1002. Anti-prime Sequences

Total:

3346

Accepted:

1220

Rating:

3.5/5.0(54 votes)

012 345

	Time Limit: 3sec    Memory Limit:32MB 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 Copy sample input to clipboard 1 10 21 10 31 10 540 60 70 0 0 Sample Output 1,3,5,4,2,6,9,7,8,101,3,5,4,6,2,10,8,7,9No 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

AC code

#include 
    
     #include 
     
      #include 
      
       using namespace std;int ans[1000],m,n,d,len;//len为sequence的长度，就是m-n+1bool flag,vis[1001],composite[10000]={1,1};//flag用于判断是否生成anti-prime sequence//筛选素数,composite[i]=1当且仅当i不是素数void InitComposite(){    for(int i=2;i<1001;i++)    {        if(!composite[i])        {            for(int j=2;j*i<10000;j++)                composite[i*j]=1;        }    }    //for(int i=0;i<100;i++) cout<
       <<":"<
        
         <<" ";}//检查sum是否是composite numberbool CheckSum(int idx){    int sum=ans[idx];    for(int i=idx-1;i>-1&&i>idx-d;i--)    {        sum+=ans[i];        if(!composite[sum]) return 0;    }    return 1;}//depth记录当前有多少个数成功加入sequencevoid DFS(int depth){    //当depth大于len时，代表找到anti-prime sequence    if(depth==len)    {        flag=1;        return ;    }    else    {        for(int i=n;i<=m;i++)        {            if(!vis[i])            {                ans[depth]=i;                if(CheckSum(depth))                {                    vis[i]=1;                    //ans中第depth个数已经确定，往depth+1处搜索                    DFS(depth+1);                    vis[i]=0;//不要漏了这个！！！                    if(flag) return ;                }            }        }    }    return ;}int main(){    InitComposite();    while(scanf("%d%d%d",&n,&m,&d) && m)    {        memset(vis,0,sizeof(vis));        len=m-n+1;        flag=0;        DFS(0);        if(flag)        {            for(int i=0;i