E. Klever Permutation

合集 - codeforces(31)

1.B. Rudolf and 1212024-03-162.C. Rudolf and the Ugly String2024-03-163.A. Rudolf and the Ticket2024-03-164.D. Rudolf and the Ball Game2024-03-185.G. Rudolf and Subway2024-04-066.C. Choose the Different Ones!2024-02-097.B. Following the String2024-02-09

8.E. Klever Permutation2024-02-09

9.D. Find the Different Ones!2024-02-0910.B. YetnotherrokenKeoard2023-12-1011.C. Removal of Unattractive Pairs2023-12-1012.D. Jumping Through Segments2023-12-1113.E. Good Triples2023-12-1314.F. Shift and Reverse2023-12-1315.C. Nikita and LCM2024-05-2916.B. Prefiquence2024-05-0317.E. Vlad and a Pair of Numbers2024-04-0718.D. XOR Construction2024-03-2719.C. Non-coprime Split2024-01-2720.D. Divide and Equalize2024-06-0221.E. Block Sequence2024-06-0222.D. In Love2024-06-1123.C. Theofanis' Nightmare2024-06-2224.D. Yet Another Monster Fight2024-06-2225.C. Count Triangles2024-07-1126.E. Count Paths2024-07-1327.Collecting Numbers II2024-07-3128.C. Even Subarrays2024-08-0729.C. Perform Operations to Maximize Score2024-08-2030.F. Color Rows and Columns2024-08-1631.C. To Become Max2024-09-10

收起

题解

假设 a1 a2 a3 ... ak  ak+1 ak+2 ... an是符合要求的数组，

那么我们可以推断出:

a(k+1)=a(1)+1;

a(k+2)=a(2)-1;

...

a(2k+1)=a(k+1)+1;

...

因此我们知晓奇数位的数要比较小，偶数的位置要比较大；又题目说明一定有解，所以我们假定a1=1,a2=n再递推出其余各项。

Code

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int a[N];
int main(){
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while (t--){
        int n,k,cnt=-1;
        cin>>n>>k;
        int sum1=n,sum2=1;
        for (int i=1;i<=k;i++){
            for (int j=i;j<=n;j+=k)
                if (i%2==1)    a[j]=sum1--;
                else a[j]=sum2++;
        }
        for (int i=1;i<=n;i++) 
            if (i==1) cout<<a[i];
            else cout<<" "<<a[i];
        cout<<endl;
    }
    return 0;
}