poj Antiprime 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

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;
}