Minimize the error

In the first sample case, we cannot perform any operations on A or B. Therefore the minimum possible error E?=?(1?-?2)²?+?(2?-?3)²?=?2.

In the second sample case, we are required to perform exactly one operation on A. In order to minimize error, we increment the first element of A by 1. Now, A?=?[2,?2]. The error is now E?=?(2?-?2)²?+?(2?-?2)²?=?0. This is the minimum possible error obtainable.

In the third sample case, we can increase the first element of A to 8, using the all of the 5 moves available to us. Also, the first element of B can be reduced to 8 using the 6 of the 7 available moves. Now A?=?[8,?4] and B?=?[8,?4]. The error is now E?=?(8?-?8)²?+?(4?-?4)²?=?0, but we are still left with 1 move for array B. Increasing the second element of B to 5 using the left move, we get B?=?[8,?5] and E?=?(8?-?8)²?+?(4?-?5)²?=?1.

题解：摘自codeforce

The problem can be interpreted as follows: array B is fixed and a total of k₁?+?k₂?=?K operations allowed on A. Let the array C be defined as Ci?=?|Ai?-?Bi| Now this is just a simple greedy problem where value of is to be minimized after exactly K subtraction/addition operations spread over the elements of array C.

1. Till E is non-zero, the largest element is chosen and 1 is subtracted from it. This is done as currently we want to maximize the reduction in error per operation and decreasing an element x by 1 reduces error by x²?-?(x?-?1)²?=?2·x?-?1.
2. Once all the elements become zero, we can use the remaining moves to alternately increase and decrease the same element till we run out of moves.

This can be implemented using a priority queue or by sorting the array C and iterating over it.

Expected complexity: O(N·K·log(N)) or O(N·log(N))

感受：好题。和wanfly1交流赛的B有相同的特点，不过需要思考得多一点。将k1和k2看作是同一种操作，然后固定一个序列，转化为对序列c得操作。贪心体现在使用一次操作，最多能减少2*x-1，与x有关，所以要不断找最大值来进行操作。好题！！！！，同时学习了一波代码。

 1 #pragma warning(disable:4996)
 2 #include<bitset>
 3 #include<queue>
 4 #include<vector>
 5 #include<string>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<iostream>
 9 #include<algorithm>
10 using namespace std;
11 typedef long long ll;
12 
13 int n, k1, k2;
14 priority_queue<ll> p;
15 
16 int main()
17 {
18     while (cin >> n >> k1 >> k2) {
19         vector<ll> a(n), b(n), arr(n);
20         int k = k1 + k2;
21         for (int i = 0; i < n; i++) cin >> a[i];
22         for (int i = 0; i < n; i++) {
23             cin >> b[i];
24             arr[i] = abs(a[i] - b[i]);
25             p.push(arr[i]);
26         }
27         while (k > 0) {
28             ll tp = p.top();
29             p.pop();
30             p.push(abs(tp - 1));  //一定要取绝对值，当最大值是0的时候。
31             k--;
32         }
33         ll ans = 0;
34         while (!p.empty()) {
35             ll tp = p.top();
36             p.pop();
37             ans += tp * tp;
38         }
39         cout << ans << endl;
40     }
41 
42 }