[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

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

巴特西

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

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