題干：

After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has?n?balls and?m?baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from?1?to?m, correspondingly. The balls are numbered with numbers from?1?to?n.

Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which??is minimum, where?i?is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.

For every ball print the number of the basket where it will go according to Valeric's scheme.

Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.

Input

The first line contains two space-separated integers?n,?m?(1?≤?n,?m?≤?105)?— the number of balls and baskets, correspondingly.

Output

Print?n?numbers, one per line. The?i-th line must contain the number of the basket for the?i-th ball.

Examples

Input

4 3

Output

2 1 3 2

Input

3 1

Output

1 1 1

題目大意：

? ? ?有n個球，m個籃子， ?要把這n個球放進這些籃子，首先放籃子中求最少的籃子，若數(shù)量相同再放距離中間籃子最近的，若距離相同放籃子編號小的。

解題報告：

? ?這個題解法很多啊，比較常見的一種就是打表，把i對應位置取模的值打表一下，然后直接輸出biao[i]。第二種方法是優(yōu)先隊列。第三種set，或者線段樹貌似都可以。

AC代碼：

#include<bits/stdc++.h> using namespace std;int main() {int n,m;cin>>n>>m;for(int i = 0; i<n; i++) printf("%d\n",(m+i%m)%2 ? ((m+i%m+1)/2) : (m-(i%m))/2);return 0 ; }

set做法還未看：

? ?其實就是按照規(guī)定的順序建立set或者優(yōu)先隊列，然后每次取出隊首，做出處理再放入容器。

#include <iostream> #include <stdio.h> #include <math.h> #include <algorithm> #include <queue> #include <stack> #include <vector> #include <string> #include <string.h> #include <map> #include <set> using namespace std; #define maxn 100005*3 #define inff 1<<30 struct node {int val,dis,id;node(int val,int dis,int id):val(val),dis(dis),id(id){}friend bool operator <(node x,node y){if(x.val<y.val)return true;else if(x.val == y.val){if(x.dis<y.dis)return true;else if(x.dis==y.dis){if(x.id<y.id)return true;else return false;}else return false;}else return false;} }; int abs(int x) {if(x<0)return -x;else return x; } int n,m; int a[maxn]; int main() {set<node>s;int i;while(scanf("%d%d",&n,&m)!=EOF){s.clear();int mid1,mid2;if(m%2!=0){mid1=mid2=(m+1)/2;}else{mid1=m/2;mid2=mid1+1;}for(i=1;i<=m;i++){node tx(0,min(abs(mid1-i),abs(mid2-i)),i);s.insert(tx);}for(i=1;i<=n;i++){node tx=*s.begin();s.erase(s.begin());a[i]=tx.id;tx.val++;s.insert(tx);}for(i=1;i<=n;i++){printf("%d\n",a[i]);}}return 0; }

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