POJ3070Fibonacci

最新推荐文章于 2018-05-23 22:39:34 发布

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最新推荐文章于 2018-05-23 22:39:34 发布

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本文链接：https://blog.youkuaiyun.com/jc514984625/article/details/71090996

本文介绍了一种高效计算斐波那契数列中任一项最后四位数字的方法，利用矩阵快速幂运算实现，适用于大整数计算场景。

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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

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:

#include <cstdio>
#include <iostream>

using namespace std;

const int MOD = 10000;

int fast_mod(int n)
{
    int t[2][2] = {1, 1, 1, 0};
    int ans[2][2] = {1, 0, 0, 1};
    int tmp[2][2];

    while(n)
    {
        if(n & 1)
        {
            for(int i = 0; i < 2; ++i)
                for(int j = 0; j < 2; ++j)
                    tmp[i][j] = ans[i][j];
            ans[0][0] = ans[1][1] = ans[0][1] = ans[1][0] = 0;

            for(int i = 0; i < 2; ++i)
            {
                for(int j = 0; j < 2; ++j)
                {
                    for(int k = 0; k < 2; ++k)
                        ans[i][j] = (ans[i][j] + tmp[i][k] * t[k][j]) % MOD;
                }
            }
        }


        for(int i = 0; i < 2; ++i)
            for(int j = 0; j < 2; ++j)
                tmp[i][j] = t[i][j];
        t[0][0] = t[1][1] = 0;
        t[0][1] = t[1][0] = 0;
        for(int i = 0; i < 2; ++i)
        {
            for(int j = 0; j < 2; ++j)
            {
                for(int k = 0; k < 2; ++k)
                    t[i][j] = (t[i][j] + tmp[i][k] * tmp[k][j]) % MOD;
            }
        }

        n >>= 1;
    }
    return ans[0][1];
}

int main()
{
    int n;
    while(scanf("%d", &n) && n != -1)
    {
        printf("%d\n", fast_mod(n));
    }
    return 0;
}