I - Increasing and Decreasing

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

Notice：Don't output extra spaces at the end of one line.

Given n,x,yn,x,y, please construct a permutation of length nn, satisfying that:

- The length of LIS(Longest Increasing Subsequence) is equal to xx.
- The length of LDS(Longest Decreasing Subsequence) is equal to yy.

If there are multiple possible permutations satisfying all the conditions, print the lexicographically minimum one.

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

Each test case contains one line, which contains three integers n,x,y(1≤n≤105,1≤x,y≤n)n,x,y(1≤n≤105,1≤x,y≤n).
OutputFor each test case, the first line contains ``YES'' or ``NO'', indicating if the answer exists. If the answer exists, output another line which contains nn integers, indicating the permutation.Sample Input

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

题意：

　　按照字典顺序求长度为n，最长上升子序列为x，最长下降子序列为y的最小排列。

思路：

　　分段

　　将排列分成 x 段，

　　保证前一段为单调上升序列，

　　数最多的段个数为 y （即最长递减数列

#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;
}