hdu 4291 A Short problem

A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 779    Accepted Submission(s): 296

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

 

Recommend

liuyiding

//矩阵快速运算 
//| 3 1 |
//| 1 0 |
由这个矩阵进行幂运算 ，今天上数据结构课时，老师讲些无聊的线性表，就回想了下这题和斐波那契的矩阵运算，在纸上才发现 递推都可以用矩阵快速运算的 、O(∩_∩)O~
//可是上次网赛没发现呀
// 当然由于数比较大，需要找出循环节、、然后就 0Ms 过了
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 500002
using namespace std;
__int64 Matrix(__int64 n,__int64 M)
{
    __int64 a[2][2]={3,1,1,0};
    __int64 b[2][2]={1,0,0,1};
    __int64 c[2][2];
    for(;n>0;n=n>>1)
    {
        if(n&1)
        {
         c[0][0]=(b[0][0]*a[0][0]%M+b[0][1]*a[1][0]%M)%M;
         c[1][0]=(b[1][0]*a[0][0]%M+b[1][1]*a[1][0]%M)%M;
         c[0][1]=(b[0][0]*a[0][1]%M+b[0][1]*a[1][1]%M)%M;
         c[1][1]=(b[1][0]*a[0][1]%M+b[1][1]*a[1][1]%M)%M;
         memcpy(b,c,sizeof(c));
        }
         c[0][0]=(a[0][0]*a[0][0]%M+a[0][1]*a[1][0]%M)%M;
         c[1][0]=(a[1][0]*a[0][0]%M+a[1][1]*a[1][0]%M)%M;
         c[0][1]=(a[0][0]*a[0][1]%M+a[0][1]*a[1][1]%M)%M;
         c[1][1]=(a[1][0]*a[0][1]%M+a[1][1]*a[1][1]%M)%M;
        memcpy(a,c,sizeof(c));
    }
    return b[1][0];
}
int main()
{
    __int64 n;
    __int64 tmp;
    while(scanf("%I64d",&n)!=EOF)
    {
        if(!n) printf("0\n");
        else
        {
            tmp=Matrix(n,183120);
            tmp=Matrix(tmp,222222224);
            printf("%I64d\n",Matrix(tmp,1000000007));
        }
    }
    return 0;
}