A Short problem【矩阵快速幂+循环节】

A Short problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2920 Accepted Submission(s): 1023

Problem Description
　　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given n (1 <= n <= 1018), You should solve for
g(g(g(n))) mod 109 + 7

　　where
g(n) = 3g(n - 1) + g(n - 2)

g(1) = 1

g(0) = 0

Input
　　There are several test cases. For each test case there is an integer n in a single line.
　　Please process until EOF (End Of File).

Output
　　For each test case, please print a single line with a integer, the corresponding answer to this case.

Sample Input
0
1
2

Sample Output
0
1
42837

注意找循环节的方法。
要是不是线性的，就要用map< int int>这样来找到第一个重复的。。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
typedef long long ll;
int n;
ll u;
ll MOD = 1000000007;
ll k[3]= {183120, 222222224, 1000000007};//每次的取余，循环节
//据金桔所说，线性的规律循环节通常是跑到头：
//ll a = 0, b = 1;
//for(ll i=3;;i++)
//{
//    ll c= 3*a+b;
//    a  = b;
//    b = c;
//    if(a==0&&b==1)//因为线性终会跑到头
//    {
//        cout<<i<<endl;//循环节
//        break;
//    }
//}//跑完之后的答案就是之前的一个要取模的数，把这个数字赋值给MOD，然后再跑，就找到了第二个……
struct mat
{
    ll m[4][4];
} init, q;
ll mod(ll x)
{
    return (x+MOD)%MOD;
}
void init_mat()
{
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            init.m[i][j] = 0;
            q.m[i][j] = 0;
        }
    }
}

mat operator *(mat a, mat b)
{
    mat ret;
    ll x;
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            x = 0;
            for(int k=0; k<n; k++)
            {
                x += mod((ll)(a.m[i][k]*b.m[k][j]));
                x %=MOD;
            }
            ret.m[i][j] = x;
        }
    }
    return ret;
}

mat mat_pow(mat a, ll x)
{
    mat ret;
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            if(j==i)
                ret.m[i][j] = 1;
            else
                ret.m[i][j] = 0;
        }
    }
    while(x)
    {
        if(x&1)
            ret = ret * a;
        a = a * a;
        x>>=1;
    }
    return ret;
}
ll k_pow(ll a, ll b)
{
    ll ret = 1;
    while(b)
    {
        if(b&1)
            ret = mod(ret * a);
        a = mod(a * a);
        b>>=1;
    }
    return ret;
}
int main()
{
    n = 2;
    while(~scanf("%lld", &u))
    {
        u%=240;//第四个循环节，不加就超时
        if(u==0)
            cout<<0<<endl;
        else if(u==1)
            cout<<1<<endl;
        else
        {
            init_mat();
            q.m[0][0] = 1;
            q.m[1][0] = 0;
            init.m[0][0] = 3;
            init.m[0][1] = 1;
            init.m[1][0] = 1;
            mat a;
            for(int i=0; i<3; i++)
            {
                MOD = k[i];
                a = mat_pow(init, u-1);
                a = a*q;
                u = a.m[0][0];
            }
            printf("%lld\n", mod(u));
        }
    }
    return 0;
}