UVA 11129 An antiarithmetic permutation

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 permutationp 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 (pi, pj, pk) 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 n from 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题意：给定一个包含了0到n - 1的序列。。要使得这个序列中每个长度大于2的子序列都不是等差数列。。

思路：对于一个等差的序列。如0 1 2 3 4 5 我们可以这样做，把他分离成2部分等差子序列0 2 4和1 3 5然后组合成一个新的序列0 2 4 1 3 5。这样做的话，可以保证前半部分无法和后半部分组合成等差数列。可以证明，一个序列的等差是k，首项为a1,序列为,a1, a1 +k, a1 + 2k, a1 + 3k .... a1 + (n - 1)k.分成的两部分为a1, a1 + 2k , a1 + 4k ....、 a1, a1 + k, a1 + 3k, a1 + 5k...如此一来，任意拿前面和后面组成序列的话。后面和后面的差都是2k的倍数，前面和后面的都是2k + 1的倍数。这样是成不了等差的。。如此一来。我们只要把序列一直变换，变换到个数小于等于2即可。

代码：

#include <stdio.h>

int n, i;
int num[10005], copy[10005];

void solve(int start, int end) {
	int i = start, j;
	if (start >= end - 2) return;
	for (j = start; j < end; j ++) copy[j] = num[j];
	for (j = start; j < end; j += 2, i ++) num[i] = copy[j];
	for (j = start + 1; j < end; j += 2, i ++) num[i] = copy[j];
	solve(start, (start + end + 1) / 2);
	solve((start + end + 1) / 2, end);
}
int main() {

	while (~scanf("%d", &n) && n) {		
		for (i = 0; i < n; i ++)
			num[i] = i;
		solve(0, n);
		printf("%d:", n);
		for (int i = 0; i < n; i ++)
			printf(" %d", num[i]);
		printf("\n");
	}
	return 0;
}

补充：软件开发 , C++ ,