POJ3070Fibonacci(矩阵快速幂+高效）

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11587		Accepted: 8229

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:

.

题意：就是求斐波那契数列，只是有点大，递推也会超时，矩阵快速幂0ms,搞定；

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int N = ;
 const int mod = ;
 struct Mat
 {
     int mat[][];
 };
 Mat operator * (Mat a, Mat b)
 {
     Mat c;
     memset(c.mat, , sizeof(c.mat));
     for(int k = ; k < N; k++)
     {
         for(int i = ; i < N; i++)
         {
             for(int j = ; j < N; j++)
             {
                 c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j] % mod) % mod;
             }
         }
     }
     return c;
 }
 Mat operator ^ (Mat a, int k)
 {
     Mat c;
     for(int i = ; i < N; i++)
         for(int j = ; j < N; j++)
             c.mat[i][j] = (i == j);
     while(k)
     {
         if(k & )
             c = c * a;
         a = a * a;
         k >>= ;
     }
     return c;
 }
 int main()
 {
     int n;
     while(scanf("%d", &n) != EOF)
     {
         if(n == -)
             break;
         if(n == )
             printf("0\n");
         else if(n == )
             printf("1\n");
         else
         {
             Mat a;
             a.mat[][] = ;
             a.mat[][] = ;
             a.mat[][] = ;
             a.mat[][] = ;
             a = a ^ (n - );
             printf("%d\n",a.mat[][]);
         }
     }

     return ;
 }