HDU----(4291)A Short problem(快速矩阵幂)

A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1716    Accepted Submission(s): 631

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

 

此题出得比较精妙，

分析:假设g(g(g(n)))=g(x),x可能超出范围,但是由于mod 10^9+7,所以可以求出x的循环节

求出x的循环节后,假设g(g(g(n)))=g(x)=g(g(y)),即x=g(y),y也可能非常大,但是由x的循环节可以求出y的循环节

如何求循环节点：

/*采用事先处理自己可以求出来*/
LL work(LL mod){  
  LL a=0,b=1;
    for(LL i=2;;++i)
    {    
      a=(b*3+a)%mod;
        a=a^b;
        b=a^b;
        a=a^b;
        if(a == 0 && b == 1)  return i;
 }

所以依次将mod1带入得到mod2=222222224;

然后将mod2带入得到mod3=183120;

然后就是快速矩阵了。

代码：

//#define LOCAL
#include<iostream>
#include<cstdio>
#include<cstring>
#define LL __int64
using namespace std;
const int mod1 =1000000007;
const int mod2=222222224;
const int mod3=183120;

LL mat[2][2];
LL ans[2][2];
LL n;

void Matrix(LL a[][2],LL b[][2],LL mod)
{
    LL cc[2][2]={0};
    for(int i=0;i<2;i++)
    {
      for(int j=0;j<2;j++)
      {
          for(int k=0;k<2;k++)
        {
            cc[i][j]=(cc[i][j]+a[i][k]*b[k][j])%mod;
        }
      }
    }
    for(int i=0;i<2;i++)
    {
      for(int j=0;j<2;j++)
      {
          a[i][j]=cc[i][j];
      }
    }
}

void pow(LL w,LL mod)
{
  while(w>0)
  {
    if(w&1) Matrix(ans,mat,mod);
     w>>=1;
     if(w==0)break;
     Matrix(mat,mat,mod);
  }
}
void input(LL w,LL mod)
{
     mat[0][0]=3;
     mat[0][1]=mat[1][0]=1;
     mat[1][1]=0;
     ans[0][0]=ans[1][1]=1;
     ans[0][1]=ans[1][0]=0;
     pow(w,mod);
//     printf("%I64d
",ans[0][0]);
}
void work(int i,__int64 w)
{
  if(i==4||w==0||w==1)
  {
      if(w==0)
         printf("0
");
      else if(w==1)
          printf("1
");
      else
      printf("%I64d
",ans[0][0]);
      return ;
  }
  LL mod;
  if(i==1)mod=mod3;
  else if(i==2)mod=mod2;
  else if(i==3)mod=mod1;
  input(w-1,mod);
  work(i+1,ans[0][0]);
}
int main()
{
  #ifdef LOCAL
   freopen("test.in","r",stdin);
  #endif
  while(scanf("%I64d",&n)!=EOF)
     work(1,n);
 return 0;
}