Finite Encyclopedia of Integer Sequences(找规律)

6617: Finite Encyclopedia of Integer Sequences

时间限制: 1 Sec  内存限制: 128 MB
提交: 375  解决: 91
[提交] [状态] [讨论版] [命题人:admin]

题目描述

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.

输入

Input is given from Standard Input in the following format:
K N

输出

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.

样例输入

3 2

样例输出

2 1 

提示

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).

来源/分类

ABC077&ARC084 

思路：

1、k为偶数，序列为：k/2、k、k、k.......，共n个数。

2、k为奇数，序列为：(k+1)/2、(k+1)/2、(k+1)/2、(k+1)/2.......再往前推2/n个。

#include <bits/stdc++.h>
using namespace std;
int n,k;
int a[300100];
int main()
{
    scanf("%d%d",&k,&n);
    if(k&1)
    {
        for(int i=1; i<=n; i++)
            a[i]=(k+1)/2;
        int last=n;//以下全是往前推n/2个序列的过程，last表示当前序列的最后一位！
        for(int i=1; i<=n/2; i++)
            if(a[last]==1) last--;
            else
            {
                a[last]--;
                for(int j=last+1; j<=n; j++)
                    a[j]=k;
                last=n;
            }
        for(int i=1; i<=last; i++)
            printf("%d ",a[i]);
    }
    else
    {
        printf("%d ",k/2);
        for(int i=2; i<=n; i++)
            printf("%d ",k);
    }
    return 0;
}