B. Equal Candies

There are nnn boxes with different quantities of candies in each of them. The iii-th box has aia_iai candies inside.

You also have nnn friends that you want to give the candies to, so you decided to give each friend a box of candies. But, you don’t want any friends to get upset so you decided to eat some (possibly none) candies from each box so that all boxes have the same quantity of candies in them. Note that you may eat a different number of candies from different boxes and you cannot add candies to any of the boxes.

What’s the minimum total number of candies you have to eat to satisfy the requirements? Problem tags ：greedy, math，sorting, *800.

Input：The first line contains an integer t(1≤t≤1000)t (1≤t≤1000)t(1≤t≤1000) — the number of test cases. The first line of each test case contains an integer n(1≤n≤50)n (1≤n≤50)n(1≤n≤50) — the number of boxes you have. The second line of each test case contains nnn integers a1,a2,…,an(1≤ai≤107)a_1,a_2,…,a_n (1≤a_i≤10^7)a1,a2,…,an(1≤ai≤107) — the quantity of candies in each box.

Output：For each test case, print a single integer denoting the minimum number of candies you have to eat to satisfy the requirements.

5
5
1 2 3 4 5
6
1000 1000 5 1000 1000 1000
10
1 2 3 5 1 2 7 9 13 5
3
8 8 8
1
10000000

Note：For the first test case, you can eat 111
candy from the second box, 222 candies from the third box, 333 candies from the fourth box and 444 candies from the fifth box. Now the boxes have [1,1,1,1,1][1,1,1,1,1][1,1,1,1,1] candies in them and you ate 0+1+2+3+4=100+1+2+3+4=100+1+2+3+4=10 candies in total so the answer is 101010.

For the second test case, the best answer is obtained by making all boxes contain 555 candies in them, thus eating 995+995+0+995+995+995=4975995+995+0+995+995+995=4975995+995+0+995+995+995=4975 candies in total.

题意不再赘述，都是简单英文。主要是学习下如何拓宽思路。有两种常用解法。一求最小值有两种方法新人喜欢sort，实际上可以在每次输入时使用min保存最小值。
下面分别是解法1/2的C++实例代码，节省精力就不添加注释了：

#include<bits/stdc++.h>
using namespace std;int main() {int t; cin>>t;while(t--) { int sum = 0;int n;cin >> n;vector <int> a(n);for(int i = 0; i < n; i++) {cin >> a[i];sum += a[i];}sort(a.begin(),a.end());cout << sum - a[0] * n << endl;}
}

#include "bits/stdc++.h"
using namespace std;int main() {int t; cin >> t;while(t--) {int n; cin >> n;vector<int> a(n);int mn = INT_MAX, ans = 0;for(int i = 0; i < n; ++i) {cin >> a[i];mn = min(mn, a[i]);}for(int i = 0; i < n; ++i) {ans += a[i] - mn;}cout << ans << "\n";}
}

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