Permutation p is an ordered set of integers p 1, p 2, …, p n, consisting of n distinct positive integers not larger than n. We’ll denote as n the length of permutation p 1, p 2, …, p n.
Your task is to find such permutation p of length n, that the group of numbers |p 1 - p 2|, |p 2 - p 3|, …, |p n - 1 - p n| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
input
3 2
output
1 3 2
input
3 1
output
1 2 3
input
5 2
output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Solution:
#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>
using namespace std;
int a[1234567];
int main() {
int n, k;
scanf("%d %d", &n, &k);
a[0] = 1;
int low = 1, high = 1;
for (int i = 1; i <= k; i++) {
if (i % 2 == 1) {
a[i] = ++high;
} else {
a[i] = --low;
}
}
for (int i = k + 1; i < n; i++) {
if (a[k] == low) {
a[i] = a[i - 1] - 1;
} else {
a[i] = a[i - 1] + 1;
}
}
int smallest = a[0];
for (int i = 1; i < n; i++) {
if (a[i] < smallest) {
smallest = a[i];
}
}
for (int i = 0; i < n; i++) {
if (i > 0) printf(" ");
printf("%d", a[i] - smallest + 1);
}
printf("\n");
return 0;
}
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