Uva--11129（分治，数学）

2014-07-27 14:13:29

Problem A: An antiarithmetic permutation

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 is no subsequence of it forming an arithmetic progression of length bigger than 2, i.e. there are no three indices 0 ≤ i < j < k < n such that (p_i, p_j, p_k) forms an arithmetic progression.

For example, the sequence (2, 0, 1, 4, 3) is an antiarithmetic permutation of 5. The sequence (0, 5, 4, 3, 1, 2) is not an antiarithmetic permutation of 6 as its first, fifth and sixth term (0, 1, 2) form an arithmetic progression; and so do its second, fourth and fifth term (5, 3, 1).

Your task is to generate an 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 each nfrom input, produce one line of output containing an (any will do) antiarithmetic permutation of n in the format 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

思路：分治思想，每次把奇数位置和偶数位置的数分开来，最后出现的排列就可以实现提述要求。

/*************************************************************************
    > File Name: l.cpp
    > Author: Nature
    > Mail: 564374850@qq.com
    > Created Time: Sat 26 Jul 2014 12:58:48 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

int n,v[10005],pos;

void Work(int p,int cnt,int d){
    if(cnt == 1){
        v[pos++] = p;
        return;
    }
    int n1 = ceil(1.0 * cnt / 2);
    int n2 = cnt / 2;
    Work(p,n1,2 * d);
    Work(p + d,n2,2 * d);
}

int main(){
    while(scanf("%d",&n) == 1 && n){
        pos = 0;
        Work(0,n,1);
        printf("%d:",n);
        for(int i = 0; i < n; ++i)
            printf(" %d",v[i]);
        printf("\n");
    }
    return 0;
}