Codeforces Round #277.5 (Div. 2)-A. SwapSort （sort）_html/css_WEB-ITnose

SwapSort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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,?...,?an?-?1 (?-?109?≤?ai?≤?109), where ai 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 ai and aj. 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

55 2 5 1 4

output

20 34 2

input

610 20 20 40 60 60

output

input

2101 100

output

10 1

题意：给一组数，问怎么可以实现用最少次的交换，实现升序排序，输出交换过程。

解题思路：当时比赛竟然脑残了，WA了两发，貌似和逆序数很像，对于每个元素，都找他后面的最大值，若最大值和当前i位置的值不等，即交换，保存一下交换记录，最后输出即可。

AC代码：

#include #include #include #include using namespace std;int a[3010];int b[120];int vis[120];typedef pair P;P m[3010];int main(){    int n;    cin>>n;    for(int i = 0; i >a[i];    int sum = 0,ans = 0;    int i,j;    for(i = 0; i  a[j])           {                x = a[j];                y = j;           }        }        if(y != i)        {            m[sum].first = i;            m[sum++].second = y;            swap(a[i],a[y]);            ans++;        }    }    cout