https://ac.nowcoder.com/acm/contest/338/C

題解：驗(yàn)算一下，發(fā)現(xiàn)其實(shí)只有1和-1兩個(gè)答案。奇數(shù)為-1，偶數(shù)為1；

演算過(guò)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)

參考文章：https://blog.csdn.net/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 int using 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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