牛客A 斐波拉契
链接:https://www.nowcoder.com/acm/contest/181/A
来源:牛客网
设f[i]表示斐波那契数论的第i项
f[1]=1,f[2] =1,f[i] = f[i - 1] + f[i - 2]
给定一个n
求
f[1]=1,f[2] =1,f[i] = f[i - 1] + f[i - 2]
给定一个n
求
输入描述:
一个整数n
输出描述:
一个整数,表示答案
备注:
分析:第六个恒等式
AC代码:
#include <map> #include <set> #include <stack> #include <cmath> #include <queue> #include <cstdio> #include <vector> #include <string> #include <bitset> #include <cstring> #include <iomanip> #include <iostream> #include <algorithm> #define ls (r<<1) #define rs (r<<1|1) #define debug(a) cout << #a << " " << a << endl using namespace std; typedef long long ll; const ll maxn = 1e3 + 10; const double eps = 1e-8; const ll mod = 1e9 + 7; const ll inf = 1e9; const double pi = acos(-1.0); int main() { std::ios::sync_with_stdio(false); string s; cin >> s; ll n = s[s.length()-1] - '0'; if(n&1) { cout << -1 << endl; } else { cout << 1 << endl; } return 0; }