Codeforces Round #753 (Div. 3) C. Minimum Extraction（最小抽离）

题目翻译：

Yelisey has an array a of n integers.

数组a中有n个整数

If a has length strictly greater than 1, then Yelisei can apply an operation called minimum extraction to it:

a如果的长度严格大于1，那么可以对其实行最小抽离操作

First, Yelisei finds the minimal（最小的） number m in the array. If there are several identical minima, Yelisey can choose any of them.
Then the selected minimal element is removed from the array. After that, m is subtracted from each remaining element

该操作规则：

1.首先，找到a中最小的数m，如果有多个最小值，只要任意选其中一个

2.然后，从a中去除该最小元素，然后使剩余的每一个元素减去m

Thus, after each operation, the length of the array is reduced by 1.

因此，在操作若干次后，a的长度会减到1

For example, if a=[1,6,−4,−2,−4]a=[1,6,−4,−2,−4], then the minimum element in it is a3=−4a3=−4, which means that after this operation the array will be equal to a=[1−(−4),6−(−4),−2−(−4),−4−(−4)]=[5,10,2,0]a=[1−(−4),6−(−4),−2−(−4),−4−(−4)]=[5,10,2,0].

Since Yelisey likes big numbers, he wants the numbers in the array a to be as big as possible.

要求：a中的数尽可能的大

Formally speaking, he wants to make the minimum of the numbers in array a to be maximal possible (i.e. he want to maximize a minimum). To do this, Yelisey can apply the minimum extraction operation to the array as many times as he wants (possibly, zero). Note that the operation cannot be applied to an array of length 1.

即：a中的最小值尽可能的大

为了使a的最小值尽可能的大，可以对a进行最小抽离操作（0或多次）

Help him find what maximal（最大值） value can the minimal element of the array have after applying several (possibly, zero) minimum extraction operations to the array.

问题：找出对a进行最小提取操作若干次（或者0次）后a中最小值的最大值

Input

The first line contains an integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The next 2t lines contain descriptions of the test cases.

In the description of each test case, the first line contains an integer n(1≤n≤2⋅1051≤n≤2⋅105) — the original length of the array a. The second line of the description lists n space-separated integers ai (−109≤ai≤109−109≤ai≤109) — elements of the array a.

It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.

输入：

第一行case个数

下面每两行一组，每组第一行记录数组整数个数，第二行是具体数组

Output

Print t lines, each of them containing the answer to the corresponding test case. The answer to the test case is a single integer — the maximal possible minimum in aa, which can be obtained by several applications of the described operation to it.

输出：每一个case的经过最小抽离操作后最小值的最大值

问题概述：

数组a有n个数，操作：去掉最小值后让其余每个元素减去最小值

求经过若干次操作后最小值的最大值

思路：

多写几个找规律，发现排序后第n个相邻两个相减就是对应操作n次后的最小值

代码:

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;long long  minimum(vector<long long>& nums)
{long long  n = nums.size();long long  maxmin = INT64_MIN;sort(nums.begin() + 1, nums.end());if (nums.size() == 2){return nums[1];}maxmin = nums[1];for (long long i = 1; i < n - 1; i++){maxmin = max(maxmin, nums[i + 1] - nums[i]);}return maxmin;}
int main()
{long long m;cin >> m;long long len;long long elem;vector<vector<long long>>vec;vector<long long>temp;while (m--){cin >> len;temp.push_back(len);for (long long i = 0; i < len; i++){cin >> elem;temp.push_back(elem);}vec.push_back(temp);temp.clear();}for (long long i = 0; i < vec.size(); i++){cout << minimum(vec[i]) << endl;}
}