Codeforces Round #398 (Div. 2) A. Snacktower 模拟

A. Snacktower

题目连接：

http://codeforces.com/contest/767/problem/A

Description

According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time n snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.

Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.

However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.

Write a program that models the behavior of Ankh-Morpork residents.

Input

The first line contains single integer n (1 ≤ n ≤ 100 000) — the total number of snacks.

The second line contains n integers, the i-th of them equals the size of the snack which fell on the i-th day. Sizes are distinct integers from 1 to n.

Output

Print n lines. On the i-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the i-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.

Sample Input

3
3 1 2

Sample Output

3

2 1

Hint

题意

有个奇怪的人，他会每天吃东西，他会先吃大的。比如第一天就会吃n，第二天吃n-1，第三天吃n-2。
如果这一天不是这个东西的话，他会一直等着这个东西出现，然后一口气吃掉。

题解：

大概就是模拟一下题意吧。
我翻译的题意实际上比较乱啦，感觉只有自己看得懂……

嘛，不懂就看我代码吧。

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5+7;
int vis[maxn],a[maxn],b[maxn],flag;
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    flag = n;
    for(int i=1;i<=n;i++){
        b[a[i]]=1;
        while(b[flag]){
            printf("%d ",flag);
            flag--;
        }
        printf("\n");
    }
}