poj 3187 Backward Digit Sums（暴力）

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.

题意：给出排列1-n的范围n和要得到的m，问字典序最小的排列方式。

思路：用next_permutation函数进行全排列，对于每个排列进行计算直到得到和为m时结束。打印此时得到的排列结果

AC代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <queue>
#include<algorithm>
using namespace std;

int main()
{
    long long n,m;
    long long a[15],b[15];
    while(~scanf("%lld%lld",&n,&m))
    {
        for(int i=0; i<n; i++)
            a[i]=i+1;
        while(1)
        {
            for(int i=0; i<n; i++)
                {
                    b[i]=a[i];
                }
            int flag=n;
            while(flag!=1)
            {
                for(int i=0; i<flag; i++)
                    b[i]=b[i]+b[i+1];
                flag--;
            }
            if(b[0]==m)
            {
                for(int i=0; i<n-1; i++)
                    printf("%lld ",a[i]);
                printf("%lld
",a[n-1]);
                break;
            }
            next_permutation(a,a+n);
        }
    }
    return 0;
}