C. Diverse Permutation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard 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.

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?<?n?≤?105).

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.

用n個數1~n，每一個數僅僅能用一次。組成差值的絕對值有k個數。為1~k。

輸出任一個方案。

構造題，我是這樣構造的，取前k+1個數。第一個數取1，先+k。后一個數-(k-1),在后一個數+k-2.......這樣從兩頭往

中間靠攏。既取完了k+1個數。又構造了1~k的差值絕對值，至于k+1后的嘛，每次+1即可了。

代碼：

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int maxn=100000+1000; int ans[maxn]; int main() {int n,k;ans[1]=1;scanf("%d%d",&n,&k);if(k==1){for(int i=1;i<=n;i++)ans[i]=i;}else{for(int i=2;i<=k+1;i++){if(i%2)ans[i]=ans[i-1]-(k-i+2);elseans[i]=ans[i-1]+(k-i+2);}int cur=1;for(int i=k+2;i<=n;i++){ans[i]=k+1+cur;cur++;}}for(int i=1;i<n;i++)printf("%d ",ans[i]);printf("%d\n",ans[n]);return 0; }

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