Codeforces Round #Pi (Div. 2)567A Lineland Mail(模拟)

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 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 of ndistinct 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

注意用long long 就好了～ 输出每个点离其他点的最大值和最小值～

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int maxn = 1e5 + 5;
typedef long long ll;
ll a[maxn];
int main(int argc, char const *argv[])
{
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; ++i)
		scanf("%I64d", &a[i]);
	printf("%I64d %I64d\n", a[1] - a[0], a[n - 1] - a[0]);
	for(int i = 1; i < n - 1; ++i)
		printf("%I64d %I64d\n", min(a[i] - a[i - 1], a[i + 1] - a[i]), max(a[i] - a[0], a[n - 1] - a[i]));
	printf("%I64d %I64d\n", a[n - 1] - a[n - 2], a[n - 1] - a[0]);
	return 0;
}