Finite Encyclopedia of Integer Sequences

Posted 2020-12-27 lemon-jade

Time limit : 2sec / Memory limit : 256MB

Score : 800 points

Problem Statement

In Finite Encyclopedia of Integer Sequences (FEIS), all integer sequences of lengths between 1 and N (inclusive) consisting of integers between 1 and K (inclusive) are listed.

Let the total number of sequences listed in FEIS be X. Among those sequences, find the (X⁄2)-th (rounded up to the nearest integer) lexicographically smallest one.

Constraints

• 1≤N,K≤3×105
• N and K are integers.

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Input

Input is given from Standard Input in the following format:

K N

Output

Print the (X⁄2)-th (rounded up to the nearest integer) lexicographically smallest sequence listed in FEIS, with spaces in between, where X is the total number of sequences listed in FEIS.

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Sample Input 1

Copy

3 2

Sample Output 1

Copy

2 1 

There are 12 sequences listed in FEIS: (1),(1,1),(1,2),(1,3),(2),(2,1),(2,2),(2,3),(3),(3,1),(3,2),(3,3). The (12⁄2=6)-th lexicographically smallest one among them is (2,1).

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Sample Input 2

Copy

2 4

Sample Output 2

Copy

1 2 2 2

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Sample Input 3

Copy

5 14

Sample Output 3

Copy

3 3 3 3 3 3 3 3 3 3 3 3 2 2 

 

题解：偶数的时候很好弄，输出k/2,k...k，奇数的时候，对与k/2.k/2 ...k/2 这个序列 里中间最近，与中间相差n/2 （证明 or 官方题解）个，即往前模拟

code:

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 3e5 + 10;
const int mod = 1e9 + 7;
int a[N];
int main()
{
    int n,k;
    scanf("%d%d",&k,&n);
    if(k&1)
    {
        for(int i = 1;i <= n;i++)
            a[i] = (k+1)/2;
        int m = n;
        for(int i = 1;i <= n/2;i++)
        {
            if(a[m] == 1)
                m--;
            else
            {
                a[m]--;
                for(int j = m + 1;j <= n;j++)
                    a[j] = k;
                m = n;
            }
        }
        for(int i = 1;i <= m;i++)
            printf("%d%c",a[i],i == m?‘
‘:‘ ‘);
    }
    else
    {
        for(int i = 1;i <= n;i++)
            printf("%d%c",i == 1?k/2:k,i == n?‘
‘:‘ ‘);
    }
    return 0;
}