POJ 3070 Fibonacci (矩阵快速幂)

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10440		Accepted: 7421

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0

9

999999999

1000000000

-1

Sample Output

0

34

626

6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

解题思路：

　　关键是会求解转移矩阵，剩下的就是扔模板了。。

 

代码：

 1 # include<cstdio>

 2 # include<iostream>

 3 # include<fstream>

 4 

 5 using namespace std;

 6 

 7 # define MOD 10000

 8 

 9 typedef long long LL;

10 

11 struct matrix

12 {

13     LL a[2][2];

14     void init()

15     {

16         a[0][0] = 1;

17         a[0][1] = 1;

18         a[1][0] = 1;

19         a[1][1] = 0;

20     }

21 };

22 

23 int n;

24 

25 matrix fun ( matrix aa, matrix bb )

26 {

27     matrix cc;

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

29     {

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

31         {

32             cc.a[i][j] = 0;

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

34             {

35                 cc.a[i][j]+=(aa.a[i][k]*bb.a[k][j]);

36             }

37             cc.a[i][j]%=MOD;

38         }

39     }

40 

41     return cc;

42 }

43 

44 

45 

46 matrix My_power( matrix aa,int k )

47 {

48     matrix  ans;

49     ans.init();

50     while ( k >= 1 )

51     {

52         if ( k&1 )

53         {

54             ans = fun(ans,aa);

55         }

56         k/=2;

57         aa = fun(aa,aa);

58     }

59 

60     return ans;

61 

62 }

63 

64 

65 int main(void)

66 {

67 

68     int n;

69     while ( scanf("%d",&n)!=EOF )

70     {

71         if ( n==-1 )

72             break;

73         if ( n==0 )

74         {

75             printf("0\n");

76             continue;

77         }

78         matrix aa;

79         aa.init();

80         aa = My_power(aa,n-1);

81         printf("%lld\n",aa.a[0][1]%MOD);

82     }

83 

84     return 0;

85 }