解题思路:

方法一: 斐波那契递归。
代码:

use std::io::{self, *};
fn fabonacii(n:u32) -> u32 {
    if n == 1 || n == 2 { return 1;}
    fabonacii(n-1) + fabonacii(n-2)
}
fn main() {
    let stdin = io::stdin();
    for line in stdin.lock().lines() {
        let n = line.unwrap().trim().parse::<u32>().unwrap_or(0);
        println!("{}",fabonacii(n));
    }
}

方法二:数组存储递归数据。
可以减少调栈,但是得多耗费一点空间。
代码:

use std::io::{self, *};
fn main() {
    let stdin = io::stdin();
    let mut v = vec![1;31];
    for i in 2..31 {
        v[i] = v[i-1] + v[i-2];
    }
    for line in stdin.lock().lines() {
        let n = line.unwrap().trim().parse::<usize>().unwrap_or(0);
        println!("{}",v[n-1]);
    }
}

方法三:打表(老实憨厚)

代码:

use std::io::{self, *};
fn main() {
    let stdin = io::stdin();
    for line in stdin.lock().lines() {
        let n = line.unwrap().trim().parse::<u32>().unwrap_or(0);
        println!("{}",match n {
            1|2 => 1,
            3 =>2,
            4 =>3,
            5 =>5,
            6 =>8,
            7 =>13,
            8 =>21,
            9 =>34,
            10 =>55,
            11 =>89,
            12 =>144,
            13 =>233,
            14 =>377,
            15 =>610,
            16 =>987,
            17 =>1597,
            18 =>2584,
            19 =>4181,
            20 =>6765,
            21 =>10946,
            22 =>17711,
            23 =>28657,
            24 =>46368,
            25 =>75025,
            26 =>121393,
            27 =>196418,
            28 =>317811,
            29 =>514229,
            30 =>832040,
            31 =>1346269,
            _ =>0,
        });
    }
}