poj 3070 Fibonacci 矩阵快速幂

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.

.

题解：

按题目要求很容易写出矩阵

F[n]    0        1  1  

F[n-1] 0        1  0

直接上矩阵快速幂即可

 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 typedef long long ll;
 const int mod=;
 int n;
 struct matrix
 {
     ll a[][];
     matrix(){for(int i=;i<=;i++)for(int j=;j<=;j++)a[i][j]=;}
     matrix(ll b[][]){for(int i=;i<=;i++)for(int j=;j<=;j++)a[i][j]=b[i][j];}
     inline matrix operator *(matrix p){
         matrix tmp;
         for(int i=;i<=;i++)
             for(int j=;j<=;j++){
                 tmp.a[i][j]=;
                 for(int k=;k<=;k++)
                     tmp.a[i][j]+=a[i][k]*p.a[k][j],tmp.a[i][j]%=mod;
             }
         return tmp;
     }
 };
 ll work(){
     if(n==)return ;
     if(n==||n==)return ;
     n-=;
     ll t[][]={{,,},{,,},{,,}};ll sum[][]={{,,},{,,},{,,}};
     matrix S=matrix(t),T=matrix(sum);
     while(n){
         if(n&)S=S*T;
         T=T*T;n>>=;
     }
     return S.a[][];
 }
 int main()
 {
     while(scanf("%d",&n))
        {
             if(n==-)break;
             printf("%lld\n",work());
         }
     return ;
 }