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

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

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #define LL long long
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int l,r,d,haveans=;
 bool mk[];
 bool mrk[];
 int s[];
 int p[],len;
 inline void getp()
 {
     for (int i=;i<=;i++)
     {
         if (!mk[i])
         {
             p[++len]=i;
             for (int j=*i;j<=;j+=i)mk[j]=;
         }
     }
 }
 inline void dfs(int now)
 {
     if (now==r+)
     {
         for (int i=l;i<r;i++)printf("%d,",s[i]);
         printf("%d\n",s[r]);
         haveans=;
         return;
     }
     for (int i=l;i<=r;i++)
     {
         if (mrk[i])continue;
         int sum=i,mrk2=;
         for (int j=now-;j>=max(l,now-d+);j--)
         {
             sum+=s[j];
             if(!mk[sum]){mrk2=;break;}
         }
         if (!mrk2)continue;
         mrk[i]=;
         s[now]=i;
         dfs(now+);
         if (haveans)return;
         s[now]=;
         mrk[i]=;
     }
 }
 int main()
 {
     getp();
     while (~scanf("%d%d%d",&l,&r,&d)&&l+r+d)
     {
         memset(mrk,,sizeof(mrk));
         haveans=;
         dfs(l);
         if (!haveans)puts("No anti-prime sequence exists.");
     }
 }

poj 2034