Gym 100971B 水&愚

Description

standard input/output
Announcement

 

• Statements

  A permutation of n numbers is a sequence of integers from 1 to n where each number is occurred exactly once. If a permutation p₁, p₂, ..., p_n has an index i such that p_i = i, this index is called a fixed point.

  A derangement is a permutation without any fixed points.

  Let's denote the operation swap(a, b) as swapping elements on positions a and b.

  For the given permutation find the minimal number of swap operations needed to turn it into derangement.

Input

The first line contains an integer n(2 ≤ n ≤ 200000) — the number of elements in a permutation.

The second line contains the elements of the permutation — n distinct integers from 1 to n.

Output

In the first line output a single integer k — the minimal number of swap operations needed to transform the permutation into derangement.

In each of the next k lines output two integers a_i and b_i(1 ≤ a_i, b_i ≤ n) — the arguments of swap operations.

If there are multiple possible solutions, output any of them.

Sample Input

Input

6
6 2 4 3 5 1

Output

1
2 5

题意：使得i！=pi  移动最少的次数  并输出的交换的两者的位置

题解：找到需要交换的所谓的位置 若个数为偶数,则直接两两交换
     若为奇数 则最后一个和前一个位置交换 
     当为的第一个位置时 注意特判。

 #include<bits/stdc++.h>
 using namespace std;
 int n;
 int a[];
 //int where[200005];
 int aa[];
 int main()
 {
     scanf("%d",&n);
     for(int i=;i<=n;i++)
         {
             scanf("%d",&a[i]);
             //where[a[i]]=i;
         }
     int ans=;
     for(int i=;i<=n;i++)
     {
         if(a[i]==i)
         {
             ans++;
             aa[ans]=i;
         }
     }
     if(ans%==)
     {
         cout<<ans/<<endl;
         for(int i=;i<ans;i+=)
         cout<<aa[i]<<" "<<aa[i+]<<endl;
     }
     else
     {
         cout<<ans/+<<endl;
         for(int i=;i<ans;i+=)
         cout<<aa[i]<<" "<<aa[i+]<<endl;
         if(aa[ans]==)
             cout<<"1 2"<<endl;
         else
             cout<<aa[ans]-<<" "<<aa[ans]<<endl;
     }
     return ;
 }

#include<bits/stdc++.h>
using namespace std;
int n;
int a[200005];
//int where[200005];
int aa[200005];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            //where[a[i]]=i;
        }
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]==i)
        {
            ans++;
            aa[ans]=i;
        }
    }
    if(ans%2==0)
    {
        cout<<ans/2<<endl;
        for(int i=1;i<ans;i+=2)
        cout<<aa[i]<<" "<<aa[i+1]<<endl;
    }
    else
    {
        cout<<ans/2+1<<endl;
        for(int i=1;i<ans;i+=2)
        cout<<aa[i]<<" "<<aa[i+1]<<endl;
        if(aa[ans]==1)
            cout<<"1 2"<<endl;
        else
            cout<<aa[ans]-1<<" "<<aa[ans]<<endl;
    }
    return 0;
}