Codeforces Round #277.5 (Div. 2)A——SwapSort

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#模拟 #codeforces

本文详细阐述了排序算法的应用及其实现过程，包括输入数组的排序、交换元素位置以达到非降序排列，并提供了示例代码。通过比较排序前后数组状态，实现了在最多n次交换内完成排序的任务。

In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are performed consecutively, one after another.

Note that in this problem you do not have to minimize the number of swaps — your task is to find any sequence that is no longer than n.

Input

The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of array elements. The second line contains elements of array: a₀, a₁, ..., a_n - 1 ( - 10⁹ ≤ a_i ≤ 10⁹), where a_i is the i-th element of the array. The elements are numerated from 0 to n - 1 from left to right. Some integers may appear in the array more than once.

Output

In the first line print k (0 ≤ k ≤ n) — the number of swaps. Next k lines must contain the descriptions of the k swaps, one per line. Each swap should be printed as a pair of integers i, j (0 ≤ i, j ≤ n - 1), representing the swap of elements a_i and a_j. You can print indices in the pairs in any order. The swaps are performed in the order they appear in the output, from the first to the last. It is allowed to print i = j and swap the same pair of elements multiple times.

If there are multiple answers, print any of them. It is guaranteed that at least one answer exists.

Sample test(s)

Input

5
5 2 5 1 4

Output

2
0 3
4 2

Input

6
10 20 20 40 60 60

Output

Input

2
101 100

Output

1
0 1

排个序，然后和排序前对比，不一样就往后找到应该在这一位上的数，然后交换

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int a[3030];
int b[3030];

struct node
{
	int x, y;
}pp[3030];

int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		int cnt = 0;
		for (int i = 0; i < n; ++i)
		{
			scanf("%d", &a[i]);
			b[i] = a[i];
		}
		sort(b, b + n);
		int x, y;
		for (int i = 0; i < n; ++i)
		{
			if (a[i] == b[i])
			{
				continue;
			}
			x = i;
			for (int j = i + 1; j < n; ++j)
			{
				if (a[j] == b[i])
				{
					y = j;
					break;
				}
			}
			++cnt;
			pp[cnt].x = x;
			pp[cnt].y = y;
			a[x] ^= a[y];
			a[y] ^= a[x];
			a[x] ^= a[y];
		}
		printf("%d\n", cnt);
		for (int i = 1; i <= cnt; ++i)
		{
			printf("%d %d\n", pp[i].x, pp[i].y);
		}
	}
	return 0;
}