# Backward Digit Sums POJ - 3187

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.

Input

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

Output

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.

Sample Input

4 16

Sample Output

3 1 2 4

Hint

Explanation of the sample:

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

FJ和他的牛喜欢玩心理游戏。他们将1到n（1<=n<=10）之间的数字按一定的顺序写下来，然后对相邻的数字求和，生成一个新的列表，其中少一个数字。它们重复这个过程，直到只剩下一个数字。例如，游戏的一个实例（n=4）可能如下所示：

3 1 2 4

4 3 6

7 9

16

1 #include <cstdio>
2 #include <fstream>
3 #include <algorithm>
4 #include <cmath>
5 #include <deque>
6 #include <vector>
7 #include <queue>
8 #include <string>
9 #include <cstring>
10 #include <map>
11 #include <stack>
12 #include <set>
13 #include <sstream>
14 #include <iostream>
15 #define mod 1000000007
16 #define eps 1e-6
17 #define ll long long
18 #define INF 0x3f3f3f3f
19 using namespace std;
20
21 int n,he;
22 int sz[15];
23 int qh()
24 {
25     int s=n;
26     int num[15];
27     for(int i=0;i<s;i++)
28     {
29         num[i]=sz[i];
30     }
31     while(s!=1)
32     {
33         for(int i=0;i<s-1;i++)
34         {
35             num[i]=num[i]+num[i+1];
36         }
37         s--;
38     }
39     return num[0];
40 }
41 int main()
42 {
43     scanf("%d %d",&n,&he);
44     for(int i=0;i<n;i++)
45     {
46         sz[i]=i+1;
47     }
48     do
49     {
50        if(qh()==he)
51        {
52             for(int i=0;i<n-1;i++)
53             {
54                 printf("%d ",sz[i]);
55             }
56             printf("%d\n",sz[n-1]);
57            break;
58        }
59     }while(next_permutation(sz,sz+n));
60
61 }

Backward Digit Sums POJ - 3187

(0)
(0)