题目传送门：https://codeforces.com/problemset/problem/767/A
题目大意：
给一个长度为(n)的排列(A)，每次遇到剩余数字的最大项后，将缓存区内从最大项开始的连续递减的数逆序输出；无法输出的数加入或者继续留存在缓存区；无任何输出则输出空行
读入的时候维护剩余数字中的最大值即可
/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
int f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
template<typename T>inline T read(T x){
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=1e5;
bool vis[N+10];
int main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n=read(0),Max=n;
for (int i=1;i<=n;i++){
vis[read(0)]=1;
while (vis[Max]) printf("%d ",Max--);
putchar('
');
}
return 0;
}