/acm/contest/338/C
题解：验算一下，发现其实只有1和-1两个答案。奇数为-1，偶数为1；
演算过程：
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)
参考文章：/weixin_43272781/article/details/82858423
/**@Author: STZG*@Language: C++*/#include <bits/stdc++.h>#include<iostream>#include<algorithm>#include<cstdlib>#include<cstring>#include<cstdio>#include<string>#include<vector>#include<bitset>#include<queue>#include<deque>#include<stack>#include<cmath>#include<list>#include<map>#include<set>//#define DEBUG#define RI register intusing namespace std;typedef long long ll;//typedef __int128 lll;const int N=100000+10;const int MOD=1e9+7;const double PI = acos(-1.0);const double EXP = 1E-8;const int INF = 0x3f3f3f3f;ll t,n,m,k,q,ans;int a;char str;int main(){#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);#endifscanf("%lld",&n);if(n%2==0){cout << 1 << endl;}else{cout << -1 << endl;}//cout << "Hello world!" << endl;return 0;}
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