Somethings about Floors,

Somethings about Floors,

Question content: There are N levels (N> = 0) in a stair. How many steps are there from the bottom to the top of every step of level 1 or Level 2?

Input requirements: Only one row of input, and only one number N(If N> 20, N = N % 21 ensures that the range of N is: 0 <= N <= 20. When the value after the modulo operation is 0, output 1), Representing the level of the stair.

Output requirements: There is only one line. The output goes from the bottom to the top, and there is a line break behind it.

Example:

Input: 3

Output: 3

 

Hint:

Problem breakdown.

Analysis:This question should be solved by recursive thinking. If there is only one level of stairs, there is only one way to walk; if there are two levels of stairs, there are two ways to walk ;... if there are N (N> 2) level stairs, because one can go up level one or two, then we can consider steps from N-1 level and N-2 level stairs to level N, then F (N) = F (N-1) + F (N-2 ). My code:

#include <stdio.h>int main() {    int f0, f1, a;    int b, n, i;    scanf("%d", &n);    n = n % 21;    if (n == 0 || n == 1) {        printf("%d\n", 1);        return 0;    }    f0 = 1;    f1 = 2;    for (i = 3; i <= n; i++) {        a = f0 + f1;        f0 = f1;        f1 = a;    }    printf("%d\n", f1);    return 0;}

Answer:

// from younglee// solve the problem in two different way, with recursion and no recursion.#include<stdio.h>#include<string.h>#define N 100 #define RECUR 1#define MODE 0 int dealWithRecursion(int f);int dealWithNoRecursion(int f);// to save the result.int arr[N]; int main(void) {    int floor;    scanf("%d", &floor);    floor %= 21;    if (MODE == RECUR) {        printf("%d\n", dealWithRecursion(floor));    } else {        memset(arr, 0, sizeof(arr));        printf("%d\n", dealWithNoRecursion(floor));    }    return 0;} int dealWithRecursion(int f) {    if (f == 1 || f == 0) return 1;    return (dealWithRecursion(f - 1) + dealWithRecursion(f - 2));} int dealWithNoRecursion(int f) {    if (arr[f] != 0) return arr[f];    int result;    if (f == 0 || f == 1) result = 1;    else result = dealWithNoRecursion(f - 1) + dealWithNoRecursion(f - 2);    arr[f] = result;    return result;}

 

Here we use two methods: recursion and non-recursion.