Codeforces Round #344 (Div. 2) C. Report (脑洞题)

网友投稿 273 2022-11-29

Codeforces Round #344 (Div. 2) C. Report (脑洞题)

C. Report

time limit per test

memory limit per test

input

output

Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are ncommodities produced by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers. Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers in non-descending or non-ascending order and then passes the report to the manager i + 1, or directly to Blake (if this manager has number i = m).

Employees of the "Blake Technologies" are preparing the report right now. You know the initial sequence ai of length n and the description of each manager, that is value ri

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of commodities in the report and the number of managers, respectively.

The second line contains n integers ai (|ai| ≤ 109) — the initial report before it gets to the first manager.

Then follow m lines with the descriptions of the operations managers are going to perform. The i-th of these lines contains two integersti and ri (

, 1 ≤ ri ≤ n), meaning that the i-th manager sorts the first ri numbers either in the non-descending (if ti) or non-ascending (if ti) order.

Output

Print n integers — the final report, which will be passed to Blake by manager number m.

Examples

input

3 11 2 32 2

output

2 1 3

input

4 21 2 4 32 31 2

output

2 4 1 3

Note

In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2 1. The report got to Blake in this form.

In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4 2 1. After the second manager the report changed to: 2 4. This report was handed over to Blake.

题意:给你n个数,然后m个操作,如果op=1,要求让前x个数按照非递减排序。如果op=2,y要求让前x个数按照非递增排序。

题解:去除一些后续的无效操作。把有效操作记录下来。不必要把每个数都排序一遍。因为有些数的位置是不变的。

最后倒序遍历一遍就可以了。(这题想不出来可以套线段树的。。。)

代码:

#pragma comment(linker, "/STACK:102400000,102400000")//#include#include#include#include#include#include#include#include#include#include#include#include#include using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;typedef vector vi;#define mst(a) memset(a, 0, sizeof(a))#define M_P(x,y) make_pair(x,y)#define in freopen("in.txt","r",stdin) #define rep(i,j,k) for (int i = j; i <= k; i++) #define per(i,j,k) for (int i = j; i >= k; i--) #define lson l , mid , rt << 1 #define rson mid + 1 , r , rt << 1 | 1 const int lowbit(int x) { return x&-x; } const double eps = 1e-8; const int INF = 1e9+7; const ll inf =(1LL<<62) ;const int MOD = 1e9+7; const ll mod = (1LL<<32);const int N =2e5+7; const int M=100010;const ll MAX=1e18;//const int maxn=1001; template inline void getmax(T1 &a, T2 b) {if (b>a)a = b;} template inline void getmin(T1 &a, T2 b) {if (b>n>>m; for(int i=0;i>a[i],b[i]=a[i]; for(int i=0;i>op>>x; while(cnt > 0 && x > r[cnt-1]) --cnt; t[cnt] = op, r[cnt] = x ; cnt++; } //cout<<"cnt="<=r[i];--j) { if(t[i-1] == 1) a[j]= b[b_right--]; //递增 else a[j]=b[b_left++]; //递减 } } for(int i=0;i

版权声明:本文内容由网络用户投稿,版权归原作者所有,本站不拥有其著作权,亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容,请联系我们jiasou666@gmail.com 处理,核实后本网站将在24小时内删除侵权内容。

上一篇:Codeforces Round #333 (Div. 2) C. The Two Routes (最短路)
下一篇:SpringBoot中使用MongoDB的连接池配置
相关文章

 发表评论

暂时没有评论,来抢沙发吧~