C. Almost Equal （ Codeforces Round #580 (Div. 2) ）
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output
standard output
You are given integer?nn. You have to arrange numbers from?11?to?2n2n,using each of them exactly once,on the circle,so that the following condition would be satisfied: For every?nn?consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard?2n2n?numbers differ not more than by?11. For example,choose?n=3n=3. On the left you can see an example of a valid arrangement:?1+4+5=101+4+5=10,?4+5+2=114+5+2=11,?5+2+3=105+2+3=10,?2+3+6=112+3+6=11,?3+6+1=103+6+1=10,?6+1+4=116+1+4=11,any two numbers differ by at most?11. On the right you can see an invalid arrangement: for example,?5+1+6=125+1+6=12,and?3+2+4=93+2+4=9,?99?and?1212?differ more than by?11.
Input
The first and the only line contain one integer?nn?(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 output?2n2n?numbers?— numbers from?11?to?2n2n?in 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,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(); }
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