hdu 4291(矩阵快速幂 + 循环节)

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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;}

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