1.杨辉三角推导
2.最优性剪枝：当s大于sum时 return
3.逻辑剪枝：dfs顺序从小到大 那么第一个符合s == sum条件的s就是字典序最小的
最优解
复杂度 < O(n!)

1 1 hack
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;
ll sum;
ll a[15];
ll mx = 1e15;
bool st[15];
ll dic[15];

void dfs(int cnt, ll s)
{
    if(cnt >= n)
    {
        if(s == sum)
        {
            mx = 0;
            for(int i = 1;i <= n;i++)cout << dic[i] << " ";
            return;
        }
    }
    if(s >= sum)return;

    for(int i = 1;i <= n;i++)
    {
        if(!st[i])
        {
            st[i] = true;
            dic[cnt + 1] = i;
            dfs(cnt + 1, s + (ll)i * a[cnt + 1]);
            if(mx != 1e15)return;
            st[i] = false;
        }
    }
}

void init()
{
    int t = n - 1;//从0层开始算的层数
    ll fz = 1, fm = 1;
    a[1] = a[n] = 1;
    for(int i = 2;i <= n / 2;i++)
    {
        fz *= t - i + 2;
        fm *= i - 1;
        a[i] = a[n - i + 1] = fz / fm;
    }
    if(n % 2 && n != 1){
        fz *= t - (n / 2 + 1) + 2;
        fm *= n / 2;
        a[n / 2 + 1] = fz / fm;
    }
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> sum;
    init();
//  for(int i = 1;i <= n;i++)cout << a[i] << endl;
    for(int i = 1;i <= n;i++)
    {
        st[i] = true;
        dic[1] = i;
        dfs(1, i);
        st[i] = false;
    }
    return 0;
}