uva-11129-分治

　　题意：对于一个0-(N-1)的等差数列，求其中一个排列Q，使得任意0<=i,j,k<n 时,ai,aj,ak不是等差数列.

解法:考虑0-(N-1)这个等差数列,假设公差为K,那么数列如下

A0+0K,A0+1K,A0+2K,A0+3K..........+A0+(N-1)K.

如果我们把A0+OK,A0+2K,A0+4K.......组成一个等差数列Q,这个等差数列的公差为2K.

同理，把A0+1K,A0+3K,A0+5K+........组成一个等差数列R,这个等差数列的公差为2K,让R排在Q的后面.这样，从Q中选取俩个,从R中选取一个组成ai,aj,ak,都不会组成等差数列.

#include <string>
#include<iostream>
#include<map>
#include<memory.h>
#include<vector>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<math.h>
#include<iomanip>
#include<bitset>

namespace cc
{
    using std::cout;
    using std::endl;
    using std::cin;
    using std::map;
    using std::vector;
    using std::string;
    using std::sort;
    using std::priority_queue;
    using std::greater;
    using std::vector;
    using std::swap;
    using std::stack;
    using std::bitset;


    constexpr int N = 10010;

    int a[N];
    int aa[N];

    void init(int n)
    {
        for (int i = 0;i < n;i++)
            a[i] = i;
    }
    void dfs(int s, int e)
    {
        if (s >= e - 2)
            return;
        for (int i = s;i < e;i++)
        {
            aa[i] = a[i];
        }
        int ss = s;
        for (int j = s;j < e;j += 2)

            a[ss++] = aa[j];

        int next = ss;
        for (int j = s + 1;j < e;j += 2)
            a[ss++] = aa[j];
        dfs(s, next);
        dfs(next, e);
    }
    void solve()
    {
        int n;
        while (cin >> n && n)
        {
            init(n);
            dfs(0,n);
            cout << n << ":";
            for (int i = 0;i < n;i++)
                cout << " " << a[i];
            cout << endl;
        }
    }

};


int main()
{

#ifndef ONLINE_JUDGE
    freopen("d://1.text", "r", stdin);
#endif // !ONLINE_JUDGE
    cc::solve();
    return 0;
}