poj 3070 Fibonacci 矩阵快速幂

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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

 

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 using namespace std;
 8 typedef long long ll;
 9 const int mod=10000;
10 int n;
11 struct matrix
12 {
13     ll a[3][3];
14     matrix(){for(int i=1;i<=2;i++)for(int j=1;j<=2;j++)a[i][j]=0;}
15     matrix(ll b[3][3]){for(int i=1;i<=2;i++)for(int j=1;j<=2;j++)a[i][j]=b[i][j];}
16     inline matrix operator *(matrix p){
17         matrix tmp;
18         for(int i=1;i<=2;i++)
19             for(int j=1;j<=2;j++){
20                 tmp.a[i][j]=0;
21                 for(int k=1;k<=2;k++)
22                     tmp.a[i][j]+=a[i][k]*p.a[k][j],tmp.a[i][j]%=mod;
23             }
24         return tmp;
25     }
26 };
27 ll work(){
28     if(n==0)return 0;
29     if(n==1||n==2)return 1;
30     n-=2;
31     ll t[3][3]={{0,0,0},{0,1,1},{0,0,0}};ll sum[3][3]={{0,0,0},{0,1,1},{0,1,0}};
32     matrix S=matrix(t),T=matrix(sum);
33     while(n){
34         if(n&1)S=S*T;
35         T=T*T;n>>=1;
36     }
37     return S.a[1][1];
38 }
39 int main()
40 {
41     while(scanf("%d",&n))
42        {
43             if(n==-1)break;
44             printf("%lld\n",work());
45         }
46     return 0;
47 }

 

 

 

 

 

 

 

 

posted @ 2017-07-09 21:41  PIPIBoss  阅读(140)  评论(0)  编辑  收藏  举报