uva 12470(矩阵快速幂)

题意：公式f(n) = f(n - 1) + f(n - 2) + f(n - 3)，给出n，f(1) = 0，f(2) = 1， f(3) = 2，要求得出f(n)。

题解：普通的矩阵快速幂模板题。

#include <stdio.h>
#include <string.h>
const int MOD = 1000000009;
struct Mat {
    long long g[3][3];
}ori, res;
long long n;

Mat multiply(Mat x, Mat y) {
    Mat temp;
    for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++) {
            temp.g[i][j] = 0;
            for (int k = 0; k < 3; k++)
                temp.g[i][j] = (temp.g[i][j] + x.g[i][k] * y.g[k][j]) % MOD;
        }
    return temp;
}

void calc(long long n) {
    while (n) {
        if (n & 1)
            ori = multiply(ori, res);
        n >>= 1;
        res = multiply(res, res);
    }
}

int main() {
    while (scanf("%lld", &n) && n) {
        if (n == 1 || n == 2 || n == 3) {
            printf("%lld\n", n - 1);
            continue;
        }
        memset(ori.g, 0, sizeof(ori));
        memset(res.g, 0, sizeof(res));
        ori.g[0][0] = 2;
        ori.g[0][1] = 1;
        ori.g[0][2] = 0;
        res.g[0][0] = res.g[1][0] = res.g[2][0] = 1;
        res.g[0][1] = res.g[1][2] = 1;
        calc(n - 3);
        printf("%lld\n", ori.g[0][0]);
    }
    return 0;
}