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

999999999

1000000000

-1

Sample Output

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:

Source

Stanford Local 2006

大意——给你一个求解Fibonacci数列的公式。问：给出一个数n，要你用这个公式计算出F(n)的后四位。

思路——题目已经告诉我们用矩阵连乘求Fibonacci数，问题是n非常大。假设直接矩阵乘n-1次，肯定TLE。因此我们能够用二分求高速幂：

复杂度分析——时间复杂度:O(n),空间复杂度:O(1)

附上AC代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               using namespace std;typedef unsigned int UI;typedef long long LL;typedef unsigned long long ULL;typedef long double LD;const double PI = 3.14159265;const double E = 2.71828182846;const int MOD = 10000;struct matrix{ int mat[2][2];} res, base; // 定义一个结构体封装矩阵matrix mul(matrix a, matrix b); // 矩阵乘法运算int fast_mod(int x); // 二分求高速幂// a^n = (a^(n/2))^2 当n为偶数// a^n = a*(a^(n/2))^2 当n为奇数int main(){ ios::sync_with_stdio(false); int num; while (cin >> num && num != -1) { cout << fast_mod(num) << endl; } return 0;}matrix mul(matrix a, matrix b){ matrix temp; for (int i=0; i<2; ++i) for (int j=0; j<2; ++j) { temp.mat[i][j] = 0; for (int k=0; k<2; ++k) temp.mat[i][j] = (temp.mat[i][j]+a.mat[i][k]*b.mat[k][j])%MOD; } return temp;}int fast_mod(int x){ base.mat[0][0]=base.mat[0][1]=base.mat[1][0]=1; base.mat[1][1]=0; // 原始矩阵 res.mat[0][0]=res.mat[1][1]=1; res.mat[0][1]=res.mat[1][0]=0; // 单位矩阵 while (x) { if (x & 1) // 相当于模2 { res = mul(res, base); } base = mul(base, base); x >>= 1; // 相当于除2 } return res.mat[0][1];}

转载于:https://www.cnblogs.com/gavanwanggw/p/6871952.html

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