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2013
11-10

POJ 2244 Eeny Meeny Moo [解题报告] Java

Eeny Meeny Moo

问题描述 :

Surely you have made the experience that when too many people use the Internet simultaneously, the net becomes very, very slow.

To put an end to this problem, the University of Ulm has developed a contingency scheme for times of peak load to cut off net access for some cities of the country in a systematic, totally fair manner. Germany’s cities were enumerated randomly from 1 to n. Freiburg was number 1, Ulm was number 2, Karlsruhe was number 3, and so on in a purely random order.

Then a number m would be picked at random, and Internet access would first be cut off in city 1 (clearly the fairest starting point) and then in every mth city after that, wrapping around to 1 after n, and ignoring cities already cut off. For example, if n=17 and m=5, net access would be cut off to the cities in the order [1,6,11,16,5,12,2,9,17,10,4,15,14,3,8,13,7]. The problem is that it is clearly fairest to cut off Ulm last (after all, this is where the best programmers come from), so for a given n, the random number m needs to be carefully chosen so that city 2 is the last city selected.

Your job is to write a program that will read in a number of cities n and then determine the smallest integer m that will ensure that Ulm can surf the net while the rest of the country is cut off.

输入:

The input will contain one or more lines, each line containing one integer n with 3 <= n < 150, representing the number of cities in the country.

Input is terminated by a value of zero (0) for n.

输出:

For each line of the input, print one line containing the integer m fulfilling the requirement specified above.

样例输入:

3
4
5
6
7
8
9
10
11
12
0

样例输出:

2
5
2
4
3
11
2
3
8
16

解题代码:

//* @author: [email protected]
import java.util.*;
public class Main
{
 public static void main(String[] args)
 {
  Scanner in=new Scanner(System.in);
  while(true)
   {
	int n=in.nextInt();
	if(n==0)break;
	int i,j,m=1;
	while(true)
	{
          i=0;
	  j=n-1;
	  while(j!=0)
	   {
		i=(i+m-1)%j;
		j--;
		if(i==0) break;
	   }
	  if(j==0)
	   {
	    System.out.println(m);
	    break;
	   }
	   m++;
	}
    }
  }
}

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  2. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”

  3. a是根先忽略掉,递归子树。剩下前缀bejkcfghid和后缀jkebfghicd,分拆的原则的是每个子树前缀和后缀的节点个数是一样的,根节点出现在前缀的第一个,后缀的最后一个。根节点b出现后缀的第四个位置,则第一部分为四个节点,前缀bejk,后缀jkeb,剩下的c出现在后缀的倒数第2个,就划分为cfghi和 fghic,第3部分就为c、c