CodeForces 686B Little Robber Girl's Zoo-优快云博客

本文介绍了一个有趣的排序问题：通过特定的交换操作将动物园中的动物按身高非递减顺序排列。文章提供了一种简单直观的解决方案，并附带了完整的C++实现代码。

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Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the order of non-decreasing height. However, the animals were so scared that they couldn't stay in the right places.

The robber girl was angry at first, but then she decided to arrange the animals herself. She repeatedly names numbers l and r such thatr - l + 1 is even. After that animals that occupy positions between l and r inclusively are rearranged as follows: the animal at position lswaps places with the animal at position l + 1, the animal l + 2 swaps with the animal l + 3, ..., finally, the animal at position r - 1 swaps with the animal r.

Help the robber girl to arrange the animals in the order of non-decreasing height. You should name at most 20 000 segments, since otherwise the robber girl will become bored and will start scaring the animals again.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — number of animals in the robber girl's zoo.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the height of the animal occupying the i-th place.

Output

Print the sequence of operations that will rearrange the animals by non-decreasing height.

The output should contain several lines, i-th of the lines should contain two space-separated integers li and ri (1 ≤ li < ri ≤ n) — descriptions of segments the robber girl should name. The segments should be described in the order the operations are performed.

The number of operations should not exceed 20 000.

If the animals are arranged correctly from the start, you are allowed to output nothing.

Examples

input
4
2 1 4 3

output
1 4

input
7
36 28 57 39 66 69 68

output
1 4
6 7

input
5
1 2 1 2 1

output
2 5
3 4
1 4
1 4

Note

Note that you don't have to minimize the number of operations. Any solution that performs at most 20 000 operations is allowed.

没看清题，晕。。。。只要操作不超过20000就好，不需要找到最小的操作次数，所以两层循环，从头到尾遍历，不符合就输出，直到全部符合。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    int a[110];
    while(scanf("%d",&n)!=EOF)
    {
        for (int i = 0; i < n; ++i)
	   {
		scanf("%d", &a[i]);
	   }
	   for (int i = n - 1; i > 0; --i)
	   {
		  for (int j = 0; j < i; ++j)
		  {
			if (a[j] > a[j + 1])
			{
				swap(a[j], a[j + 1]);
				printf("%d %d\n", j + 1, j + 2);
			}
		  }
	   }
    }
    return 0;
}