anti prime

sicily 1002. Anti-prime Sequences

Constraints

Time Limit: 3 secs, Memory Limit: 32 MB

Description

 

Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number. For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10. This is also the lexicographically first such sequence. We can extend the definition by defining a degree d anti-prime sequence as one where all consecutive subsequences of length 2,3,...,d sum to a composite number. The sequence above is a degree 2 anti-prime sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11. The lexicographically .rst degree 3 anti-prime sequence for these numbers is 1,3,5,4,6,2,10,8,7,9.

 

Input

 

Input will consist of multiple input sets. Each set will consist of three integers, n, m, and d on a single line. The values of n, m and d will satisfy 1 <= n < m <= 1000, and 2 <= d <= 10. The line 0 0 0 will indicate end of input and should not be processed.

 

Output

 

For each input set, output a single line consisting of a comma-separated list of integers forming a degree danti-prime sequence (do not insert any spaces and do not split the output over multiple lines). In the case where more than one anti-prime sequence exists, print the lexicographically first one (i.e., output the one with the lowest first value; in case of a tie, the lowest second value, etc.). In the case where no anti-prime sequence exists, output No anti-prime sequence exists.

 

Sample Input

1 10 2
1 10 3
1 10 5
40 60 7
0 0 0

Sample Output

1,3,5,4,2,6,9,7,8,10
1,3,5,4,6,2,10,8,7,9
No anti-prime sequence exists.
40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54


这个题目初看起来好像很麻烦，暴力搜索的话，时间复杂度会很高。但是，再仔细推敲一下，你会发现能满足条件的序列其实非常有限，有很多的搜索是不会进行的。

因此暴力未必不可行，放手一试，深度优先搜索，果然过了。

static bool used[MAXV];//标记当前的数字是否已被使用。
static int result[MAXV];
static int m,n,d,t;

//检查当前位置 pos，放置 value是否合法。
bool Check(int pos,int value)
{
  int sum = value;
  for(int i=2;i <= d && (pos - (i-1))>=0;i++)
  {
     sum += result[pos - (i-1)];
     if(isPrime[sum])
         return false;
  }
  return true;
}



int dfs(int pos)
{
  if(pos > m-n)
      return 1; 

  int ret;
  for(int i = n;i <= m;++i)
  {
      if(used[i])
         continue;

      if(pos >= 2 && !Check(pos,i))
          continue;      

      used[i]=1;
      result[pos] = i;
      ret = dfs(pos+1);
      if(ret)
          return 1;
      used[i] = 0;
    }
   
    return 0;
}