Codeforces 938.B Run For Your Prize

最新推荐文章于 2018-08-01 09:11:19 发布

转载最新推荐文章于 2018-08-01 09:11:19 发布 · 144 阅读

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原文链接：http://www.cnblogs.com/zbtrs/p/8451879.html

本文介绍了一道名为“RunForYourPrize”的算法题目，解析了最优解的思路并给出了实现代码。该题要求计算两人合作收集所有奖品所需的最短时间。

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You and your friend are participating in a TV show "Run For Your Prize".

At the start of the show n prizes are located on a straight line. i-th prize is located at position ai. Positions of all prizes are distinct. You start at position 1, your friend — at position 106 (and there is no prize in any of these two positions). You have to work as a team and collect all prizes in minimum possible time, in any order.

You know that it takes exactly 1 second to move from position x to position x + 1 or x - 1, both for you and your friend. You also have trained enough to instantly pick up any prize, if its position is equal to your current position (and the same is true for your friend). Carrying prizes does not affect your speed (or your friend's speed) at all.

Now you may discuss your strategy with your friend and decide who will pick up each prize. Remember that every prize must be picked up, either by you or by your friend.

What is the minimum number of seconds it will take to pick up all the prizes?

Input

The first line contains one integer n (1 ≤ n ≤ 105) — the number of prizes.

The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 106 - 1) — the positions of the prizes. No two prizes are located at the same position. Positions are given in ascending order.

Output

Print one integer — the minimum number of seconds it will take to collect all prizes.

Examples

input

Copy

3
2 3 9

output

input

Copy

2
2 999995

output

Note

In the first example you take all the prizes: take the first at 1, the second at 2 and the third at 8.

In the second example you take the first prize in 1 second and your friend takes the other in 5 seconds, you do this simultaneously, so the total time is 5.

题目大意：你在1位置，朋友在10^6位置，每次花1单位时间挪动1格，有n个礼物，问最少花多少时间拿到所有礼物.

分析：这题目水的......求一下礼物距你和朋友的距离，取个min,然后排序，取最大的就好了.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include <cmath>

using namespace std;

typedef long long LL;

int n,a[100010];

int main()
{
    scanf("%d",&n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d",&a[i]);
        a[i] = min(a[i] - 1,1000000 - a[i]);
    }
    sort(a + 1,a + 1 + n);
    printf("%d\n",a[n]);

    return 0;
}