[暑假集训--数论]poj2034 Anti-prime Sequences

Posted 2020-10-01 zhber

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

 

需要一个 l 到 r 的排列，使得任意一个长度是2,3,...,k的子串当中子串和都不是质数

直接搜就是了

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #define LL long long
 5 using namespace std;
 6 inline LL read()
 7 {
 8     LL x=0,f=1;char ch=getchar();
 9     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
10     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
11     return x*f;
12 }
13 int l,r,d,haveans=0;
14 bool mk[100010];
15 bool mrk[100010];
16 int s[100010];
17 int p[100010],len;
18 inline void getp()
19 {
20     for (int i=2;i<=10000;i++)
21     {
22         if (!mk[i])
23         {
24             p[++len]=i;
25             for (int j=2*i;j<=10000;j+=i)mk[j]=1;
26         }
27     }
28 }
29 inline void dfs(int now)
30 {
31     if (now==r+1)
32     {
33         for (int i=l;i<r;i++)printf("%d,",s[i]);
34         printf("%d\n",s[r]);
35         haveans=1;
36         return;
37     }
38     for (int i=l;i<=r;i++)
39     {
40         if (mrk[i])continue;
41         int sum=i,mrk2=1;
42         for (int j=now-1;j>=max(l,now-d+1);j--)
43         {
44             sum+=s[j];
45             if(!mk[sum]){mrk2=0;break;}
46         }
47         if (!mrk2)continue;
48         mrk[i]=1;
49         s[now]=i;
50         dfs(now+1);
51         if (haveans)return;
52         s[now]=0;
53         mrk[i]=0;
54     }
55 }
56 int main()
57 {
58     getp();
59     while (~scanf("%d%d%d",&l,&r,&d)&&l+r+d)
60     {
61         memset(mrk,0,sizeof(mrk));
62         haveans=0;
63         dfs(l);
64         if (!haveans)puts("No anti-prime sequence exists.");
65     }
66 }

poj 2034