Anti-prime Sequences

Anti-prime Sequences

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 3355		Accepted: 1531

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 danti-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
题意：在【2，d】长度的连续序列的和都要为合数。
思路：DFS。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<stdlib.h>
 5 #include<string.h>
 6 #include<queue>
 7 #include<stack>
 8 #include<math.h>
 9 using namespace std;
10 typedef long long LL;
11 bool prime[20000]= {0};
12 int tt[10000];
13 bool cm[1005];
14 int ts=0;
15 bool check(int n,int m);
16 int dfs(int n,int m,int d,int kk,int pp);
17 int main(void)
18 {
19     int i,j,k;
20     for(i=2; i<=1000; i++)
21     {
22         if(!prime[i])
23         {
24             for(j=i; (i*j)<=20000; j++)
25             {
26                 prime[i*j]=true;
27             }
28         }
29     }
30     int n,m;
31     while(scanf("%d %d %d",&n,&m,&k),n!=0&&m!=0&&k!=0)
32     {
33         memset(cm,0,sizeof(cm));
34         ts=0;
35         int uu=dfs(0,m-n+1,k,n,m);
36         if(uu)
37         {
38             printf("%d",tt[0]);
39             for(i=1; i<(m-n+1); i++)
40             {
41                 printf(",%d",tt[i]);
42             }
43             printf("\n");
44         }
45         else printf("No anti-prime sequence exists.\n");
46     }
47 }
48 bool check(int n,int m)
49 {
50     int i,j;
51
52
53         LL sum=tt[m];
54         for(i=m-1; i>=max(n,0); i--)
55         {
56             sum+=tt[i];
57             if(!prime[sum])
58                 return false;
59         }
60         return true;
61 }
62 int dfs(int n,int m,int d,int kk,int pp)
63 {
64     int i;
65     if(ts)return 1;
66     if(n==m)
67     {
68
69             bool cc=check(n-d,m-1);
70             if(!cc)
71             {
72                 return 0;
73             }
74         ts=1;
75         return 1;
76     }
77     else
78     {
79         bool cc=check(n-d,n-1);
80         if(cc)
81         {
82             for(i=kk; i<=pp; i++)
83             {
84                 if(ts)return 1;
85                 if(!cm[i])
86                 {
87                     tt[n]=i;
88                     cm[i]=true;
89                     int uu=dfs(n+1,m,d,kk,pp);
90                     cm[i]=false;
91                     if(uu)return 1;
92                 }
93             }
94         }
95         else return 0;
96     }
97     return 0;
98 }