codeforces A. Slightly Decreasing Permutations 题解

最新推荐文章于 2018-10-09 16:24:34 发布

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最新推荐文章于 2018-10-09 16:24:34 发布

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本文介绍了一种算法，用于生成长度为n且下降系数为k的排列。通过两种不同的实现方式，展示了如何根据输入参数n和k构造符合要求的排列序列。

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Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1, p2, ..., pn.

The decreasing coefficient of permutation p1, p2, ..., pn is the number of such i (1 ≤ i < n), that pi > pi + 1.

You have numbers n and k. Your task is to print the permutation of length n with decreasing coefficient k.

Input

The single line contains two space-separated integers: n, k (1 ≤ n ≤ 105, 0 ≤ k < n) — the permutation length and the decreasing coefficient.

Output

In a single line print n space-separated integers: p1, p2, ..., pn — the permutation of length n with decreasing coefficient k.

If there are several permutations that meet this condition, print any of them. It is guaranteed that the permutation with the sought parameters exists.

Sample test(s)

input
5 2

output
1 5 2 4 3

这种题目难就难在下标处理。当然题目本身不难。

void SlightlyDecreasingPermutations()
{
	int n, k;
	cin>>n>>k;
	int *A = new int[n];
	
	for (int i = 0; i < n - k - 1; i++)
	{
		A[i] = i + 1;
	}
	for (int i = n - k - 1, j = n; i < n; i++, j--)
	{
		A[i] = j;
	}
	for (int i = 0; i < n; i++)
	{
		cout<<A[i]<<' ';
	}
	delete [] A;
}

void SlightlyDecreasingPermutations_2()
{
	int n, k;
	cin>>n>>k;
	int *A = new int[n];

	for(int i = 0; i < n; ++i) A[i] = i+1;
	reverse(A,A+k+1);

	for(int i = 0; i < n; ++i) cout<<A[i]<<' ';
}