（体会不同的编程思想的效率差异）A and B and Compilation Errors （看完题先思考如何解决更好）

最新推荐文章于 2022-09-26 08:52:19 发布

Markfound

最新推荐文章于 2022-09-26 08:52:19 发布

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本文介绍了一种高效算法，通过对比三次编译错误的集合，确定被修复的具体错误编号。利用求和的方法避免了逐个元素对比带来的性能瓶颈。

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A and B are preparing themselves for programming contests.

B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.

Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.

However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.

Can you help B find out exactly what two errors he corrected?

Input

The first line of the input contains integer n (3 ≤ n ≤ 10⁵) — the initial number of compilation errors.

The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the errors the compiler displayed for the first time.

The third line contains n - 1 space-separated integers b₁, b₂, ..., b_n - 1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.

The fourth line contains n - 2 space-separated integers с₁, с₂, ..., с_n - 2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.

Output

Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.

Example

Input

5
1 5 8 123 7
123 7 5 1
5 1 7

Output

8
123

Input

6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5

Output

1
3

Note

In the first test sample B first corrects the error number 8, then the error number 123.

In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.

题目的大意是：

给你三组数据，要你每次判断出比上一次减少了哪个数。

#include<iostream>
using namespace std;

int main()
{
	int n;
	while(cin>>n)
	{
		int i;
		int sum_1=0,sum_2=0;
		int T;
		for(i=0;i<n;i++)
		{
			cin>>T;
			sum_1+=T;
		}
		int j;
		int n1=n-1;
		for(j=0;j<2;j++)
		{
			for(i=0;i<n1;i++)
			{
				cin>>T;
				sum_2+=T;
			}
		cout<<sum_1-sum_2<<endl;
		sum_1=sum_2;
		sum_2=0;
		n1--;
		}
		
	}
	return 0;
}

直接用和减和，比一个数一个数的比较效率高太多了，用一个数一个数比较的方法会超秒。