NYOJ-148 fibonacci数列(二)

fibonacci数列(二)

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 3

 

描述

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.

 

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:

.

 

 

输入

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.

输出

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

样例输入

0

9

1000000000

-1

样例输出

0

34

6875

 1 /*

 2 //代码一：找取模后的周期---因为是加法，所以完全可以根据取模后的周期推倒正确的结果

 3 #include <iostream>

 4 #include <cstdio>

 5 using namespace std;

 6 

 7 const int MAX = 15000 + 1;

 8 int fib[MAX];

 9 

10 int fun(int *fib, int mod)

11 {

12     fib[0] = 0;

13     fib[1] = 1;

14     int i = 2;

15     do

16     {

17         fib[i] = (fib[i - 1] + fib[i - 2]) % mod;

18         ++i;

19     }while(fib[i - 1] != 1 || fib[i - 2] != 0);

20     return i - 2;

21 }

22 

23 int main()

24 {

25     int T = fun(fib, 10000);

26     //cout << T;     T = 15000;

27     int n;

28     while(scanf("%d", &n) && n != -1)

29     {

30         printf("%d\n",fib[n % T]);

31     }

32     return 0;

33 }

34 

35 

36 */

37 

38 //代码二：二分幂

39 #include <cstdio>

40 #include <iostream>

41 

42 using namespace std;

43 

44 void fun(int a1[][2], int a2[][2])     // 相当于 a1 = a1 * a2

45 {

46     int c[2][2];

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

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

49         {

50             c[i][j] = 0;

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

52                 c[i][j] = (c[i][j] + a1[i][k] * a2[k][j])%10000;

53         }

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

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

56             a1[i][j] = c[i][j];

57 }

58 

59 int main()

60 {

61     int n;

62     while(scanf("%d", &n) && n != -1)

63     {

64         int a[2][2] = {{1,1}, {1,0}};

65         int b[2][2] = {{1,0}, {0,1}};     //初始化为单位矩阵，保存fib的值

66         if(n == -1) break;

67         while(n)                         //求 a 的 n 次幂

68         {

69             if(n & 1)

70                 fun(b, a);

71             fun(a, a);

72             n >>= 1;

73         }

74         printf("%d\n", b[1][0]); //最后n必然从1变为0，所以最后一次总要执行 fun(b, a);

75     }

76     return 0;

77 }