Poj3187backward Digit sums[Yang Hui Triangle]

Backward Digit Sums

Time limit:1000ms		Memory Limit: 65536K
Total Submissions: 6350		Accepted: 3673

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to n (1 <= n <=) 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 was 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 has started playing a more difficult game, in which they try to determine the starting Sequenc E 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 so leads to the given sum. If There is multiple solutions, choose the one that's 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 is and possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

Source

Usaco 2006 February Gold & Silver The number of numbers in a full arrangement contributes to sum is Yang Hui triangle n-1 columnYang Hui Triangle is starting from 0Combinatorial number formula recursiondfs+ Pruning

////main.cpp//poj3187////Created by Candy on 9/12/16.//copyright©2016 Candy. All rights reserved.//#include<iostream>#include<cstdio>using namespacestd;Const intn= the;intn,sum,c[n],vis[n],ans[n],flag=0;voidDfsintDintNow ) {    if(flag)return; if(d==n&&now==sum) {flag=1; for(intI=0; i<n;i++) printf ("%d", Ans[i]);return;} if(d==n)return;  for(intI=1; i<=n;i++){        if(Vis[i])Continue; if(now+i*c[d]<=sum) {Ans[d]=i; vis[i]=1; DFS (d+1, now+i*C[d]); Vis[i]=0; }    }}intMainintargcConst Char*argv[]) {scanf ("%d%d",&n,&sum); N--; c[0]=1;  for(intI=1; i<=n;i++) c[i]=c[i-1]* (n-i+1)/i; N++; DFS (0,0); return 0;}

Poj3187backward Digit sums[Yang Hui Triangle]