Codeforces Round #Pi (Div. 2) A. Lineland Mail 水题

A. Lineland Mail
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/567/problem/A

Description

All cities of Lineland are located on the Ox coordinate axis. Thus, each city is associated with its position xi — a coordinate on the Oxaxis. 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 Input

4
-5 -2 2 7

Sample Output

3 12
3 9
4 7
5 12

HINT

题意

对于每个点，输出离这个点最近的点的距离，和离这个点最远的点的距离

题解：

最近的点就是相邻的点，最远的点就是边界上的点

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************
ll a[maxn];
int main()
{
    int n=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    a[n+1]=infll;
    a[0]=-infll;
    sort(a,a+n);
    for(int i=1;i<=n;i++)
    {
        printf("%lld %lld
",min(a[i]-a[i-1],a[i+1]-a[i]),max(a[i]-a[1],a[n]-a[i]));
    }
}