[hdu] Anti-prime Sequences

Anti-prime Sequences
Time Limit : 6000/3000ms (Java/Other) Memory Limit : 60000/30000K (Java/Other)

Total Submission(s) : 8 Accepted Submission(s) : 5

Problem 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

Source
PKU

[cpp]
// 8.16.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"

#include<cstdio>
#include<cstring>
#define MAX 10500
int n,m,d;
int flag[MAX],res[MAX],prime[MAX+1];
bool fflag;
void Prime()
{
    prime[1]=1;
    for(int i=2;i*i<=MAX;i++)
    {
        if(!prime[i])
        {
            for(int j=i<<1;j<=MAX;j+=i)
                prime[j]=1;
        }
    }
}
bool check(int k)
{
    if(k==0) return true;
    int sum=res[k];
    for(int i=k-1;i>=0&&i>=k-d+1;i--)
    {
        sum+=res[i];
        if(!prime[sum]) return false;
    }
    return true;
}
void DFS(int k)//k表示当前res数组所装元素的大小
{
    if(k==m-n+1)
    {
        fflag=true;
        for(int i=0;i<=m-n;i++)
        {
            if(i==0) printf("%d",res[i]);
            else printf(",%d",res[i]);
        }
        return ;
    }
    if(fflag)return ;
    for(int i=n;i<=m;i++)
    {
        if(!flag[i])
        {
            if(fflag)return ;
             res[k]=i;
             if(check(k))
             {
                flag[i]=true;
                DFS(k+1);
                if(fflag)return ;
                flag[i]=false;
             }
        }
    }
}
int main()
{
    Prime();
    while(scanf("%d%d%d",&n,&m,&d),(n||m||d))
    {
        fflag=false;
        memset(flag,false,sizeof(flag));
        DFS(0);
        if(!fflag) printf("No anti-prime sequence exists.");
        puts("");

    }
    return 0;
}

补充：软件开发 , C++ ,