P1118 [USACO06FEB]數字三角形Backward Digit Su…

自為風月馬前卒發表於2017-05-15

題目描述

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.

有這麼一個遊戲:

寫出一個1~N的排列a[i],然後每次將相鄰兩個數相加,構成新的序列,再對新序列進行這樣的操作,顯然每次構成的序列都比上一次的序列長度少1,直到只剩下一個數字位置。下面是一個例子:

3 1 2 4

4 3 6

7 9 16 最後得到16這樣一個數字。

現在想要倒著玩這樣一個遊戲,如果知道N,知道最後得到的數字的大小sum,請你求出最初序列a[i],為1~N的一個排列。若答案有多種可能,則輸出字典序最小的那一個。

[color=red]管理員注:本題描述有誤,這裡字典序指的是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]

輸入輸出格式

輸入格式:

兩個正整數n,sum。

輸出格式:

輸出包括1行,為字典序最小的那個答案。

當無解的時候,請什麼也不輸出。(好奇葩啊)

輸入輸出樣例

輸入樣例#1:
4 16
輸出樣例#1:
3 1 2 4

說明

對於40%的資料,n≤7;

對於80%的資料,n≤10;

對於100%的資料,n≤12,sum≤12345。

楊輝三角+暴力!

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 using namespace std;
 6 int n,num;
 7 const int PT[][13] = //先打一張楊輝三角表
 8 {
 9     { 1 } , // N = 1 
10     { 1 , 1 } , // N = 2
11     { 1 , 2 , 1 } , // N = 3
12     { 1 , 3 , 3 , 1 } , // N = 4
13     { 1 , 4 , 6 , 4 , 1 } , // N = 5
14     { 1 , 5 , 10, 10, 5 , 1 } , // N = 6
15     { 1 , 6 , 15, 20, 15, 6 , 1 } , // N = 7
16     { 1 , 7 , 21, 35, 35, 21, 7 , 1 } , // N = 8
17     { 1 , 8 , 28, 56, 70, 56, 28, 8 , 1 } , // N = 9
18     { 1 , 9 , 36, 84,126,126, 84, 36, 9 , 1 } , // N = 10 
19     { 1 , 10, 45,120,210,252,210,120, 45, 10 , 1 } , // N = 11
20     { 1 , 11, 55,165,330,462,462,330,165, 55 ,11 , 1 } }; // N = 12 
21 int vis[15];
22 int ans[15];
23 int flag;
24 int dfs(int p,int k,int now)//第p個數,值為k,和是now 
25 {
26     
27     if(now>num)return 0;
28     if(p==n)
29     {
30         if(now==num)
31         {
32             ans[p]=k;
33             return 1;
34         }
35         else 
36         return 0;    
37     }
38     vis[k]=1;
39     for(int i=1;i<=n;i++)
40     {
41         if(vis[i]==0&&dfs(p+1,i,now+PT[n-1][p]*i))
42         {
43             ans[p]=k;
44              return 1;
45         }    
46     }
47     vis[k]=0;
48     return 0;
49 }
50 int main()
51 {
52     
53     scanf("%d%d",&n,&num);
54     if(dfs(0,-1,0))
55         for(int i=1;i<=n;i++)
56             printf("%d ",ans[i]);
57     return 0;
58 }

 

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