Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as nthe length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Print n integers forming the permutation. If there are multiple answers, print any of them.
3 2
1 3 2
3 1
1 2 3
5 2
1 3 2 4 5
By |x| we denote the absolute value of number x.
题意:
从1-n的数中构造出一个序列,使得相邻数的差的绝对值的不同值的个数恰好为k,构造出这样的序列。
思路:
可以发现这样的一件事,当我们有n个数,k=n-1的时候,例如n=5,k=4
我们可以把1放第一个,第二个为1+4=5,第三个是5-3=2,第四个是2+2=4,第五个是4-1=3。这样刚好用了n个数不重复。
所以我们可以想到,前面的我们用递增的序列,保证前面差值均为1,后面再从差值为n-1开始往后按照上面的发现来构造,就一定能构造出这样的序列来。
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
int a[maxn];
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k))
{
int remain=k;
int i;
int f=1;
for(i=1;i<=n-k;i++)
{
a[i]=i;
}
for(;i<=n;i++)
{
if(f)
{
f=0;
a[i]=a[i-1]+remain;
remain--;
}
else
{
f=1;
a[i]=a[i-1]-remain;
remain--;
}
}
for(int i=1;i<=n;i++)
{
if(i==n)
printf("%d\n",a[i]);
else
printf("%d ",a[i]);
}
}
return 0;
}