【D - ECJTU_ACM 11级队员2012年暑假训练赛（2）】

斐波那契数列计算

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原文链接：http://www.cnblogs.com/ismdeep/archive/2012/07/10/2584777.html

本文介绍了一种高效计算斐波那契数列中任意项最后四位数字的方法，使用矩阵快速幂运算来减少计算复杂度，适用于大整数输入。

D - D

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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_nmod 10000).

Sample Input

Sample Output

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:

 1 // Project name : D ( Fibonacci ) 
 2 // File name    : main.cpp
 3 // Author       : Izumu
 4 // Date & Time  : Tue Jul 10 14:02:23 2012
 5 
 6 #include <iostream>
 7 #include <stdio.h>
 8 using namespace std;
 9 
10 
11 int p[30][4]={1,1,1,0};
12 
13 void mm(int * ret,int * a,int * b)
14 {
15     int x[4],y[4],i;
16     for(i=0;i<4;i++)
17     {
18          x[i]=a[i];
19          y[i]=b[i];
20      }    
21      ret[0] = (x[0]*y[0] + x[1]*y[2]) % 10000;
22      ret[1] = (x[0]*y[1] + x[1]*y[3]) % 10000;
23      ret[2] = (x[2]*y[0] + x[3]*y[2]) % 10000;
24      ret[3] = (x[2]*y[1] + x[3]*y[3]) % 10000;    
25 }
26 
27 int main()
28 {
29     int i,n;
30     int a[4]={0,1,1,0};
31     for(i=1;i<30;i++)
32     {
33          mm(p[i],p[i-1],p[i-1]);
34      }
35     while(1)
36     {
37          scanf("%d",&n);
38         if(n==-1) break;
39          a[0]=1;a[1]=0;a[2]=0;a[3]=1;
40         
41         for(i=0;i<30;i++)
42         {
43             if(n&(1<<i))
44             {
45                  mm(a,a,p[i]);
46             }
47         }
48          cout << a[1] << endl;
49     }
50 
51     return 0;
52 }
53 
54 
55 // end 
56 // ism

转载于:https://www.cnblogs.com/ismdeep/archive/2012/07/10/2584777.html