洛谷 P1118 [USACO06FEB]数字三角形Backward Digit Su…

题目描述

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.

有这么一个游戏:

写出一个1~N的排列a[i],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1,直到只剩下一个数字位置。下面是一个例子:

3 1 2 4

4 3 6

7 9

16
最后得到16这样一个数字。

现在想要倒着玩这样一个游戏,如果知道N,知道最后得到的数字的大小sum,请你求出最初序列a[i],为1~N的一个排列。若答案有多种可能,则输出字典序最小的那一个。

管理员注:本题描述有误,这里字典序指的是1,2,3,4,5,6,7,8,9,10,11,12

而不是1,10,11,12,2,3,4,5,6,7,8,9[/color]

输入输出格式

输入格式:
两个正整数n,sum。

输出格式:
输出包括1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。(好奇葩啊)

输入输出样例

输入样例#1:
4 16
输出样例#1:
3 1 2 4
说明

对于40%的数据,n≤7;

对于80%的数据,n≤10;

对于100%的数据,n≤12,sum≤12345。

(标题有点长啊)
分析:最后的结果其实是一个杨辉三角。(拿样例来说,第4层的杨辉三角是1 3 3 1; 1 3+3 1+3 2+1 4=16),我们可以枚举全排列,然后带入杨辉三角系数中计算,这还是会超时。我们每一步都要判断一下,当当前值已经大于要求的值,就可以剪枝掉。

代码:

var
 b:array [1..12] of boolean;
 a:Array [1..12] of longint;
 f:array [0..12,0..12] of longint;
 n,m,i,j:longint;
procedure dfs(x:longint);
 var i,t,j:longint;
begin
if x=n+1 then
 begin
  t:=0;
  for i:=1 to n do
   t:=t+f[n,i]*a[i];
  if t=m then
   begin
    for i:=1 to n-1 do
     write(a[i],' ');
     write(a[n]);
    halt;
   end
  else exit;
 end;
 for i:=1 to n do
  if b[i]=false then
   begin
    b[i]:=true;
    a[x]:=i;
    t:=0;
    for j:=1 to x do
     t:=t+f[n,j]*a[j];
    if t<=m then dfs(x+1);
    b[i]:=false;
   end;
end;

begin
 readln(n,m);
 f[1,1]:=1;
 for i:=2 to n do
  for j:=1 to i do
   f[i,j]:=f[i-1,j]+f[i-1,j-1];
 dfs(1);
end.

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