HDU 4291 A Short problem

　　矩阵快速幂。最外层的取模是1e9+7，第二层的取模 L1 = （1e9+7的循环节），第三层的取模是 L2 = (L1的循环节)

暴力找出L2, L1, 然后矩阵快速幂即可。。。

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   x < y ? x : y
#define Max(x, y)   x < y ? y : x
#define E(x)    (1 << (x))
#define iabs(x) (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()    freopen("data.in", "r", stdin)
#define Write()   freopen("data.out", "w", stdout);

const double eps = 1e-8;
typedef long long LL;
const int inf = ~0u>>2;

using namespace std;

int mod;
int l2 = 183120;
int l1 = 222222224;

struct Mat {
    LL mat[3][3];
    void init() {
        mat[0][0] = 3; mat[0][1] = 1;
        mat[1][0] = 1; mat[1][1] = 0;
    }
}a;

Mat operator * (Mat a, Mat b) {
    Mat c;
    CL(c.mat, 0);
    int i, j, k;
    REP(k, 2) {
        REP(i, 2) {
            if(a.mat[i][k] <= 0)    continue;
            REP(j, 2) {
                if(b.mat[k][j] <= 0)    continue;
                c.mat[i][j] = (c.mat[i][j] + (a.mat[i][k] * b.mat[k][j])%mod)%mod;
            }
        }
    }
    return c;
}

Mat operator ^ (Mat a, LL k) {
    Mat c;
    int i, j;
    REP(i, 2)    REP(j, 2)    c.mat[i][j] = (i == j);
    for(; k; k >>= 1) {
        if(k&1)    c = c*a;
        a = a*a;
    }
    return c;
}

LL solve(LL n, LL m) {
    a.init();
    mod = m;
    a = a^n;
    return a.mat[0][1];
}

int main() {
    //freopen("data.in", "r", stdin);
    LL n;
    while(cin >> n) {
        cout << solve(solve(solve(n, l2), l1), 1e9 + 7) << endl;
    }
    return 0;
}