zoukankan      html  css  js  c++  java
  • C. Almost Equal (模拟打表) ( Codeforces Round #580 (Div. 2) )

    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 (1n1051≤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,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();
        }
    所遇皆星河
  • 相关阅读:
    np背包问题【算法:折半枚举】
    数字游戏【后缀积问题,一个数学分析问题】
    bfs求最短路径
    利用费马小定理求逆元
    [蓝桥杯2016初赛]剪邮票【全排列,连通块】
    np背包问题【算法:折半枚举】
    火星救[数学,一点前缀和]
    bfs求最短路径
    检测一个正整数是否是2的N次方
    IIS与asp.net身份认证
  • 原文地址:https://www.cnblogs.com/Shallow-dream/p/11619227.html
Copyright © 2011-2022 走看看