POJ 3070 Matrix Fast Power

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12457		Accepted: 8851

Description

In the Fibonacci integer sequence, f0 = 0, f1 = 1, and fn = F N −1 + Fn −2 for n ≥2. For example, the first ten terms of the Fibonacci sequence is:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal was to compute the last 4 digits of Fn.

Input

The input test file would contain multiple test cases. Each of the test case consists of a 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 is all zeros, print ' 0 '; Otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

Hint

As a reminder, matrix multiplication is associative, and the product of the A/2x2 matrices is given by

.

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

.

Source

Stanford Local 2006Test instructions: Solving the problem of Fibonacci series; the key to the first question of the Matrix fast power is that the structure of the matrix is fast power after the structure of the Matrix http://www.cnblogs.com/frog112111/archive/2013/05/19/308 7648.html

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <queue>5#include <stack>6#include <map>7 #defineMoD 100008 using namespacestd;9 structMatrixTen { One     intm[5][5]; A } ANS,EXM; -  - structMatrix Matrix_mulit (structMatrix AA,structmatrix BB) the { -     structmatrix there; -      for(intI=0;i<2; i++) -     { +          for(intj=0;j<2; j + +) -         { +there.m[i][j]=0; A              for(intk=0;k<2; k++) atthere.m[i][j]= (there.m[i][j]+aa.m[i][k]*bb.m[k][j]%mod)%MoD; -         } -     } -     returnthere; - } - intMatrix_quick (intgg) in { -exm.m[0][0]=exm.m[0][1]=exm.m[1][0]=1; toexm.m[1][1]=0; +ans.m[0][0]=ans.m[1][1]=1;  -ans.m[0][1]=ans.m[1][0]=0; the       while(GG) *      { $          if(gg&1)  Panax Notoginseng          { -ans=Matrix_mulit (ANS,EXM);  the          } +EXM =matrix_mulit (EXM, EXM); AGG >>=1; the     } +      returnans.m[0][0]; - }  $ intN; $ intMain () - { -      while(SCANF ("%d", &n)! =EOF) the     { -         if(n==-1)Wuyi           Break;  theprintf"%d\n", Matrix_quick (n)); -     } Wu     return 0; -}

POJ 3070 Matrix Fast Power