Codeforces Round #Pi (Div. 2) A B

【比赛链接】 click here~~

今早发现昨天的比赛居然掉分了，顿时无语了，发一下AB题解吧，感觉前两道还是很好做的，不过这时间卡在半夜，水了两道，实在是困的不行了，囧~~

A. Lineland Mail

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

All cities of Lineland are located on the Ox coordinate axis. Thus, each city is associated with its position xi —
a coordinate on theOx axis. No two cities are located at a single point.

Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).

Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.

For each city calculate two values ​​mini and maxi,
where mini is the minimum
cost of sending a letter from the i-th city to some other city, and maxi is
the the maximum cost of sending a letter from the i-th city to some other city

Input

The first line of the input contains integer n (2 ≤ n ≤ 105)
— the number of cities in Lineland. The second line contains the sequence ofn distinct integers x1, x2, ..., xn ( - 109 ≤ xi ≤ 109),
where xi is the x-coordinate
of the i-th city. All the xi's
are distinct and follow inascending order.

Output

Print n lines, the i-th
line must contain two integers mini, maxi,
separated by a space, where mini is
the minimum cost of sending a letter from the i-th city, and maxi is
the maximum cost of sending a letter from the i-th city.

Sample test(s)

input

4
-5 -2 2 7

output

3 12
3 9
4 7
5 12

input

2
-1 1

output

2 2
2 2

【题意】给出一个n个元素的已经排序的数列，求每个元素在剩下的元素之间任取一个元素求和的最小和最大值，因为是已经排好序，所以直接比较

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=4e5+10;
LL num
;
int main(){
    int n,m;
    cin>>n;
    for(int i=0; i<n; ++i)cin>>num[i];
    int minn,maxx;
    for(int i=0; i<n; ++i){
        if(i&&i<n-1) minn=min(num[i]-num[i-1],num[i+1]-num[i]);
        else if(i==0) minn=num[1]-num[0];
        else minn=num[n-1]-num[n-2];
        maxx=max(num[i]-num[0],num[n-1]-num[i]);
        cout<<minn<<" "<<maxx<<endl;
    } return 0;
}

B. Berland National Library

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Berland National Library has recently been built in the capital of Berland. In addition, in the library you can take any of the collected works of Berland leaders, the library has a reading
room.

Today was the pilot launch of an automated reading room visitors' accounting system! The scanner of the system is installed at the entrance to the reading room. It records the events of the form "reader entered room", "reader left
room". Every reader is assigned aregistration number during the registration procedure at the library — it's a unique integer from 1 to 106.
Thus, the system logs events of two forms:

"+ ri"
— the reader with registration number ri entered
the room;
"- ri"
— the reader with registration number ri left
the room.

The first launch of the system was a success, it functioned for some period of time, and, at the time of its launch and at the time of its shutdown, the reading room may already have visitors.

Significant funds of the budget of Berland have been spent on the design and installation of the system. Therefore, some of the citizens of the capital now demand to explain the need for this system and the benefits that its implementation
will bring. Now, the developers of the system need to urgently come up with reasons for its existence.

Help the system developers to find the minimum possible capacity of the reading room (in visitors) using the log of the system available to you.

Input

The first line contains a positive integer n (1 ≤ n ≤ 100)
— the number of records in the system log. Next follow n events from the system journal in the order in which the were made. Each event was written
on a single line and looks as "+ ri"
or "- ri",
where ri is an integer from 1 to 106,
the registration number of the visitor (that is, distinct visitors always have distinct registration numbers).

It is guaranteed that the log is not contradictory, that is, for every visitor the types of any of his two consecutive events are distinct. Before starting the system, and after stopping the room may possibly contain visitors.

Output

Print a single integer — the minimum possible capacity of the reading room.

Sample test(s)

input

6
+ 12001
- 12001
- 1
- 1200
+ 1
+ 7

output

3

input

2
- 1
- 2

output

2

input

2
+ 1
- 1

output

1

Note

In the first sample test, the system log will ensure that at some point in the reading room were visitors with registration numbers 1,1200 and 12001.
More people were not in the room at the same time based on the log. Therefore, the answer to the test is 3.

【题目大意】：给出一个reading room的人数的流量，求最大的容量，set的运用

代码：

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL num[400010];
char op[10];
int main()
{
   int t,n,m;
   cin>>t;
   set <int >val;
   int res=0;
   for(int i=0; i<t; ++i)
   {
       cin>>op>>m;
       switch(op[0])
       {
       case'+':
        val.insert(m);
        res=max(res,int(val.size()));
        break;
       case'-':
        if(val.find(m)==val.end()) res++;
        else {
            val.erase(m);
            res=max(res,int(val.size()));
        }
        break;
       }
   }
   cout<<res<<endl;
}