poj-3070-Fibonacci

来源:互联网 发布:元数据ios 编辑:程序博客网 时间:2024/05/17 06:40
Fibonacci
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9405 Accepted: 6676

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

099999999991000000000-1

Sample Output

0346266875
矩阵快速幂
import java.util.Scanner;import java.util.Arrays;public class Main {public static void main(String[] args){Scanner in=new Scanner(System.in);while(true){int n=in.nextInt();if(n==-1) break;int [][]a= {{1,0},{0,1}};int [][]b={{1,1},{1,0}};int [][]c=new int[2][2];while(n>0){if(n%2!=0){Arrays.fill(c[0],0);Arrays.fill(c[1],0);for(int i=0;i<2;i++)for(int j=0;j<2;j++)for(int k=0;k<2;k++)c[i][k]+=(a[i][j]*b[j][k])%10000;for(int i=0;i<2;i++)for(int j=0;j<2;j++)a[i][j]=c[i][j];}Arrays.fill(c[0],0);Arrays.fill(c[1],0);for(int i=0;i<2;i++)for(int j=0;j<2;j++)for(int k=0;k<2;k++)c[i][k]+=(b[i][j]*b[j][k])%10000;for(int i=0;i<2;i++)for(int j=0;j<2;j++)b[i][j]=c[i][j];n>>=1;}System.out.println(a[0][1]%10000);}in.close();}}


0 0