codeforces 1770B. Koxia and Permutation-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_60775360/article/details/128553036

该文描述了一个算法问题，要求构造一个长度为n的排列p，使得由数组c定义的成本最小，其中c中的每个元素ci等于p中连续k个元素的最大值和最小值之和。给定n和k，任务是生成具有最低成本的排列。当k等于1或n时，有特定的解决方案。提供的C++代码示例给出了在某些情况下的一种可能的构造方法。

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B. Koxia and Permutation
Reve has two integers n and k.

Let p be a permutation† of length n. Let c be an array of length n−k+1 such that
ci=max(pi,…,pi+k−1)+min(pi,…,pi+k−1).
Let the cost of the permutation p be the maximum element of c.

Koxia wants you to construct a permutation with the minimum possible cost.

† A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array), and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Input
Each test consists of multiple test cases. The first line contains a single integer t (1≤t≤2000) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers n and k (1≤k≤n≤2⋅105).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.

Output
For each test case, output n integers p1,p2,…,pn, which is a permutation with minimal cost. If there is more than one permutation with minimal cost, you may output any of them.

Example
inputCopy
3
5 3
5 1
6 6
outputCopy
5 1 2 3 4
1 2 3 4 5
3 2 4 1 6 5
Note
In the first test case,

c1=max(p1,p2,p3)+min(p1,p2,p3)=5+1=6.
c2=max(p2,p3,p4)+min(p2,p3,p4)=3+1=4.
c3=max(p3,p4,p5)+min(p3,p4,p5)=4+2=6.
Therefore, the cost is max(6,4,6)=6. It can be proven that this is the minimal cost.

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int a[N]={0};
void solve(){
    int n,k;
    cin>>n>>k;
    if(k==1||k==n){
        for(int i=1;i<=n;i++)cout<<i<<' ';
        cout<<endl;
        return;
    }
    int p=n+1,l=1;
    for(int i=1;i<=n;i+=2){
        a[i]=p-l;
        a[i+1]=l;
        l++;
    }
    for(int i=1;i<=n;i++){
        cout<<a[i]<<' ';
    }
    cout<<endl;
}
int main(void){
    int t;
    cin>>t;
    while(t--){
        solve();
    }
}
//code by 01100_10111;