poj3070矩阵快速幂

Fibonacci

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

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_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <stdlib.h>

 using namespace std;
 int N;
 struct matrix
 {
        int a[][];
 }origin,res;
 matrix multiply(matrix x,matrix y)
 {
        matrix temp;
        memset(temp.a,,sizeof(temp.a));
        for(int i=;i<N;i++)
        {
                for(int j=;j<N;j++)
                {
                        for(int k=;k<N;k++)
                        {
                                temp.a[i][j]+=x.a[i][k]*y.a[k][j]%;
                                temp.a[i][j]%=;
                        }
                }
        }
        return temp;
 }
 void calc(int n)
 {
      memset(res.a,,sizeof(res.a));
       origin.a[][]=;origin.a[][]=;
       origin.a[][]=;origin.a[][]=;
      for(int i=;i<N;i++)
      res.a[i][i]=;
      while(n)
      {
              if(n&)
                     res=multiply(res,origin);
              n>>=;
              origin=multiply(origin,origin);
      }
 }
 int main()
 {
     N=;
     int n;
     while(cin>>n)
     {
         if(n==-)
         break;
         if(n)
         calc(n-);
         if(n)
         cout<<res.a[][]<<endl;
         else cout<<<<endl;
     }
 }