Codeforces Round #277.5 (Div. 2) A

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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 = 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 test(s)
input
55 2 5 1 4
output
20 34 2
input
610 20 20 40 60 60
output
0
input
2101 100
output
10 1
题意：让你在最多n次交换后使数组变成递增数列。
做法：先用另一个a数组复制原先的b数组然后把数组a排序一下，然后两个数组进行比较，如果不同则在数组b中搜索直到搜到需要的值，然后记录需要交换的两个位置。
#include <iostream>#include <cstdio>#include <climits>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <map>#include <set>#include <algorithm>#include<ctime>#define esp 1e-6#define LL long long#define inf 0x0f0f0f0fusing namespace std;int main(){    int n;    int num[3005],num2[3005],num3[3005];    int i,j,ans;    while(scanf("%d",&n)!=EOF)    {        ans=0;        for(i=0;i<n;i++)        {            scanf("%d",&num[i]);            num3[i]=num[i];            num2[i]=num[i];        }        sort(num3,num3+n);        for(i=0;i<n;i++)        {            if(num3[i]!=num[i])            {                ans++;                for(j=i+1;j<n;j++)                if(num[j]==num3[i])                {                swap(num[j],num[i]);                break;                }            }        }        printf("%d\n",ans);        for(i=0;i<n;i++)        {            if(num2[i]!=num3[i])            {                for(j=i+1;j<n;j++)                if(num2[j]==num3[i])                {                printf("%d %d\n",i,j);                swap(num2[i],num2[j]);                break;                }            }        }    }}

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