斐波那契数列

fibonacci数列(二)

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

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.

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:

.

输入

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.

输出

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).

样例输入

091000000000-1

样例输出

0346875

#include <iostream>using namespace std;#define mod 10000struct Matrix{long long temp[2][2];Matrix operator*(Matrix a){Matrix c;for (int i = 0; i < 2; i++){for (int j = 0; j < 2; j++){long long sum = 0;for (int k = 0; k < 2; k++){sum += (temp[i][k] % mod) * (a.temp[k][j] % mod) % mod;}c.temp[i][j] = sum % mod;}}return c;}}matrix;void Qpower(int n){Matrix t;t.temp[0][0] = t.temp[1][1] = 1;t.temp[0][1] = t.temp[1][0] = 0;while (n){if (n & 1){t = t * matrix;}matrix = matrix * matrix;n /= 2;}matrix = t;}void input(){int n;while (cin >> n){matrix.temp[0][0] = matrix.temp[1][0] = matrix.temp[0][1] = 1;matrix.temp[1][1] = 0;if (n == -1){break;}if (!n){cout << 0 << endl;}else{Qpower(n - 1);cout << matrix.temp[0][0] % mod << endl;}}}int main(){std::ios::sync_with_stdio(false);input();return 0;}