codeforces #314 A. Lineland Mail
给一个有序序列,问对于没一个数,和它相差最少和最多的数的位置。
/*************************************************************************
> File Name: code/cf/#314/A.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年08月06日 星期四 00时01分51秒
************************************************************************/
1#include<iostream>
2#include<iomanip>
3#include<cstdio>
4#include<algorithm>
5#include<cmath>
6#include<cstring>
7#include<string>
8#include<map>
9#include<set>
10#include<queue>
11#include<vector>
12#include<stack>
13#define y0 abc111qqz
14#define y1 hust111qqz
15#define yn hez111qqz
16#define j1 cute111qqz
17#define tm crazy111qqz
18#define lr dying111qqz
19using namespace std;
20#define REP(i, n) for (int i=0;i<int(n);++i)
21typedef long long LL;
22typedef unsigned long long ULL;
23const int inf = 0x7fffffff;
24const int N=2E5+7;
25LL a[N];
26int main()
27{
28 int n;
29 cin>>n;
30 a[0]=-inf;
31 a[n+1]=inf;
1 for ( int i = 1 ; i <= n ; i ++ )
2 {
3 cin>>a[i];
4 }
5 int mx = -1;
6 int mi = inf;
7 cout<<a[2]-a[1]<<" "<<a[n]-a[1]<<endl;
8 for ( int i = 2 ; i <= n-1 ; i++ )
9 {
10 cout<<min(abs(a[i]-a[i-1]),abs(a[i+1]-a[i]))<<" "<<max(abs(a[i]-a[1]),abs(a[n]-a[i]))<<endl;
11 }
12 cout<<a[n]-a[n-1]<<" "<<a[n]-a[1]<<endl;
13 return 0;
14}