P1962 斐波那契数列 题解

Doveqise

2019-03-05 21:41:33

Solution

这道题还没有看到我这种做法emmm就先发一下 这个东西~~(经过玄学操作)~~ 效率奇棒(๑•̀ㅂ•́)و✧ 我就分享一下(代码短的一皮) (提示:本代码只供演示,不能A此题) (才不是∵没模数) ```cpp #include<iostream> using namespace std; unsigned long long f(register unsigned long long i, register unsigned long long n1, register unsigned long long n2){return i == 1 ? n2 : f(i - 1, n2, n1 + n2);} unsigned long long fib(register unsigned long long i){return f(i, 1, 1);} signed main(){ register int n = 0; cin >> n; cout << fib(n) << endl; return 0; } ```