 Tribonacci 矩阵快速幂

题目大意：fibonacci的升级版，规则是f(n) = f(n-1) + f(n-2) + f(n-3)

解题思路：水题

#include<cstdio>
typedef long long ll;
const int N = 3;
const ll mod = 1e9 + 9;

struct Matrix{
    ll mat[N][N];
}A, B, tmp;
ll n;

void init(){ 
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++) {
            A.mat[i][j] = B.mat[i][j] = 0;
            if(i == j) 
                B.mat[i][j] = 1;
        }
    A.mat[0][0] = A.mat[1][0] = A.mat[2][0] = A.mat[0][1] = A.mat[1][2] = 1;
}

Matrix matMul(Matrix &x, Matrix &y) {
    for(int i = 0; i < N; i++) 
        for(int j = 0; j < N; j++) {
            tmp.mat[i][j] = 0;
            for(int k = 0; k < N; k++) 
                tmp.mat[i][j] = (tmp.mat[i][j] + (x.mat[i][k] * y.mat[k][j]) % mod ) % mod;
        }
    return tmp;
}

void solve() {
    while(n) {
        if(n & 1) 
            B = matMul(B,A);
        A = matMul(A,A);
        n >>= 1;
    }
}

int main() {
    while(scanf("%lld", &n) != EOF && n) {
        if(n == 3 || n == 1 || n == 2) {
            printf("%lld\n", n - 1);
            continue;
        }
        n -= 3;
        init();
        solve();
        ll ans = 0;
        ans = (ans + 2 * B.mat[0][0]) % mod;
        ans = (ans + B.mat[1][0]) % mod;
        printf("%lld\n", ans);
    }
    return 0;
}