一.Introductionhttps://app.yinxiang.com /Home.action?offer=www_menu#n=4b034a29-986d-4914-8220-eb99c2e8f0a3&ses=4&sh=2&sds=2& 贝多芬第五交响曲(命运)与第九交响曲(欢乐颂)给人不同的感觉,前者焦虑.消极,后者积极向上,原因在于:1)小调与大调:2)跳进与级进:3)主音高是否明确:4)旋律走向,下行或上行. 音乐的两个维度 时间和音高 节奏或时值 二拍:强

package Joseph;//约瑟夫环,m个人围成一圈.从第K个人开始报数,报道m数时,那个人出列,以此得到出列序列//例如1,2,3,4.从2开始报数,报到3剔除,顺序为4,3,1,2public class Joseph{ class node//结点类 {private int number;  private node next;  node(int num,node next){   this.number=num;   this.next=next;  } } private n

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1512    Accepted Submission(s): 948 Problem Description   The Joseph's problem is notoriously known. For those who are not familiar with th

题目:改进的Joseph环.一圈人报数,报数上限依次为3,7,11,19,循环进行,直到所有人出列完毕. 思路:双向循环链表模拟. 代码: #include <cstdio> #include <cstdlib> #define N 20 typedef struct node { int id; struct node *next; struct node *pre; }Node, *pNode; //双向循环链表的构建 pNode RingConstruct (int n) {

Joseph Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44650   Accepted: 16837 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n,

Joseph Time Limit: 1 Sec  Memory Limit: 64 MB Submit: 493  Solved: 311 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle e

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2240    Accepted Submission(s): 1361 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with the

题意: 给出n, k,求 分析: 假设,则k mod (i+1) = k - (i+1)*p = k - i*p - p = k mod i - p 则对于某个区间,i∈[l, r],k/i的整数部分p相同,则其余数成等差数列,公差为-p 然后我想到了做莫比乌斯反演时候有个分块加速,在区间[i, n / (n / i)],n/i的整数部分相同,于是有了这份代码. #include <cstdio> #include <algorithm> using namespace std;

  Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 50596   Accepted: 19239 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, stan

Joseph Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 47657   Accepted: 17949 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n,

 Joseph  The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, ..., n, standing in circle every mth is going to be executed and only the life of the last remaining per

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1652    Accepted Submission(s): 1031 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with th

Web scraping with Python (part II) « Jean, aka Sig(gg) Web scraping with Python (part II)

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2094    Accepted Submission(s): 1273 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with the

Joseph环,这次模固定是2.假设不是固定模2,那么一般时间效率是O(n).可是这次由于固定模2,那么能够利用2的特殊性,把时间效率提高到O(1). 规律能够看下图: watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQva2VuZGVuMjM=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast" alt=""> 具体具体解析请看大师

UVA305 - Joseph(数论 + 打表) 题目链接 题目大意:约瑟夫环问题:n个人围成一圈,每次都淘汰第m个人,问最后一个幸存下来的人的编号. 这题的意思有点不一样,它规定前面的k个人是好人,后面的k个人是坏人(2 ∗ k形成环).问最小的m是多少,可以先把后面的k个坏人淘汰再淘汰好人. 解题思路:这题有个递推式:ai = (ai - 1 + m - 1) % (2 ∗ k - (i - 1)) ai表示第i次淘汰的人的编号.后面取余的是剩下的人数. 注意这题的k仅仅可能有14种,可是測

基本上纯Joseph核心问题,只是第一步多一件.m. 然后你就可以用获得的递推公式: Win(n) 代表n当个人的中奖号码, 然后,Win(n)必须相等Win(n-1).当一个人将在下一次删除队列. 下一个出队列的人是谁呢? 假设模是mod的话,那么下一个出队号码计算为: Lose(n) = mod % n; if (Lose(n) == 0) Lose(n) = n; 这样得到公式: Win(n) - Lose(n)  = Win(n-1); Win(n) = Win(n-1) + Lose(

一:首先科普一下约瑟夫问题的数学方法 (1)  不管是用list实现还是用vector实现都有一个共同点:要模拟整个游戏过程,不仅程序写起来比較烦,并且时间复杂度高达O(nm),当n,m很大(比如上百万,上千万)的时候,差点儿是没有办法在短时间内出结果的.我们注意到原问题不过要求出最后的胜利者的序号,而不是要读者模拟整个过程.因此假设要追求效率,就要打破常规,实施一点数学策略. (2) 为了讨论方便,先把问题略微改变一下,并不影响原意:  问题描写叙述:n个人(编号0~(n-1)),从0開始报数

1.Didi Chuxing's Jean Liu on The Future of Cities  https://www.youtube.com/watch?v=G9uPGoN0dvQ 2.Didi-Chuxing's Jean Liu Charlie Rose video https://www.bloomberg.com/news/videos/2017-10-10/didi-chuxing-s-jean-liu-charlie-rose-video 3.Pioneering the S

约瑟夫环问题: 在一间房间总共有n个人(下标0-n-1),只能有最后一个人活命. 按照如下规则去杀人: 所有人围成一圈 顺时针报数,每次报到q的人将被杀掉 被杀掉的人将从房间内被移走 然后从被杀掉的下一个人重新报数,继续报q,再清除,直到剩余一人 要求模拟这个问题 #include<stdio.h> #include<malloc.h> void Joseph(int count,int doom); void Joseph(int count,int doom){ int ali

/** 题目:Joseph's Problem 链接:https://vjudge.net/problem/UVA-1363 题意:给定n,k;求k%[1,n]的和. 思路: 没想出来,看了lrj的想法才明白. 我一开始往素数筛那种类似做法想. 想k%[1,n]的结果会有很多重复的,来想办法优化. 但没走通. 果然要往深处想. 通过观察数据发现有等差数列.直接观察很难确定具体规律:此处应该想到用式子往这个方向推导试一试. lrj想法: 设:p = k/i; 则:k%i = k-i*p; 容易想到