[USACO06FEB]数字三角形

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。

[color=red]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。
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分析

运用的dfs思想,f代表这一层的杨辉三角。a代表这一层的排列。
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程序:
var
f:array[0..130,0..130]of longint;
a:array[0..130]of longint;
v:array[0..130]of boolean;
j,n,m,sum,i:longint;

procedure dfs(i:longint);
var
j:longint;
begin
    if m>sum then exit;
    if i=n+1 then
    begin
        if sum=m then
        begin
            for j:=1 to n do
            write(a[j],' ');
            halt;
        end;
        exit;
    end;
    for j:=1 to n do
    if v[j]=true then
    begin
        a[i]:=j;
        m:=m+f[n,i]*j;
        v[j]:=false;
        dfs(i+1);
        a[i]:=0;
        m:=m-f[n,i]*j;
        v[j]:=true;
    end;
end;

begin
    read(n,sum);
    for i:=1 to n do
    f[i,1]:=1;
    for i:=2 to n do
    for j:=2 to i do
    f[i,j]:=f[i-1,j]+f[i-1,j-1];
    m:=0;
    fillchar(v,sizeof(v),true);
    dfs(1);
end.
原文地址:https://www.cnblogs.com/YYC-0304/p/9499989.html