[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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int l,r,d,haveans=0;
bool mk[100010];
bool mrk[100010];
int s[100010];
int p[100010],len;
inline void getp()
{
    for (int i=2;i<=10000;i++)
    {
        if (!mk[i])
        {
            p[++len]=i;
            for (int j=2*i;j<=10000;j+=i)mk[j]=1;
        }
    }
}
inline void dfs(int now)
{
    if (now==r+1)
    {
        for (int i=l;i<r;i++)printf("%d,",s[i]);
        printf("%d
",s[r]);
        haveans=1;
        return;
    }
    for (int i=l;i<=r;i++)
    {
        if (mrk[i])continue;
        int sum=i,mrk2=1;
        for (int j=now-1;j>=max(l,now-d+1);j--)
        {
            sum+=s[j];
            if(!mk[sum]){mrk2=0;break;}
        }
        if (!mrk2)continue;
        mrk[i]=1;
        s[now]=i;
        dfs(now+1);
        if (haveans)return;
        s[now]=0;
        mrk[i]=0;
    }
}
int main()
{
    getp();
    while (~scanf("%d%d%d",&l,&r,&d)&&l+r+d)
    {
        memset(mrk,0,sizeof(mrk));
        haveans=0;
        dfs(l);
        if (!haveans)puts("No anti-prime sequence exists.");
    }
}