stairs专题

leetcode#70. Climbing Stairs

题目 You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note: Given n will be a posi

【九度】题目1388：跳台阶【LeetCode】Climbing Stairs

1、【九度】题目1388：跳台阶时间限制：1 秒内存限制：32 兆特殊判题：否提交：2435解决：995 题目描述：一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法。输入：输入可能包含多个测试样例，对于每个测试案例，输入包括一个整数n(1<=n<=70)。输出：对应每个测试案例，输出该青蛙跳上一个n级的台

codeforces--2014/1/13--B. Sereja and Stairs

B. Sereja and Stairs time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja loves integer sequences very much. He espec

【LeetCode】70. Climbing Stairs【牛客网】变态跳台阶

今天在LeetCode做到一个动态规划的题，之前在牛客网做过一道“变态跳台阶”的题，猛一看还以为是一个意思，牛客网那边能AC，LeetCode这边却不能AC，很奇怪，其实题目有细微的差别，这两题难道不是一个意思吗？ 1、跳台阶题目描述：一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种方法？与LeetCode Problem No.70 Cl

LeetCode之旅（16）-Climbing Stairs

题目描述： You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? 先验知识：斐波那契数列斐波那契数列（Fib

70. Climbing Stairs dynamic programming

You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note: Given n will be a positive

Leetcode: Climbing Stairs

题目： You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? 思路分析：设 f (n) 表示爬 n 阶楼梯的不

[leetcode]Climbing Stairs

class Solution {public:int climbStairs(int n) {// Start typing your C/C++ solution below// DO NOT write int main() functionif(n <= 2) return n;vector<int> f(n+1, 0);f[1] = 1;f[2] = 2;for(int i = 3; i

leetcode746-Min Cost Climbing Stairs

题目给你一个整数数组 cost ，其中 cost[i] 是从楼梯第 i 个台阶向上爬需要支付的费用。一旦你支付此费用，即可选择向上爬一个或者两个台阶。你可以选择从下标为 0 或下标为 1 的台阶开始爬楼梯。请你计算并返回达到楼梯顶部的最低花费。示例 1：输入：cost = [10,15,20] 输出：15 解释：你将从下标为 1 的台阶开始。支付 15 ，向上爬两个台阶，到达楼梯顶

Codeforces 1419B. Stairs

题意： x个方格可以构成多少种不同阶梯图形，要求每个组成单位都是正方形。思路：容易推出 s u m [ i ] = s u m [ i − 1 ] ∗ 2 + n o w ∗ n o w sum[i] = sum[i - 1] * 2 + now * now sum[i]=sum[i−1]∗2+now∗now，然后算能用多少个就好了。 #include <cstdio>#include

[LeetCode] 70. Climbing Stairs @ python

一．题目：爬一个n层的楼梯，每一次可以爬两层或者一层，请问到达楼梯顶部时一共有多少种走法？二．解题思路：很显然，这是一道很简单的动态规划题．dp[i]表示爬到第i层时的走法，有状态转移方程为 dp[i] = dp[i-1]+dp[i-2] 所以代码如下： class Solution(object):def climbStairs(self, n):""":type n: int:rtyp

leetcode -- 70. Climbing Stairs

题目描述题目难度：Easy You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note: Given n wi

LeetCode Climb Stairs

题目： You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? 题意：就是上楼梯的方式的种数。每次只能用1

Climbing stairs (70)

写了个top-down的divide and conquer程序，可惜超时了。 class Solution:def climbStairs(self, n: int) -> int:# divide and conquer?if n == 0:return 1if n <= 1:return self.climbStairs(n-1)if n >= 2:return self.climbSta

Dasha and Stairs CodeForces - 761A

Dasha and Stairs 题目链接： CodeForces - 761A 题意：问能否找到一段连续整数，使得偶数与奇数的个数与给出数据相同；只要两数差值<=1即可；注意特判0， 0； #include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;in

Leet Code OJ 70. Climbing Stairs [Difficulty: Easy] -python

爬楼梯问题： 70. Climbing Stairs Easy You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top

leetcode70: Climbing Stairs

要求： You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? 注意：基本动态规划问题，只要写出状态转移方程就可

climb_stairs爬楼梯递归C++

#include <iostream>using namespace std;int N;int stairs(int n){//递归出口1if( n < 0)return 0;//递归出口2if( n == 1)return 1;return stairs(n-1) + stairs(n-2);}int main(){while(cin>>N){cout<<stairs(N)<<en

CF498E Stairs and Line 题解

CF498E Stairs and Lines CF498E Stairs and Lines 肯定是状压，而且状压的就是轮廓线，我们考虑同时状压横着和竖着的情况。然后肯定需要进行矩阵快速幂复杂度还是很高的。但是我们考虑我们当前位置和之前位置进行转移的时候使用的是竖着的线，而且横着的线仅仅在当前转移出现了，意味着肯定不会影响后面的计算。那么我们不妨枚举轮廓线，这样复杂度就是直接少

leetcode刷题日记:70.Climbing Stairs(爬楼梯)

给了我们n阶楼梯让我们去爬，每次只能爬1阶或2阶，这个我们可以使用递归的方法进行解决，向上爬的过程与向下降的过程是一样的，所以我们就可以这样考虑，现在就是n阶楼梯每一次下降1阶或2阶有多少种方法刚好下降到0阶。 int climbStairs(int n){if(n==0){return 1;}else if(n<0){return 0;}return climbStairs(n-1)+clim

Leetcode-70. Climbing Stairs

题目 You are climbing a staircase. It takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Example 1: Input: n = 2Output: