Backward Digit Sums

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.

输入

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

输出

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.

样例输入

4 16

样例输出

3 1 2 4

提示

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：

我们把1-n这n个数的某个排列摆成一排，然后相邻两个数的和放到下一行，依此类推，形成一个三角形

3   1    2   4

4   3   6

7   9

16

给你n和最终得到的和k，求第一行的系列（字典序最小）

我们可以发现每个数字的计算次数是一个杨辉三角

1   1

1    2    1

1    3     3    1

1     4     6      4     1

......

所以我们只要枚举1-n的全排列，计算sum{now[i]*as(n-1,i)}是否为K就行了

#include <iostream>
#include <queue>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
typedef long long LL;
const LL INF = 0xfffffff;
const int maxn = 500;
const LL MOD = 1e9+7;
int ok[maxn], now[maxn], as[maxn][maxn], n, m, vis[maxn], flag;///poj 3187
void init()
{
    int i, j;
    as[0][0] = 1;
    for(i = 0; i <= 10; i++)
    {
        as[i][0] = as[i][i] = 1;
        for(j = 1; j < i; j++)
            as[i][j] = as[i-1][j] + as[i-1][j-1];
    }
}
void dfs(int cnt)
{
    if(flag )return ;
    if(cnt == n)
    {
        int ans = 0;
        for(int i = 0; i < n; i++)
            ans += as[n-1][i] * now[i];
        if(ans == m)
        {
            flag = 1;
           for(int i = 0; i < n; i++)
            ok[i] = now[i];
        }
        return ;
    }
    for(int i = 1; i <= n; i++)
    {
        if(!vis[i])
        {
            vis[i] = 1;
            now[cnt++] = i;
            dfs(cnt);
            vis[i] = 0;
            cnt--;
        }
    }
  //  return ;
}
int main()
{
    init();
    while(~scanf("%d %d", &n, &m))
    {
        flag = 0;
        memset(vis, 0, sizeof(vis));
        dfs(0);
        printf("%d", ok[0]);
        for(int i = 1; i < n; i++)
            printf(" %d", ok[i]);
        printf("\n");
    }
    return 0;
}