10/16/2018 GNY Region 2015 ACM Regional Running Steps

本文介绍了一个算法问题,即计算在给定的总步数下,运动员如何按照特定规则上楼梯的不同方式数量。规则包括每条腿的两步和一步数量相等,且两步数量不少于一步。文章提供了一个Java程序实现,通过组合数学方法计算可能的步数组合。

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Problem H
Running Steps

The coach wants his team members to run up the stadium steps taking either one or two steps with each stride so that:

  1. The number of two step strides taken by each leg is the same.

  2. The number of one step strides taken by each leg is the same.

  3. The number of two step strides is no smaller than the number of one step strides.

  4. Start with the left leg.

The coach wants to know for a given (necessarily even) number of steps how many different ways there are to run the steps and satisfy his rules.

For example, with six steps (three for each leg), there are 4 possibilities:

 2211, 2112, 1221, 1122 (right leg strides are in highlighted type)

With eight steps (four for each leg) there is only one possibility since there must be at least as many two step strides as one step strides:

 2222

For this problem, you will write a program that calculates the number of different ways there are to run the steps that satisfy the coach’s four criteria.

Input

The first line of input contains a single integer P, (1≤P≤1000), which is the number of data sets that follow. Each data set should be processed identically and independently.

Each data set consists of a single line of input. It contains the data set number, K, followed by an even integer which is the total number of steps to be run, S, (2≤S≤100).

Output

For each data set there is a single line of output. The single output line consists of the data set number, K, followed by a single space followed by the number of different ways of running the steps that satisfy the coach’s four criteria.

Sample Input 1Sample Output 1
5
1 6
2 8
3 10
4 12
5 60
1 4
2 1
3 9
4 37
5 40197719157

 

 


思路:只有步数是偶数的时候才能被走完。因为2步的步数一定不小于1步的步数,所以我们直接从两步开始走,但注意因为教练的要求,两步的步数必须是偶数,所以假如一开始用两步走完所有步数的次数是奇数的话,则要将初始两步的步数减一,一步的步数加二(1 * 2 - (2 * 1) = 0)。之后我们进行一个While循环,每次减少两次迈两步的步数并且加上四步一步的次数,答案加上在这种情况下的排列组合,直到一步的次数大于两步的次数。

C_m^n = C_{m - 1} ^ {n - 1} + C_{m -1} ^{n}: This is how we get the combination table

import java.util.*;
public class RunningSteps {
	static long[][] c;
	public static void combination() {
		c = new long[55][55];
		c[0][0] = 1;
		c[1][0] = 1;
		c[1][1] = 1;
		//The combination table: C^n _k: i stands for n, j stands for j;
		//That is, the ways to select j items from i items
		for (int i = 2; i <= 50; i++) {
			c[i][0] = 1;
			for (int j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
		}
	}
	public static void main(String args[]) {
		combination();
		Scanner sc = new Scanner(System.in);
		int kase = sc.nextInt();
		while (kase > 0) {
			int idx = sc.nextInt();
			int steps = sc.nextInt();
			int two_step = steps / 2;
			int one_step = 0;
//If the num of 2-steps is odd, you have to subtract one to make it even
			if (two_step % 2 == 1) {
				two_step--;
				one_step += 2;
			}
			long ans = 0;
			while (two_step >= one_step) {
//bundle two steps as a unit
				long temp = c[two_step / 2 + one_step / 2][one_step / 2];
				ans += (temp * temp);
				two_step -= 2; one_step += 4;
			}
			System.out.println(idx + " " + ans);
			kase--;
		}
		sc.close();
	}
}

 

标题基于SpringBoot+Vue的社区便民服务平台研究AI更换标题第1章引言介绍社区便民服务平台的研究背景、意义,以及基于SpringBoot+Vue技术的研究现状和创新点。1.1研究背景与意义分析社区便民服务的重要性,以及SpringBoot+Vue技术在平台建设中的优势。1.2国内外研究现状概述国内外在社区便民服务平台方面的发展现状。1.3研究方法与创新点阐述本文采用的研究方法和在SpringBoot+Vue技术应用上的创新之处。第2章相关理论介绍SpringBoot和Vue的相关理论基础,以及它们在社区便民服务平台中的应用。2.1SpringBoot技术概述解释SpringBoot的基本概念、特点及其在便民服务平台中的应用价值。2.2Vue技术概述阐述Vue的核心思想、技术特性及其在前端界面开发中的优势。2.3SpringBoot与Vue的整合应用探讨SpringBoot与Vue如何有效整合,以提升社区便民服务平台的性能。第3章平台需求分析与设计分析社区便民服务平台的需求,并基于SpringBoot+Vue技术进行平台设计。3.1需求分析明确平台需满足的功能需求和性能需求。3.2架构设计设计平台的整体架构,包括前后端分离、模块化设计等思想。3.3数据库设计根据平台需求设计合理的数据库结构,包括数据表、字段等。第4章平台实现与关键技术详细阐述基于SpringBoot+Vue的社区便民服务平台的实现过程及关键技术。4.1后端服务实现使用SpringBoot实现后端服务,包括用户管理、服务管理等核心功能。4.2前端界面实现采用Vue技术实现前端界面,提供友好的用户交互体验。4.3前后端交互技术探讨前后端数据交互的方式,如RESTful API、WebSocket等。第5章平台测试与优化对实现的社区便民服务平台进行全面测试,并针对问题进行优化。5.1测试环境与工具介绍测试
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