poj Anti-prime Sequences

Anti-prime Sequences

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 2175		Accepted: 1022

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


题意：求n到m的数中任意连续2到d的数的和是合数

DFS

#include<stdio.h>

#include<string.h>

#include<math.h>



const int MAXN=10001;

int n,m,d;

int vis[MAXN];

int num[MAXN];

int pri[MAXN];

int ans;

int flag;



void init ()

{

    memset(pri,0,sizeof(pri));

    pri[0] = pri[1] = 1;

    for ( int i = 2; i <= 100; i++ )

    {

        if ( pri[i] ) continue;

        for ( int j = 2; i * j < 10001; j++ )//这里错了

            pri[i*j] = 1;

    }

}





bool judge(int t,int step)

{

    num[step]=t;

    if(step>0)

    {

        ans=0;

        int judge=0;

        for(int j=step; j>=step-d+1; j--)

        {

            ans+=num[j];

            if(judge==1 && pri[ans]==0)

            {

                return false;

            }

            judge=1;

            if(j==0) break;

        }

    }

    return true;

}



void DFS(int step)

{

    int ans,i,t;

    if(flag) return ;

    if(step==m-n+1)//已经找到了

    {

        flag=1 ;

        return ;

    }

    for(i=n; i<=m; i++)

    {

        t=0;

        if(flag) return ;

        if(!vis[i] && judge(i,step))

        {

            vis[i]=1;

            DFS(step+1);

            vis[i]=0;

        }

    }

}



int main()

{

    int i,j;

    init();

    while(scanf("%d%d%d",&n,&m,&d))

    {

        flag=0;

        if(n==0 && m==0 && d==0)  break;

        memset(vis,0,sizeof(vis));

        DFS(0);

        if(!flag) printf("No anti-prime sequence exists.");

        else

        {

            for(i=0; i<=m-n; i++)

                if(i==0) printf("%d",num[i]);

                else printf(",%d",num[i]);

        }

        printf("\n");

    }

    return 0;

}