How to solve Almost Equal problem in Codeforces Round?

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输出标准输出：你被给定了整数 n。需要将数字从 1 到 2n（包括1和2n）按顺序排列在圆周上，每个数字使用一次，以满足以下条件：对于每个 n > n，相邻的数字之和应等于 n。

output standard output

You are given integern">nn. You have to arrange numbers from1">11to2n">2n2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:

For everyn">nnconsecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard2n">2n2nnumbers differ not more than by1">11.

For example, choosen=3">n=3n=3. On the left you can see an example of a valid arrangement:1+4+5=10">1+4+5=101+4+5=10,4+5+2=11">4+5+2=114+5+2=11,5+2+3=10">5+2+3=105+2+3=10,2+3+6=11">2+3+6=112+3+6=11,3+6+1=10">3+6+1=103+6+1=10,6+1+4=11">6+1+4=116+1+4=11, any two numbers differ by at most1">11. On the right you can see an invalid arrangement: for example,5+1+6=12">5+1+6=125+1+6=12, and3+2+4=9">3+2+4=93+2+4=9,9">99and12">1212differ more than by1">11.

Input

The first and the only line contain one integern">nn(1≤n≤105">1≤n≤1051≤n≤105).

Output

If there is no solution, output "NO" in the first line.

If there is a solution, output "YES" in the first line. In the second line output2n">2n2nnumbers— numbers from1">11to2n">2n2nin the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.

Examples input Copy

3 output Copy

YES 1 4 5 2 3 6 input Copy

4 output Copy

NO Note

Example from the statement is shown for the first example.

It can be proved that there is no solution in the second example.

cin , cout , printf ,scanf 一定不要混淆使用，关闭同步后会影响输出顺序，2个小时的BUG -_-|

#include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cstdlib> #include <map> #include <vector> #include <set> #include <queue> #include <stack> #include <cmath> // #define lson rt<<1, l, m #define rson rt<<1|1, m+1, r // #define fi first #define se second #define pb push_back #define pq priority_queue<int> #define ok return 0; #define oi(x) cout<<x<<endl; #define os(str) cout<<string(str)<<endl; #define gcd __gcd #define mem(s,t) memset(s,t,sizeof(s)) #define debug(a,n) for(int i=0;i<n;i++) cout<<a[i]<<" "; cout<<endl; #define debug1(a,n) for(int i=1;i<=n;i++) cout<<a[i]<<" "; cout<<endl; #define Debug(a,n,m) for(int i=0;i<n;i++) {for(int j=0;j<m;j++) cout<<a[i][j]<<" "; cout<<endl; } #define input(a,k) for (int i = 1; i <= (int)(k); i++) {cin>>a[i];} #define INPUT(a,n,m) for(int i=0;i<n;i++) {for(int j=0;j<m;j++) cin>>a[i][j] ; } #define TLE std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cout.precision(10); using namespace std; inline void NO() { cout<<"NO"<<endl; } inline void YES(){ cout<<"YES"<<endl;} const int mxn = 2e5+10; #define rep(k) for (int i=0;i<n;i++) #define rep1(j,k) for (int i=j;i<=k; i++) #define per(j,k) for (int i=j;i>=k; i--) #define per(k) for (int i=k-1;i>=0;i--) int a[mxn]; int main() { int n; TLE; while(cin>>n) { if(n&1) { rep1(1,n) { if(i%2) { a[i]=2*i-1; a[i+n] = 2*i; } else { a[i]=2*i; a[i+n] = 2*i-1; } } YES(); debug1(a,2*n); cout<<endl; } else NO(); }