Fibonacci(矩阵快速幂)

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24331		Accepted: 16339

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

Sample Output

解题思路：因为n很大所以不能用递推来算;

a[2] a[1] a[0] ------> a[3] a[2] a[1]

1 1 0 　　　　2 1 1

构造一个矩阵 1行3列的矩阵*（莫一个矩阵）=1行3列的矩阵 -----> 矩阵大小为3行3列

递推式= a[n]=a[n-1]]+a[n-2]; a[n+1]=a[n]+a[n-1];

后一个由第一个怎么变换到达 -----> 构造的矩阵为

1 1 0

1 0 1

0 0 0

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define ll long long
 5 #define mod(x) ((x)%MOD)
 6 using namespace std;
 7 
 8 ll n;
 9 const int maxn=3;
10 const int MOD=10000;
11 
12 struct mat{
13     int m[maxn][maxn];
14 }unit;
15 
16 mat operator *(mat a,mat b){
17     mat ret;
18     ll x;
19     for(int i=0;i<maxn;i++)
20     for(int j=0;j<maxn;j++){
21         x=0;
22         for(int k=0;k<maxn;k++){
23             x+=mod(1LL*a.m[i][k]*b.m[k][j]);
24         }
25         ret.m[i][j]=mod(x);
26     }
27     return ret;
28 }
29 
30 void init(){
31     for(int i=0;i<maxn;i++)
32         unit.m[i][i]=1;
33     return;
34 }
35 
36 mat pow_mat(mat a,ll m){
37     mat ret=unit;
38     while(m){
39         if(m&1) ret=ret*a;
40         a=a*a;
41         m=m/2;
42     }
43     return ret;
44 }
45 
46 int main(){
47     ios::sync_with_stdio(false);
48     init();
49     while(cin>>n&&n!=-1){
50         if(n==0) cout << 0 << endl;
51         else if(n==1) cout << 1 << endl;
52         else if(n==2) cout << 1 << endl;
53         else{
54             mat a,b;
55             a.m[0][0]=1,a.m[0][1]=1,a.m[0][2]=0;
56 
57             b.m[0][0]=1,b.m[0][1]=1,b.m[0][2]=0;
58             b.m[1][0]=1,b.m[1][1]=0,b.m[1][2]=1;
59             b.m[2][0]=0,b.m[2][1]=0,b.m[2][2]=0;
60 
61             b=pow_mat(b,n-2);
62             a=a*b;
63             cout << mod(a.m[0][0]) << endl;
64         }
65     }
66     return 0;
67 }

矩阵快速幂

Fibonacci(矩阵快速幂)

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