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, ..., an - 1 ( - 109 ≤ ai ≤ 109), where ai 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 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 = 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
5
5 2 5 1 4
2
0 3
4 2
6
10 20 20 40 60 60
0
2
101 100
1
0 1
题意：用少于n次的swap来排好序
分析：选择排序，每次在未排序的后面选出最小的一个然后和未排序的第一个交换，这样最多n-1次交换即可排好序，用数组记录交换的两个数的位置。
#include<stdio.h>
int n,i,j,a[],coun,temp,swap1[],swap2[];
int main()
{
scanf("%d",&n);
for(i=; i<n; i++)
scanf("%d",a+i);
for(i=; i<n-; i++)
{
int minj=i;
for(j=i+; j<n; j++)
if(a[j]<a[minj])minj=j;
if(i!=minj)
{
swap1[coun]=i;
swap2[coun]=minj; temp=a[minj];
a[minj]=a[i];
a[i]=temp; coun++;
}
}
printf("%d\n",coun);
if(coun)
for(i=; i<coun; i++)
printf("%d %d\n",swap1[i],swap2[i]);
return ;
}