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 <= 10 ¹⁸), You should solve for
g(g(g(n))) mod 10 ⁹ + 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：

 1 /*采用事先处理自己可以求出来*/

 2 LL work(LL mod){  

 3   LL a=0,b=1;

 4     for(LL i=2;;++i)

 5     {    

 6       a=(b*3+a)%mod;

 7         a=a^b;

 8         b=a^b;

 9         a=a^b;

10         if(a == 0 && b == 1)  return i;

11  }

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：

 1 //#define LOCAL

 2 #include<iostream>

 3 #include<cstdio>

 4 #include<cstring>

 5 #define LL __int64

 6 using namespace std;

 7 const int mod1 =1000000007;

 8 const int mod2=222222224;

 9 const int mod3=183120;

10 

11 LL mat[2][2];

12 LL ans[2][2];

13 LL n;

14 

15 void Matrix(LL a[][2],LL b[][2],LL mod)

16 {

17     LL cc[2][2]={0};

18     for(int i=0;i<2;i++)

19     {

20       for(int j=0;j<2;j++)

21       {

22           for(int k=0;k<2;k++)

23         {

24             cc[i][j]=(cc[i][j]+a[i][k]*b[k][j])%mod;

25         }

26       }

27     }

28     for(int i=0;i<2;i++)

29     {

30       for(int j=0;j<2;j++)

31       {

32           a[i][j]=cc[i][j];

33       }

34     }

35 }

36 

37 void pow(LL w,LL mod)

38 {

39   while(w>0)

40   {

41     if(w&1) Matrix(ans,mat,mod);

42      w>>=1;

43      if(w==0)break;

44      Matrix(mat,mat,mod);

45   }

46 }

47 void input(LL w,LL mod)

48 {

49      mat[0][0]=3;

50      mat[0][1]=mat[1][0]=1;

51      mat[1][1]=0;

52      ans[0][0]=ans[1][1]=1;

53      ans[0][1]=ans[1][0]=0;

54      pow(w,mod);

55 //     printf("%I64d\n",ans[0][0]);

56 }

57 void work(int i,__int64 w)

58 {

59   if(i==4||w==0||w==1)

60   {

61       if(w==0)

62          printf("0\n");

63       else if(w==1)

64           printf("1\n");

65       else

66       printf("%I64d\n",ans[0][0]);

67       return ;

68   }

69   LL mod;

70   if(i==1)mod=mod3;

71   else if(i==2)mod=mod2;

72   else if(i==3)mod=mod1;

73   input(w-1,mod);

74   work(i+1,ans[0][0]);

75 }

76 int main()

77 {

78   #ifdef LOCAL

79    freopen("test.in","r",stdin);

80   #endif

81   while(scanf("%I64d",&n)!=EOF)

82      work(1,n);

83  return 0;

84 }

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