题意：

约瑟夫问题的变式。先指定第m个人必须死，然后每隔k个人死一个。求最后那个死的人的编号是什么。

题目

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
2019

分析

首先不考虑第一个必须死的人是m的情况。我们把n个人编号为[0,n-1]。那么第一轮出局的就是编号为k-1的人，剩下的人的编号是[0,k−2]∪[k,n−1]。然后我们从编号为k的人开始，循环一圈给所有人重新分配编号

0----->i

1----->i+1

2----->i+2

… …

k-2----->n-2

k-1（杀死）

k----->0///倒推，从这里最容易看出来dp[i]=(dp[i-1]+k)%i;(此时i==n)

k+1----->1

… …

n-2----->i-2

n-1----->i-1

其实这个问题就是问如何用存活者在i-1个人中的编号求出存活者在i个人中的编号。我们知道，从原问题推到子问题其实是把所有人的编号-k，那么从子问题推到原问题就是把人的编号+k，但要注意，此时+k可能会大于当前人的规模i，所以要对i取膜。

AC代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int M=1e4+10;
int dp[M];
int n,m,k;int main()
{while(~scanf("%d%d%d",&n,&k,&m)&&(m||n||k)){memset(dp,0,sizeof(dp));for(int i=2;i<n;i++)dp[i]=(dp[i-1]+k)%i;dp[n]=(dp[n-1]+m)%n;printf("%d\n",dp[n]+1);}return 0;
}

And Then There Was One POJ - 3517(变形约瑟夫环+规律)相关推荐

poj 3517
题目链接 http://poj.org/problem?id=3517 题意约瑟夫环要求最后删掉的那个人是谁: 方法理解递推公式就行了考虑这样一组数据 k ...
POJ 3517 And Then There Was One（约瑟夫环模板）
链接:传送门题意:典型约瑟夫环问题约瑟夫环模板题:n个人( 编号 1-n )在一个圆上,先去掉第m个人,然后从m+1开始报1,报到k的人退出,剩下的人继续从1开始报数,求最后剩的人编号 /**** ...
(顺序表的应用5.4.2)POJ 1591 M*A*S*H(约瑟夫环问题的变形——变换步长值)
/** POJ_1591_2.cpp** Created on: 2013年10月31日* Author: Administrator*/ #include <iostream> #inc ...
poj 2240 Bellman-Flod 求环
http://poj.org/problem?id=2240 深刻体现了自己代码能力有问题外加改模板能力有问题.外加Debug有问题.以后做到: 1.算法原理能够轻易弄出来. 2.代码模板自己收集各种 ...
poj 1012 Joseph（约瑟夫环求每次出圈人的序号）
http://poj.org/problem?id=1012 有k个好人和k个坏人,前k个是好人后k个是坏人.模拟约瑟夫环,每次数到m的数要被杀死,然后他后面的人从1开始报数.重复这个过程.要求输出最 ...
2018年全国多校算法寒假训练营练习比赛（第一场）D. N阶汉诺塔变形（找规律）
链接:https://www.nowcoder.com/acm/contest/67/D 来源:牛客网题目描述相信大家都知道汉诺塔问题.那么现在对汉诺塔问题做一些限制,成为一个新的玩法. 在一个底 ...
1875 丢手绢约瑟夫环变形枚举
1875 丢手绢基准时间限制:1 秒空间限制:131072 KB 分值: 20 难度:3级算法题收藏关注六一儿童节到了,小朋友们在玩丢手绢的游戏.总共有C个小朋友,编号从1到C,他们站成一个 ...
约瑟夫环 poj 3750 小孩报数问题模拟
Language: Default 小孩报数问题 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10071 Accept ...
POJ 1012 Joseph 题解
POJ 1012 Joseph 题解题目解读: The Joseph's problem is notoriously known. For those who are not familiar w ...

And Then There Was One POJ - 3517(变形约瑟夫环+规律)

题意：

题目