HDU 4291 A Short problem（矩阵+循环节）

A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2711    Accepted Submission(s): 951

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

 

Source

2012 ACM/ICPC Asia Regional Chengdu Online

 

/*
 * Author: lyucheng
 * Created Time:  2017年05月20日 星期六 16时40分42秒
 * File Name: HDU-4291-A_Short_problem.cpp
 */
 /*
 * 题意:让你求g(g(g(n)))mod 1e9+7,其中g(n)=3*g(n-1)+g(n-2)
 *
 *
 * 思路:g(n)可以通过矩阵快速幂求出来,但是干后分别求出各自的循环节,能得到第一个循环节是222222224,
 *      第二个循环节是183120
 * */
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <time.h>
#define LL long long
#define maxn 3  
using namespace std;
LL n;
LL mod;
/********************************矩阵快速幂**********************************/
class Matrix {
    public:
        LL a[maxn][maxn];

        void init() {
            memset(a,0,sizeof(a));
        }
        
        Matrix operator +(Matrix b) {
            Matrix c;
            for (int i = 0; i < 2; i++)
                for (int j = 0; j < 2; j++)
                    c.a[i][j] = (a[i][j] + b.a[i][j]) % mod;
            return c;
        }

        Matrix operator +(LL x) {
            Matrix c = *this;
            for (int i = 0; i < 2; i++)
                c.a[i][i] += x;
            return c;
        }

        Matrix operator *(Matrix b)
        {
            Matrix p; 
            p.init();
            for (int i = 0; i < 2; i++)
                for (int j = 0; j < 2; j++)
                for (int k = 0; k < 2; k++)
                    p.a[i][j] = (p.a[i][j] + (a[i][k]*b.a[k][j])%mod) % mod;
            return p;
        }

        Matrix power(LL t) {
            Matrix ans,p = *this;
            ans.init();
            ans.a[0][0]=1;
            ans.a[1][1]=1;
            while (t) {
                if (t & 1)
                    ans=ans*p;
                p = p*p;
                t >>= 1;
            }
            return ans;
        }
}unit,init;
/********************************矩阵快速幂**********************************/
int main(int argc, char* argv[])
{
    // freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout;
    while(scanf("%lld",&n)!=EOF){
        if(n<2){
            printf("%lld
",n);
            continue;
        }
        //首先求最里面的g(n)
        mod=183120;
        unit.init();
        unit.a[0][0]=1,unit.a[0][1]=0,unit.a[1][0]=0,unit.a[1][1]=0;
        init.a[0][0]=3,init.a[0][1]=1,init.a[1][0]=1,init.a[1][1]=0;
        init=init.power(n-1);
        unit=unit*init;
        n=unit.a[0][0];
        
        if(n>=2){//很重要,要不然当n<2的时候矩阵乘法就会死循环
            mod=222222224;
            unit.init();
            unit.a[0][0]=1,unit.a[0][1]=0,unit.a[1][0]=0,unit.a[1][1]=0;
            init.a[0][0]=3,init.a[0][1]=1,init.a[1][0]=1,init.a[1][1]=0;
            init=init.power(n-1);
            unit=unit*init;
            n=unit.a[0][0];
        }
        
        if(n>=2){//很重要,要不然当n<2的时候矩阵乘法就会死循环
            mod=1000000007;
            unit.init();
            unit.a[0][0]=1,unit.a[0][1]=0,unit.a[1][0]=0,unit.a[1][1]=0;
            init.a[0][0]=3,init.a[0][1]=1,init.a[1][0]=1,init.a[1][1]=0;
            init=init.power(n-1);
            unit=unit*init;
            n=unit.a[0][0];
        }
        printf("%lld
",n);
    }
    return 0;
}