问题 F: Finite Encyclopedia of Integer Sequences
时间限制: 1 Sec 内存限制: 128 MB
提交: 304 解决: 62
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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).
1.k为偶时：因为序列总数x=
，那(X⁄2)-th (rounded up to the nearest integer) lexicographically smallest sequence一定是{k/2,k,k,k...}
2.k为奇时：可以证明(不贴了)，(X⁄2)-th (rounded up to the nearest integer) lexicographically smallest sequence是{k/2,k/2,k/2,...}的前[n/2](取下整)个序列,(即{k/2,k/2,k/2,...}再向前挪[n/2]就是答案)；
#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
int n,i,m;
int len;
int a[300010];
scanf("%d%d",&m,&n);
if(m%2==0)
{
printf("%d",m/2);
for(i=2;i<=n;i++)
{
printf(" %d",m);
}
printf("\n");
}
else
{
int t;
if(n%2==1)
{
t=(n-1)/2;
}
else
{
t=(n-1)/2+1;
}
for(i=1;i<=n;i++)
{
a[i]=m/2+1;
}
int len=n;
while(t--)
{
if(a[len]==1)
{
len--;
}
else
{
a[len]--;
for(i=len+1;i<=n;i++)
{
a[i]=m;
}
len=n;
}
}
for(i=1;i<=len;i++)
{
if(i!=1)
{
printf(" ");
}
printf("%d",a[i]);
}
printf("\n");
}
return 0;
}