And Then There Was One

Time limit: 3.000 seconds

Let’s play a stone removing game.

Initially, n stones are arranged on a circle and numbered 1,...,n clockwise (Figure 1). You are also given two numbers k and m. From this state, remove stones one by one following the rules explained below, until only one remains. In step 1, remove stone m. In step 2, locate the k-th next stone clockwise from m and remove it. In subsequent steps, start from the slot of the stone removed in the last step, make k hops clockwise on the remaining stones and remove the one you reach. In other words, skip (k −1) remaining stones clockwise and remove the next one. Repeat this until only one stone is left and answer its number.

For example, the answer for the case n = 8, k = 5, m = 3 is 1, as shown in Figure 1.


Initial state: Eight stones are arranged on a circle.

Step 1: Stone 3 is removed since m = 3.

Step 2: You start from the slot that was occupied by stone 3. You skip four stones 4, 5, 6 and 7 (since k = 5), and remove the next one, which is 8.

Step 3: You skip stones 1, 2, 4 and 5, and thus remove 6. Note that you only count stones that are still on the circle and ignore those already removed. Stone 3 is ignored in this case.

Steps 4-7: You continue until only one stone is left. Notice that in later steps when only a few stones remain, the same stone may be skipped multiple times.

For example, stones 1 and 4 are skipped twice in step 7.

Final State: Finally, only one stone, 1, is on the circle. This is the final state, so the answer is 1.
Input
The input consists of multiple datasets each of which is formatted as follows.
n k m
The last dataset is followed by a line containing three zeros. Numbers in a line are separated by a single space. A dataset satisfies the following conditions.

2 ≤ n ≤ 10000,1 ≤ k ≤ 10000,1 ≤ m ≤ n

The number of datasets is less than 100.
Output
For each dataset, output a line containing the stone number left in the final state. No extra characters such as spaces should appear in the output.
Sample Input
8 5 3

100 9999 98

10000 10000 10000

0 0 0
Sample Output
1

93

题意就是约瑟夫问题,看人家的博客吧。

传送门1:http://www.cnblogs.com/shanyr/p/5213631.html

传送门2:http://blog.csdn.net/hyczms/article/details/43817799

公式写上就可以A了。

代码:

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+;
int a[N];
int main(){
int n,k,m,cnt;
while(~scanf("%d%d%d",&n,&k,&m)&&n){
a[]=;
for(int i=;i<=n;i++)
a[i]=(a[i-]+k)%i;
cnt=(a[n]+m-k+)%n;
if(cnt<=) cnt+=n;
printf("%d\n",cnt);
}
return ;
}

UVALive 3882.And Then There Was One-约瑟夫问题(递推)的更多相关文章

  1. Joseph POJ - 1012 约瑟夫环递推

    题意:约瑟夫环  初始前k个人后k个人  问m等于多少的时候 后k个先出去 题解:因为前k个位置是不动的,所以只要考虑每次递推后的位置在不在前面k个就行 有递推式 ans[i]=(ans[i-1]+m ...

  2. UVALive 3882 - And Then There Was One【约瑟夫问题】

    题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_ ...

  3. UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There Was One / POJ 3517 And Then There Was One / Aizu 1275 And Then There Was One (动态规划,思维题)

    UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There W ...

  4. LA 3882 And Then There Was One[约瑟夫问题的变形]

    And Then There Was One UVALive - 3882 Sample Input   Sample Output //设f[i]为(原约瑟夫问题)第i次要删除的标号 #includ ...

  5. LA 3882 经典约瑟夫环问题的数学递推解法

    就是经典约瑟夫环问题的裸题 我一开始一直没理解这个递推是怎么来的,后来终于理解了 假设问题是从n个人编号分别为0...n-1,取第k个, 则第k个人编号为k-1的淘汰,剩下的编号为  0,1,2,3. ...

  6. 「模拟8.23」one递推,约瑟夫

    前置芝士约瑟夫问题 这样大概就是板子问题了 考场的树状数组+二分的60分暴力??? 1 #include<bits/stdc++.h> 2 #define int long long 3 ...

  7. LA 3882 - And Then There Was One(约瑟夫 递归)

    看题传送门 题目大意: N个数排成一圈,第一次删除m,以后每k个数删除一次,求最后一被删除的数. 如果这题用链表或者数组模拟整个过程的话,时间复杂度都将高达O(nk),而n<=10000,k&l ...

  8. UVALive - 3882:And Then There Was One

    约瑟夫环 f[i]表示有i个人先处理第k个人,最后被处理的人是谁 #include<cstdio> #include<cstdlib> #include<algorith ...

  9. UVALive - 6577 Binary Tree 递推+找规律

    题目链接: http://acm.hust.edu.cn/vjudge/problem/48421 Binary Tree Time Limit: 3000MS 问题描述 Binary Tree is ...

随机推荐

  1. [剑指Offer] 36.两个链表的第一个公共结点

    题目描述 输入两个链表,找出它们的第一个公共结点. [思路]找出两个链表的长度,然后让长的走两个链表的长度差,然后再一起走(因为两个链表用公共的尾部). /* struct ListNode { in ...

  2. 【题解】JLOI2013卡牌游戏

    这题最开始是用 \(n^{4}\)的算法水过的,之后才想出的\(n^{3}\)正解.首先,\(n^{4}\) 应该是很容易想到的:设状态 \(f[i][j][k]\) 为有 \(i\) 个人,庄家为 ...

  3. 【题解】JSOI2010满汉全席

    ~bzoj1823 第一次接触2-SAT——SAT,即适定性(Satisfiability)的缩写.像名称所说,即满足需求的可能性问题,而k-SAT即每个人有k种需求,已经证明k>2时是一个NP ...

  4. 某ISP的流氓行径 劫持用户HTTP请求插入js代码

    最近公司搞的项目有用户反应点击任意链接后偶尔会跳到一个“莫名奇妙”的网站………… 喏,就是这个咯.

  5. [Leetcode] Binary tree maximum path sum求二叉树最大路径和

    Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. ...

  6. COGS1752. [BOI2007]摩基亚Mokia CDQ

    CDQ的板子题 #include<cstdio> #include<cstring> #include<iostream> #include<algorith ...

  7. 用原生JavaScript做个简单的回到顶部

    很多网页在下方都会放置一个“返回顶部”按钮,尤其是页面底部没有导航的网页,这样可以帮助访客重新找到导航或者重温一遍广告(想得真美).随着近几年来 JavaScript 的应用日渐广泛,滑动效果无处不在 ...

  8. POJ2289:Jamie's Contact Groups(二分+二分图多重匹配)

    Jamie's Contact Groups Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/ ...

  9. Codeforces Round #525 (Div. 2)E. Ehab and a component choosing problem

    E. Ehab and a component choosing problem 题目链接:https://codeforces.com/contest/1088/problem/E 题意: 给出一个 ...

  10. Google File System中文版

    英文原文地址: Google File system 译文原文地址: The Google File System中文版 Google File System中文版 摘要 我们设计并实现了Google ...