poj_3070Fibonacci(矩阵快速幂)

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12732		Accepted: 9060

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

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:

.

Source

Stanford Local 2006

 

 //快速幂矩阵
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 struct Mat{
     int mat[][];
 };
 const int Mod = ;
 Mat operator *(Mat a, Mat b)
 {
     Mat c;
     memset(c.mat,,sizeof(c.mat));
     for(int i = ; i < ; i++)
     {
         for(int j = ; j < ; j++)
         {
             for(int k = ; k < ; k++)
             {
                 c.mat[i][j] = (c.mat[i][j]+(a.mat[i][k]*b.mat[k][j])%Mod)%Mod;
             }
         }
     }
     return c;
 }
 Mat multi(int n)
 {
     Mat c;
     memset(c.mat,,sizeof(c.mat));
     c.mat[][] = c.mat[][] = ;
     Mat a;
     memset(a.mat,,sizeof(a.mat));
     a.mat[][] = a.mat[][] = a.mat[][] = ;
     while(n)
     {
         if(n&) c = c*a;
         a = a*a;
         n>>=;
     }
     return c;
 }
 int main()
 {
     int n;
     while(~scanf("%d",&n))
     {
         if(n==-) return ;
         Mat ans = multi(n);
         printf("%d\n",ans.mat[][]);
     }
     return ;
 }