#P1118. [USACO06FEB] Backward Digit Sums G/S

    ID: 118 Type: RemoteJudge 1000ms 125MiB Tried: 3 Accepted: 2 Difficulty: 3 Uploaded By: Tags>搜索2006USACO枚举深度优先搜索,DFS

[USACO06FEB] Backward Digit Sums G/S

题目描述

FJ and his cows enjoy playing a mental game. They write down the numbers from 11 toN(1N10) N(1 \le N \le 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=4N=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 NN. 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.

输入格式

共一行两个正整数 n,sumn,sum

输出格式

输出包括一行,为字典序最小的那个答案。

当无解的时候,请什么也不输出。

题目大意

有这么一个游戏:

写出一个 1n1\sim n 的排列 aa,然后每次将相邻两个数相加,构成新的序列,再对新序列进行这样的操作,显然每次构成的序列都比上一次的序列长度少 11,直到只剩下一个数字位置。

下面是一个例子:

  • 3,1,2,43,1,2,4
  • 4,3,64,3,6
  • 7,97,9
  • 1616

最后得到 1616 这样一个数字。

现在想要倒着玩这样一个游戏,如果知道 nn,以及最后得到的数字的大小 sumsum,请你求出最初序列 aia_i(应该是一个 1n1\sim n 的排列)。若答案有多种可能,则输出字典序最小的那一个。

我们称序列 a=a1,a2,,ana=\lang a_1,a_2,\cdots,a_n\rang 的字典序小于序列 b=b1,b2,,bnb=\lang b_1,b_2,\cdots,b_n\rang 的字典序,当且仅当存在一个位置 pp,满足 a1=b1,a2=b2,,ap1=bp1,ap<bpa_1=b_1,a_2=b_2,\cdots,a_{p-1}=b_{p-1},a_p<b_p

4 16
3 1 2 4

提示

  • 对于 40%40\% 的数据,1n71\le n\le 7
  • 对于 80%80\% 的数据,1n101\le n \le 10
  • 对于 100%100\% 的数据,1n121\le n \le 121sum123451\le sum\le 12345