https://vjudge.net/contest/388843#problem/I

InputThe first line contains an integerT(1≤T≤100)T(1≤T≤100), indicating the number of test cases.

Each test case contains one line, which contains three integersn,x,y(1≤n≤105,1≤x,y≤n)

4
10 1 10
10 10 1
10 5 5
10 8 8

Sample Output

YES
10 9 8 7 6 5 4 3 2 1
YES
1 2 3 4 5 6 7 8 9 10
YES
1 2 3 5 4 10 9 8 7 6
NO

Sponsor

題意：

　　按照字典順序求長度為n，最長上升子序列為x，最長下降子序列為y的最小排列。

思路：

　　分段

　　將排列分成 x 段，

　　保證前一段為單調上升序列，

　　數最多的段個數為 y （即最長遞減數列

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
#define debug(x) cout<<"debug:"<<#x<<" = "<<x<<endl;

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=1e6+10;
const int mod=1e9+7;

vector<int> v;
int main()
{
	int T = 0;
	int n = 0, x = 0, y = 0;
    cin >> T;
    while(T--)
	{
        cin >> n >> x >> y;
        int s = sqrt(n); 
        int base = n / s;
        if(base * s != n)
		{
			base += 1;
		}

		int tell = base + s;
		if(tell <= x+y && x+y <= n+1) 
        {
        	int p = 0;
            cout << "YES" <<endl;
            v.clear();
            int D = 0;
            for(int i = 0;i<x;i++)
			{
                D = min(n-x+1+i , y);
                p = n-D+1;
                while(p <= n)
                {
                	 v.push_back(p);
                	 p += 1;
				}
                n -= D;
            }
            p = v.size()-1;
            while(p >= 0)
			{
				cout << v[p];
				if(p)cout<<" ";
				p -= 1;
			}
            cout <<endl;
        }
        else
        {
        	cout << "NO" << endl;
		}
        /*
        if(tell <= n+1 && tell <= x+y)
		{
            cout << "YES" << endl;
            v.clear();
            int temp = 0;
            for(int i = 0;i<x;i++)
			{
				temp = min(n+x-i+1 , y);
                for(int j = n-temp+1; j <= n; j ++)
				{
                	v.push_back(j);
				}
                n -= temp;
            }
            int p = v.size() - 1;
            for(int i = 0;i<=p;i++)
            {
            	cout << v[p-i] <<" ";
			}
        }
        else
        {
        	cout << "NO" <<endl;
		}
		*/
    }

    return 0;
}