湖南大学第十四届ACM程序设计新生杯（重现赛）C：Sleepy Kaguya_houraisan☆kaguya is the princess who lives in lite-优快云博客

探讨了斐波那契数列中一个有趣的数学性质，即对于任意正整数k，表达式F[k+1]*F[k+1]-F[k]*F[k+2]的值总是等于(-1)^k。这一规律不仅简化了复杂计算，还揭示了斐波那契数列的内在规律。

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链接：https://ac.nowcoder.com/acm/contest/338/C
来源：牛客网

题目描述

Houraisan☆Kaguya is the princess who lives in Literally House of Eternity. However, she is very playful and often stays up late. This morning, her tutor, Eirin Yagokoro was going to teach her some knowledge about the Fibonacci sequence. Unfortunately, the poor princess was so sleepy in class that she fell asleep. Angry Eirin asked her to complete the following task:

This sequence can be described by following equations:
1.F[1]=F[2]=1

2.F[n]=F[n-1]+F[n-2] (n>2)

Now, Kaguya is required to calculate F[k+1]*F[k+1]-F[k]*F[k+2] for each integer k that does not exceed 10^18.

Kaguya is so pathetic. You have an obligation to help her.

(I love Houraisan Kaguya forever!!!)

image from pixiv,id=51208622

输入描述:

Input
Only one integer k.

输出描述:

Output
Only one integer as the result which is equal to F[k+1]*F[k+1]-F[k]*F[k+2].

示例1

输入

复制

输出

复制

说明

F[2]=1,F[3]=2,F[4]=3

2*2-1*3=1

备注:

0 < k ≤ 1^18

If necessary, please use %I64d instead of %lld when you use "scanf", or just use "cin" to get the cases.

The online judge of HNU has the above feature, thank you for your cooperation.

思路：直接计算超时，想着展开化简，超时，原来是找规律题，答案为（-1）的n次方。

题解:

首先注意K的范围

F[n+1]*F[n+1]-F[n]*F[n+2]=(-1)2 1.n=1时，结论显然成立 2. 假设n=k时结论成立，即F[k+1]*F[k+1]-F[k]*F[k+2]=(-1)2 则 F[k+2]*F[k+2]=F[k+2]*F[k]+F[k+2]*F[k+1] =F[k+1]2-(-1)2+F[k+2]*F[k+1] =F[k+1]*(F[k+1]+F[k+2])+(-1)k+1 =F[k+1]*F[k+3]+(-1)k+1 故n=k+1时结论仍然成立

#include <iostream>
using namespace std;
int main()
{
    long long k;
    cin>>k;
    if(k&1)
    {
        cout<<"-1"<<endl;
    }
    else
    {
        cout<<"1"<<endl;
    }
    return 0;
}