湖南大学第十四届ACM程序设计大赛 C Sleepy Kaguya

链接：https://ac.nowcoder.com/acm/contest/338/C
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

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

2

输出

复制

1

说明

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

2*2-1*3=1

备注:

0 < k ≤ 10^18

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.

题目大意：

这个题是围绕斐波那契扩展而来的 F[1]=F[2]=1,F[n]=F[n-1]+F[n-2],

输入一个数k代表第k个斐波那契数，让你求 F[k+1]*F[k+1]-F[k]*F[k+2]的结果，暴力肯定是不行了，斐波那契数的递增速度非常快，且数据范围k是1e18.尝试递推几次后会发现结果不是-1就是1，奇数为-1，偶数为1，后来看了热心博友STZG发的题解才明白原理，感谢申哥的指导！

递推公式：

G(k)=F[k+1]*F[k+1]-F[k]*F[k+2]

=F[k+1]*F[k+1]-F[k]*(F[k]+F[k+1])

=(F[k+1]-F[k])*F[k+1]-F[k]*F[k]

=F[k-1]*F[k+1]-F[k]*F[k]

所以G(k)=-G(k-1)

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