B. Preparing for Merge Sort(二分)

2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest, qualification stage (Online Mirror, ACM-ICPC Rules, Teams Preferred)

B. Preparing for Merge Sort

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Ivan has an array consisting of n different integers. He decided to reorder all elements in increasing order. Ivan loves merge sort so he decided to represent his array with one or several increasing sequences which he then plans to merge into one sorted array.

Ivan represent his array with increasing sequences with help of the following algorithm.

While there is at least one unused number in array Ivan repeats the following procedure:

iterate through array from the left to the right;

Ivan only looks at unused numbers on current iteration;

if current number is the first unused number on this iteration or this number is greater than previous unused number on current iteration, then Ivan marks the number as used and writes it down.

For example, if Ivan’s array looks like [1, 3, 2, 5, 4] then he will perform two iterations. On first iteration Ivan will use and write numbers [1, 3, 5], and on second one — [2, 4].

Write a program which helps Ivan and finds representation of the given array with one or several increasing sequences in accordance with algorithm described above.

Input

The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of elements in Ivan’s array.

The second line contains a sequence consisting of distinct integers a1, a2, …, an (1 ≤ ai ≤ 109) — Ivan’s array.

Output

Print representation of the given array in the form of one or more increasing sequences in accordance with the algorithm described above. Each sequence must be printed on a new line.

Examples

input

5

1 3 2 5 4

output

1 3 5

2 4

input

4

4 3 2 1

output

4

3

2

1

input

4

10 30 50 101

output

10 30 50 101

题意：给你一个长度为n的序列，用一个或几个递增序表示给定的序列。

解题思路：二分查找。

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

const int maxn = 1e5 + 5;
int a[2 * maxn],b[2 * maxn];
vector<i
4000
nt>G[2 * maxn];

int main(){
int n;
scanf("%d",&n);
for(int i = 1;i <= n;i++){
scanf("%d",&a[i]);
int pos = lower_bound(b + 1,b + 1 + n,a[i]) - b;
pos--;
b[pos] = a[i];
G[pos].push_back(a[i]);
}
for(int i = n;i >= 1;i--){
for(int j = 0;j < G[i].size();j++){
if(j == G[i].size() - 1) printf("%d\n",G[i][j]);
else printf("%d ",G[i][j]);
}
}
}
/*lower_bound(b + 1,b + 1 + n,a[i]) - b: 返回b序列中第一个大于等于a[i]的下标，若b序列中的数都小于a[i],返回 n + 1。*/

/*upper_bound(b + 1,b + 1 + n,a[i]) - b: 返回b序列中第一个大于a[i]的下标，若b序列中的数都小于a[i],返回 n + 1。*/