题解 | #不死神兔问题#

#include <iostream>
using namespace std;

int getSum(int n);

int main() {

    int n;
    cin >> n;

    cout << getSum(n) << endl;

    return 0;
}

int getSum(int n) {

    //方法二：迭代法
    if(n == 1 || n == 2)
        return 1;
    int pre_pre=1;
    int pre = 1;
    int now = 0;
    for(int i = 2; i < n; i++)
    {
        //计算当前的now,now的值为前两次之和
        now = pre+pre_pre;
        //更新pre和pre_pre的值
        pre_pre = pre;
        pre = now;
    }
    return now;
    
    // 方法一：递归法
    //时间复杂度：O(2^n)，空间复杂度o(n)
//     if(n == 1 || n == 2)
//         return 1;
//     return (getSum(n-1)+getSum(n-2));
}