Fibonacci
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 7752   Accepted: 5501

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <math.h>
 5 #include <stdlib.h>
 6 
 7 using namespace std;
 8 int N;
 9 struct matrix
10 {
11        int a[2][2];
12 }origin,res;
13 matrix multiply(matrix x,matrix y)
14 {
15        matrix temp;
16        memset(temp.a,0,sizeof(temp.a));
17        for(int i=0;i<N;i++)
18        {
19                for(int j=0;j<N;j++)
20                {
21                        for(int k=0;k<N;k++)
22                        {
23                                temp.a[i][j]+=x.a[i][k]*y.a[k][j]%10000;
24                                temp.a[i][j]%=10000;
25                        }
26                }
27        }
28        return temp;
29 }
30 void calc(int n)
31 {
32      memset(res.a,0,sizeof(res.a));
33       origin.a[0][0]=1;origin.a[0][1]=1;
34       origin.a[1][0]=1;origin.a[1][1]=0;
35      for(int i=0;i<N;i++)
36      res.a[i][i]=1;
37      while(n)
38      {
39              if(n&1)
40                     res=multiply(res,origin);
41              n>>=1;
42              origin=multiply(origin,origin);
43      }
44 }
45 int main()
46 {
47     N=2;
48     int n;
49     while(cin>>n)
50     {
51         if(n==-1)
52         break;
53         if(n)
54         calc(n-1);
55         if(n)
56         cout<<res.a[0][0]<<endl;
57         else cout<<0<<endl;
58     }
59 }
View Code

 

posted on 2013-08-21 15:13  ERKE  阅读(148)  评论(0编辑  收藏  举报