hdu 4291(矩阵快速幂 + 循环节)

题意：求s
s = g(g(g(n))) mod 1000000007
其中g(n)
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0
题解：普通的矩阵快速幂会超时，看到别人的题解是需要计算循环节得到小的MOD从而减小计算量。1000000007太大，需要计算更小的一个循环节，新技能get。

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

Mat multiply(Mat x, Mat y) {
    Mat temp;
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++) {
            temp.g[i][j] = 0;
            for (int k = 0; k < 2; 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() {
    /* 求循环节
    long long a = 0, b = 1;
    for (int i = 2;;i++) {
        a = (b * 3 + a) % 1000000007;
        a = a ^ b;
        b = a ^ b;
        a = a ^ b;
        printf("%lld %lld\n", a, b);
        if (a == 0 && b == 1)
            printf("%d\n", i);
    }
    i - 1 = 即为循环节
    */
    while (scanf("%lld", &n) == 1) {
        if (n == 0) {
            printf("0\n");
            continue;
        }
        if (n == 1) {
            printf("1\n");
            continue;
        }
        res.g[0][0] = 3; res.g[1][0] = res.g[0][1] = 1; res.g[1][1] = 0;
        ori.g[0][0] = 1; ori.g[1][0] = ori.g[0][1] = ori.g[1][1] = 0;
        MOD = (long long)183120; //g(g(g(n))) = g(g(y)) 那么183120是y的循环节
        calc(n - 1);
        n = ori.g[0][0];
        if (n != 0 && n != 1) {
            res.g[0][0] = 3; res.g[1][0] = res.g[0][1] = 1; res.g[1][1] = 0;
            ori.g[0][0] = 1; ori.g[1][0] = ori.g[0][1] = ori.g[1][1] = 0;
            MOD = (long long)222222224; //g(g(g(n))) = g(x) 那么222222224是x的循环节
            calc(n - 1);
            n = ori.g[0][0];
        }
        if (n != 0 && n != 1) {
            res.g[0][0] = 3; res.g[1][0] = res.g[0][1] = 1; res.g[1][1] = 0;
            ori.g[0][0] = 1; ori.g[1][0] = ori.g[0][1] = ori.g[1][1] = 0;
            MOD = (long long)1000000007;
            calc(n - 1);
        }
        printf("%lld\n", ori.g[0][0]);
    }
    return 0;
}