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 （ or ）个，即往前模拟

code:

#include 
     
      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?'\n':' ');    }    else    {        for(int i = 1;i <= n;i++)            printf("%d%c",i == 1?k/2:k,i == n?'\n':' ');    }    return 0;}