$AT2163\ Median\ Pyramid\ Easy$ 构造

正解:构造

解题报告:

传送门$QwQ$

考虑如果有两个相邻格子是相同数字那么它们以上这两列就都会是这列数字(显然$QwQ$?

所以考虑只要构造出第$n-1$行的中心和中心右侧($or$左侧一样的$QwQ$都等于$x$,其他格子随便填就成鸭$QwQ$

然后就做完辣?记得分类讨论几个特殊情况嗷$QwQ$

$over$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define mp make_pair
#define P pair<int,int>
#define ri register int
#define rb register bool
#define rc register char
#define t(i) edge[i].to
#define w(i) edge[i].wei
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];i;i=edge[i].nxt)

const int N=2e5+;
int n,x,a[N],tmp;
bool vis[N];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch<'' || ch>''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}

int main()
{
    n=read();x=read();if(x== || x==n*-)return printf("No\n"),;printf("Yes\n");
    if(x==){a[n]=;a[n+]=;tmp=;n=n*-;rp(i,,n)if(!a[i])a[i]=++tmp;rp(i,,n)printf("%d\n",a[i]);return ;}
    a[n]=x;a[n+]=x+;a[n-]=x-;a[n+]=x-;vis[x]=vis[x+]=vis[x-]=vis[x-]=;
    tmp=;n=n*-;rp(i,,n)if(!a[i]){while(vis[tmp])++tmp;a[i]=tmp;vis[tmp]=;}rp(i,,n)printf("%d\n",a[i]);
    return ;
}