poj Fibonacci 矩阵快速幂

题目链接：​​点我​​ 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

9

999999999

1000000000

-1

Sample Output

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:

.

大佬博客：​​点我​​ 这里应用到矩阵快速幂，矩阵快速幂是用来求解递推式的，所以第一步先要列出递推式:

f(n)=f(n-1)+f(n-2)；第二步是建立矩阵递推式，找到转移矩阵:

,简写成T * A(n-1)=A(n)，T矩阵就是那个2 乘2的常数矩阵，而

这里就是个矩阵乘法等式左边:1f(n-1)+1f(n-2)=f(n);1f(n-1)+0f(n-2)=f(n-1);

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<iomanip>
#include<cctype>
using namespace std;
const int MAXN=2e5+5;
const int INF=1<<30;
//const long long mod=1e9+7;
#define ll long long
#define edl putchar('\n')
#define sscc ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define FORLL(i,a,b) for(ll i=a;i<=b;i++)
#define ROFLL(i,a,b) for(ll i=a;i>=b;i--)
#define mst(a) memset(a,0,ssizeof(a))
#define mstn(a,n) memset(a,n,ssizeof(a))
#define zero(x)(((x)>0?(x):-(x))<eps)
ll mod=10000;
const int ssize=2;
struct Matrix {
  ll a[ssize][ssize];
  Matrix() {
    memset(a,0,sizeof(a));
  }
  void init() {
    for(int i=0; i<ssize; i++)
      for(int j=0; j<ssize; j++)
        a[i][j]=(i==j);
  }
  Matrix operator + (const Matrix &B)const {
    Matrix C;
    for(int i=0; i<ssize; i++)
      for(int j=0; j<ssize; j++)
        C.a[i][j]=(a[i][j]+B.a[i][j])%mod;
    return C;
  }
  Matrix operator * (const Matrix &B)const {
    Matrix C;
    for(int i=0; i<ssize; i++)
      for(int k=0; k<ssize; k++)
        for(int j=0; j<ssize; j++)
          C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%mod;
    return C;
  }
  Matrix operator ^ (const ll &t)const {
    Matrix A=(*this),res;
    res.init();
    ll p=t;
    while(p) {
      if(p&1)res=res*A;
      A=A*A;
      p>>=1;
    }
    return res;
  }
};

int main() {
  Matrix a,b;
  int n;
  while(scanf("%d",&n)==1&&n!=-1) {
    a.a[0][0]=1;
    a.a[0][1]=1;
    a.a[1][0]=1;
    a.a[1][1]=0;

    b.a[0][0]=1;
    b.a[1][0]=0;
        if(n<=0){
          cout<<0<<endl;
    }
    else{
      a=a^(n-1);
        b=a*b;
        cout<<b.a[0][0]<<endl;
    }
    
  }
}