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
这个题目乍一看，好像很麻烦，暴力搜索的话，时间复杂度会很高。但是，再细看题目，会发现这种序列能满足的非常有限，有很多的搜索是不会时行的。
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; }