Codeforces 1148C (thinking)

Face questions

Given a length of n are arranged in a, be exchanged every time the i-th and j-th element of a sequence, if and only if \ (2 \ Times | ij of | \ GEQ n \) , seeking for exchanging programs, so that sequence from small to large sorted

analysis

Our focus is to consider how to change the first x number of i-th position, and try not to undermine the position of the other numbers

We use first, a sequence number n as a springboard, if the distance between the n and n satisfy the conditions to the exchange, or the exchange 1, defined find (x) as x Who should exchange.

1. Find the x and (x) switching, then to Find x (x)

2.find (x) and find (i) exchange of x at this time to find (i)

3.find (i) exchange of i, then x to i

We can find such exchange will only produce in exchange for n number 1, regardless of the number of other location unchanged. We just perform the above operations on operands 2 ~ n-1, and finally we look at a [1] is not 1, then the exchange or need determination (1, n-) to

Code

#include<iostream>
#include<cstdio>
#include<algorithm> 
#include<vector> 
#define maxn 300005
using namespace std;

int n; 
vector< pair<int,int> >ans;
int a[maxn];
int pos[maxn];//值为i的数的位置 
inline int find(int x){
    if(2*(n-x)>=n) return n;
    else return 1;
} 
void change(int x,int y){
    ans.push_back(make_pair(x,y));
    swap(pos[a[x]],pos[a[y]]);
    swap(a[x],a[y]);
} 

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        pos[a[i]]=i;
    }
    for(int i=2;i<n;i++){
        int t=pos[i];
        change(t,find(t));
        if(find(t)!=find(i)) change(find(t),find(i));
        change(i,find(i)); 
    }
    if(a[1]!=1) change(1,n);
    printf("%d\n",ans.size());
    for(auto p : ans){
        printf("%d %d\n",p.first,p.second); 
    } 
}