Roman Roulette Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 286 Accepted Submission(s): 105 Problem Description The historian Flavius Josephus relates how, in the Romano-Jewish conflict of
方法一.自定义的链表实现 package com.code.yuesefu; public class YueSeFuList { public static void main(String[] args) { ;//申请一个指定长度的链表 Node n = YueSeFuList.createNodes(count); ;i<count;i++){ Node second = n.next;//第2个 n = n.next.next;//第3个 System.out.println(n.it
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace orderlyOffarray { class Program { static void Main(string[] args) { , , ); Console.WriteLine("出列顺序:"); ; i < intPers.Length; i++) { Console.WriteL
1. 循环链表(约瑟夫问题) https://github.com/BodhiXing/Data_Structure 2. Have Fun with Numbers https://pta.patest.cn/pta/test/17/exam/4/question/263 #include <iostream> #include <cstring> using namespace std; int main() { ],b[],buffer[]; ]; int i,j,len;
SRM 147 2 600PeopleCircle Problem Statement There are numMales males and numFemales females arranged in a circle. Starting from a given point, you count clockwise and remove the K'th person from the circle (where K=1 is the person at the current poin
问题描述:皇帝决定找出全国中最幸运的一个人,于是从全国选拔出 n 个很幸运的人,让这 n 个人围着圆桌进餐,可是怎么选择出其中最幸运的一个人呢?皇帝决定:从其中一个人从 1 开始报数,按顺序数到第 k 个数的人自动出局,然后下一个人从 1 开始报数,数到 k 的人出局…….如此直到最后只剩下约瑟夫一人,然后他就成为全国最幸运的人.请问约瑟夫最初的位置?(注:原问题略显暴力,故自创此趣味题目) 分析:把第一个开始报 1 的人标定为 1,然后按报数顺序依次标定其余的人为:2,3,……,n - 1,n