Codeforces Round #275 (Div. 2) C Diverse Permutation_html/css_WEB-ITnose

题目链接：diverse permutation

Diverse Permutation

time limit per test

memory limit per test

input

output

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.

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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.

Input

The single line of the input contains two space-separated positive integers n, k (1?≤?k?

Output

Print n integers forming the permutation. If there are multiple answers, print any of them.

Sample test(s)

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.

大致题意：给一个n和k。让找到一种1~n的排列，满足，任意相邻两数之间的差值的绝对值必须为k个不同的值。

解题思路：开始用next_permutation(a, a+n)直接生成排列，然后判断，结果TLE到死。暴力枚举搞不定，那只能构造了。详见代码

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint a[100005], b[100005];int main(){//     freopen("in.txt","r",stdin);     int n, k, l, f;    while(scanf("%d%d",&n,&k)!=EOF)    {        f = n - k;                         for(int i=1; i<=f; i++){     //构造n-k个差值为1的序列            printf("%d ", i);        }        f ++;        l = n;        while(f <= l){               //构造差值为2～k的序列            printf("%d ", l--);            if(f <= l)  printf("%d ", f++);            else   break;        }        printf("\n");    }    return 0;}