codeforces489A——SwapSort

Description

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 - 1from 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 = jand 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 Input

Input

5
5 2 5 1 4

Output

2
0 3
4 2

Input

6
10 20 20 40 60 60

Output

0

Input

2
101 100

Output

1
0 1
大意：让你求出交换次数以及每次交换的元素
不知道为什么用归并做WA，用vector遍历每一个A了。。

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<vector>

using namespace std;

vector<pair <int ,int> >ans;

int a[3000],n;

int main(){

    scanf("%d",&n);

    for(int i = 0; i < n ; i++)

    scanf("%d", &a[i]);

    for(int i = n - 1; i >= 0; i--){

            int p = max_element(a,a+i+1)-a;

            swap(a[p],a[i]);

            ans.push_back(make_pair(i,p));

    }

    printf("%d\n",ans.size());

    for(int i = 0; i < ans.size();i++)

        printf("%d %d\n",ans[i].first,ans[i].second);

   return 0;

}

View Code

 学到了vector的max_elememt的用法，如max_element(a,a+i+1)就是求其之间的最大元素的下标！注意是下标还要减去a，用容器的size比较简单