[POJ](3187)Backward Digit Sums ---- 穷竭搜索

Description

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.

《挑战程序设计竞赛》新知：
当搜索的状态空间比较特殊时，可以很简短的实现，比如C++的标准库提供了next_permutation这一函数，可以把n个元素共n!种不同的序列生成出来。暴力枚举~

本题解题思路：一个倒置的“杨辉三角”，枚举所以第一层的情况，直到满足最后一层等于final_sum即可。

AC代码：

#include
#include
#include
#include
using namespace std;
int main()
{
    int a[15];
    int b[15];
    ios_base::sync_with_stdio(false);
    cin.tie(NULL),cout.tie(NULL);
    int n,sum;
    while(cin>>n>>sum){
        for(int i=1;i<=n;i++)
            a[i] = i;
        do{
            for(int i=1;i<=n;i++)
                b[i] = a[i];
            int nn = n;
            while(nn>1)
            {
                for(int i=1;i1];
                nn--;
            }
            if(b[1] == sum)
            {
                for(int i=1;i<=n;i++)
                {
                    if(i!=n)
                        cout<" ";
                    else
                        cout<break;
            }
        }while(next_permutation(a+1,a+n+1));
    }
    return 0;
}