约瑟夫问题专题

 

UVa 1452 - Jump(约瑟夫环变形)

分类： 动态规划 2013-08-10 16:25  538人阅读  评论(0)  收藏  举报

目录(?)[-]

1. 题意
2. 思路
3. 代码
   1. 来自CODE的代码片 uvaa-1452-Jumpcpp

本文出自   http://blog.youkuaiyun.com/shuangde800

题目点击打开链接

题意

把1～n按逆时针顺序排成一个圆圈，从1开始没k个数字删除掉一个，知道所有数字都删完。

求最后删除的3个数。

思路

我们已经知道了，怎么可以推出最后一个被删除的编号（可参考百度百科）

http://baike.baidu.com/view/213217.htm

f(1) = 0, 表示最后还剩下一个时，这个编号为0

f(n) = (f(n-1) + m) % n

那么保存最后第1,2,3个数，一直推到第一个即可。

代码

<a target=_blank id="L1" href="http://blog.youkuaiyun.com/shuangde800/article/details/9745849#L1" rel="#L1" style="color: rgb(102, 102, 102); text-decoration: none;"> 1</a> <a target=_blank id="L2" href="http://blog.youkuaiyun.com/shuangde800/article/details/9745849#L2" rel="#L2" style="color: rgb(102, 102, 102); text-decoration: none;"> 2</a> <a target=_blank id="L3" href="http://blog.youkuaiyun.com/shuangde800/article/details/9745849#L3" rel="#L3" style="color: rgb(102, 102, 102); text-decoration: none;"> 3</a> <a target=_blank id="L4" href="http://blog.youkuaiyun.com/shuangde800/article/details/9745849#L4" rel="#L4" style="color: rgb(102, 102, 102); text-decoration: none;"> 4</a> <a target=_blank id="L5" href="http://blog.youkuaiyun.com/shuangde800/article/details/9745849#L5" rel="#L5" style="color: rgb(102, 102, 102); text-decoration: none;"> 5</a> <a target=_blank id="L6" href="http://blog.youkuaiyun.com/shuangde800/article/details/9745849#L6" rel="#L6" style="color: rgb(102, 102, 102); 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/**========================================== * This is a solution for ACM/ICPC problem * * @source: uva-1452 Jump * @author: shuangde * @blog: blog.youkuaiyun.com/shuangde800 * @email: zengshuangde@gmail.com *===========================================*/ #include<iostream> #include<cstdio> #include<algorithm> #include<vector> #include<queue> #include<cmath> #include<cstring> using namespace std ; typedef long long int64 ; const int INF = 0x3f3f3f3f ; const double PI = acos ( - 1.0 ); const int MAXN = 110 ; int n , m ; int main (){ int T ; scanf ( "%d" , & T ); while ( T -- ){ scanf ( "%d%d" , & n , & m ); int ans1 = 0 , ans2 , ans3 ; for ( int i = 2 ; i <= n ; ++ i ){ ans1 = ( ans1 + m ) % i ; if ( i == 2 ){ // 当剩下最后2个数时，编号为0，1, 推出倒数第二个删除的数当前值 ans2 = ! ans1 ; } else if ( i == 3 ){ // 当剩下最后3个数时，编号为0，1，2, 推出倒数第三个删除的数当前值 ans2 = ( ans2 + m ) % i ; bool vis [ 3 ]; memset ( vis , 0 , sizeof ( vis )); vis [ ans1 ] = vis [ ans2 ] = true ; for ( int j = 0 ; j < 3 ; ++ j ) if ( ! vis [ j ]){ ans3 = j ; break ; } } else { ans2 = ( ans2 + m ) % i ; ans3 = ( ans3 + m ) % i ; } } ans1 = ( ans1 + 1 ) % n ; ans2 = ( ans2 + 1 ) % n ; ans3 = ( ans3 + 1 ) % n ; printf ( "%d %d %d \n " , ans3 ? ans3 : n , ans2 ? ans2 : n , ans1 ? ans1 : n ); } return 0 ; }

 来自CODE的代码片

uvaa-1452-Jump.cpp