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
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 #include
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 }