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=,b=;
     for(LL i=;;++i)
     {
       a=(b*+a)%mod;
         a=a^b;
         b=a^b;
         a=a^b;
         if(a ==  && b == )  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 =;
 const int mod2=;
 const int mod3=;

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

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

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