2013
11-12

# Backward Digit Sums

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.

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

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.

4 16

3 1 2 4

Explanation of the sample:

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

//* @author: [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */
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。