Fibonacci 矩阵乘法

最新推荐文章于 2022-10-10 21:48:42 发布

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最新推荐文章于 2022-10-10 21:48:42 发布

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Fibonacci

斐波那契数题目链接
Description

In the Fibonacci integer sequence, $F_0$ = 0, $F_1$ = 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
在这里插入图片描述

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print $F_n$ mod 10000).

Sample Input

0
9
999999999
1000000000
-1
Sample Output

0
34
626
6875
Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
在这里插入图片描述

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
在这里插入图片描述

AC代码

#include<iostream>
#include<cstring>
using namespace std;
const int mod = 10000;
typedef long long ll;
ll n;
void mul(ll f[2], ll a[2][2])
{
    ll c[2];
    memset(c, 0, sizeof c);
    for(int j = 0; j < 2; j++)
        for(int k = 0; k < 2; k++)
            c[j] = (c[j] + (ll)f[k] * a[k][j]) % mod;
    memcpy(f, c, sizeof c);
}
void myself(ll a[2][2])
{
    ll c[2][2];
    memset(c, 0, sizeof c);
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            for(int k = 0; k < 2; k++)
                c[i][j] = (c[i][j] + (ll)a[i][k] * a[k][j]) % mod;
    memcpy(a, c, sizeof c);
}
int main()
{
    while(cin >> n && n != -1)
    {
        ll f[2] = {0, 1};
        ll a[2][2] = {{0, 1}, {1, 1}};
        for(; n; n >>= 1)
        {
            if(n & 1)
                mul(f, a);
            myself(a);
        }
        cout << f[0] << endl;
    }
    return 0;
}