HDU 1443( Joseph 约瑟夫问题 )

spoiler posted @ 2011年4月18日 15:36 in 规律及递归经典题 , 3450 阅读

Problem 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 every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.

Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.
 

Input
The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.
 
Output
The output file will consist of separate lines containing m corresponding to k in the input file.
 

Sample Input
3
4
0
 
Sample Output
5
30

题目分析:这个题目我目前没找到更好的办法,上网没查到,唉,我的代码跑了500MS 郁闷死了!我是用枚举+模拟的

思路:

for i : 1 -> 14

    for j : i ->

     然后分别用长度位2 * i, 周期为 j 的循环模拟,

solve( k ,m ) :

      每次都是用 kill = (m - 1)%i (这是求本次需要去除的元素的下标)

      更新: start=((start-m)%i+i)%i   ( 其实也可以写成 :start =( (start - kill -1) %i + i )%i   )

              end=((end - m ) % i + i )  % i; 同理也可以写成上面的那样!

 示例:

    1(0)  2(1)  3(2)  4(3)  5(4)  6(5)

假设  :周期为5

第一次: kill = ( 5-1)%6=4(因为数组里下标4 就是第5个元素)

然后关键地方 start=(( 0 - 5 )%6+6) %6=1  这里就是关键: 此时 1(1)  2(2)  3(3) 4(4)  6(0)

同理 end=((2-5)%6+6)%6 =3;

这个时候 我们把数组 的顺序做个变动 按下标 重新 排列!

6(0)  1(1)  2(2) 3(3)  4(4)

这样不是又重复了上述过程麽? 但是start 与end 的指向 永远都是

没变 还是指向那几个元素!

 就这样整个模拟过程的准备工作都完成了!

源程序+部分注释:

#include<iostream>
#include<cstdio>
using namespace std;
int f[15];
bool solve(int k,int m){
	bool flag=true;
	int start=0,end=k-1,kill;
	for(int i=2*k;i>k;i--){
		kill=(m-1)%i;
		//cout<<"i = "<<i<<" "<<"kill = "<<kill<<endl;
		if(kill>=start&&kill<=end){
			flag=false;
			break;
		}
		start=((start-m)%i+i)%i;
		end=((end-m)%i+i)%i;//这样是为了更新起始点,以当前点为0坐标!想象成一个圆形!
		//cout<<"start = "<<start<<" "<<" end = "<<end<<endl;
	}
	return flag;
}

int main(){
	int n;
	for(int i=1;i<=14;i++){
		for(int j=i;;j++){
			if(solve(i,j)){
				f[i]=j;
				break;
			}
		}
	}
	while(~scanf("%d",&n),n){
		printf("%d\n",f[n]);
	}
	return 0;
}

题目总结: 注意学习,模拟约瑟夫循环的过程!

三个方程:

       kill = ( m - 1 ) % n

      start = ( (start - m) % n +n )%n

     end = ((end - m )%n +n)% n;

上述的n是当前的长度,是动态的!


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