UVa 11129:an antiarithmetic permutation

Link

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_ problem&problem=2070

"Original question"

A permutation of n+1 is a bijective function of the initial n+1 natural-numbers:0, 1, ... n. A permutation p is called antiarithmetic if there isn't no subsequence of it forming an arithmetic progression of length Bigg Er than 2, i.e. there are no three indices 0≤i < J < K < n such that (pi, PJ, pk) forms a arithmetic Progressi On.

For example, the sequence (2, 0, 1, 4, 3) are an antiarithmetic permutation of 5. The sequence (0, 5, 4, 3, 1, 2) is not a antiarithmetic permutation of 6 as its A, fifth and sixth term (0, 1, 2) for m an arithmetic progression; And so does its second, fourth and fifth term (5, 3, 1).

Your task is to generate a antiarithmetic permutation of N.

Each line of the input file contains a natural number 3≤n≤10000. The last line of input contains 0 marking the "end of" input. For all n from input, produce one line of output containing of antiarithmetic permutation of N in the forma T shown below.

Sample input

3
5
6
0

Output for sample input

3:0 2 1 
5:2 0 1 4 3 6:2 4 3 5 0 1

"The main effect of the topic"

Enter N, and then by 0, 1, ... N-1 consists of a sequence of antiarithmetic, which cannot have a linear sequence of 3 elements.

"Analysis and Summary"

This column more highlights: http://www.bianceng.cn/Programming/sjjg/

This problem has been thought for a long time or no idea. So Baidu study. The original is the use of the method of partition, which has always been my weaker place.

Code

* 
 * * Uva:11129-an antiarithmetic permutation * time:0.012s * author:d_double * * * * 
 and
     
#inc lude<cstdio>  
int init[10003],tar[10003],t=10000;  
     
void ST (int l,int h)  
{  
    int t=l;      
    if (l==h) return;  
    for (int i=l;i<=h;i+=2)  
        tar[t++]=init[i];  
    for (int i=l+1;i<=h;i+=2)  
        tar[t++]=init[i];  
    for (int i=l;i<=h;i++)  
        init[i]=tar[i];   
    ST (L, (l+h)/2);  
    ST ((l+h)/2+1,h);  
}  
     
int main ()  
{     
    while (~scanf ("%d", &t) &t) {for  
        (int i=0;i<t;i++)  
           init[i]=i;  
        ST (0,t-1);  
        printf ("%d:%d", t,tar[0]);  
        for (int i=1;i<t;i++)  
            printf ("%d", Tar[i]);  
        printf ("\ n");  
    }  
    return 0;  
}