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2013
11-12

POJ 3187 Backward Digit Sums [解题报告] Java

Backward Digit Sums

问题描述 :

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this:

    3   1   2   4

4 3 6
7 9
16

Behind FJ’s back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ’s mental arithmetic capabilities.

Write a program to help FJ play the game and keep up with the cows.

输入:

Line 1: Two space-separated integers: N and the final sum.

输出:

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.

样例输入:

4 16

样例输出:

3 1 2 4

温馨提示:

Explanation of the sample:

There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

解题代码:

//* @author: [email protected]
import java.io.*;
import java.util.Arrays;
public class Main
{
	static int[] p,arr;
	static int a;
	public static void main(String[] args) throws IOException
	{
	 InputStreamReader is=new InputStreamReader(System.in);
	 BufferedReader in=new BufferedReader(is);
	 String[] ss=in.readLine().split(" ");
	 a=Integer.parseInt(ss[0]);
	 int sum=Integer.parseInt(ss[1]);
	 p=new int[a];
	 arr=new int[a];
	 for(int i=0;i< a;i++)
	 	p[i]=i+1;
	 for(int u=0;;u++)
	 {
	  if(sum==get(arr))
	  {
	   for(int i=0;i< a;i++)
		System.out.print(p[i]+" ");
	   break;
	   }
	  next();
	  }
	}

 static void next()
 {
   for(int i=a-1;i>=0;i--)
    {
      for(int j=a-1;j>i;j--)
      {
    	if(p[j]>p[i])
	{
		int temp=p[j];
		p[j]=p[i];
		p[i]=temp;
		Arrays.sort(p,i+1,a);
		return;
	}
      }
    }
  }


static int get(int[] arr)
 {
	int b=a;
	for(int i=0;i< b;i++)
		arr[i]=p[i];
	while((b--)!=0)
	{
	 for(int i=0;i< b;i++)
		arr[i]=arr[i]+arr[i+1];
	}
	return arr[0];
  }
}

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  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。