1681 B Card Trick(无限循环数组）

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#c++ #算法

这篇文章介绍了一个关于观察者如何通过观察洗牌过程预测魔术后顶部牌的数学谜题。通过理解每次洗牌的影响并应用周期性，解决者能够揭示出即使在多次复杂操作后，卡片的真实位置。关键在于理解操作的累积效应和取模运算在问题中的运用。

B. Card Trick

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Monocarp has just learned a new card trick, and can't wait to present it to you. He shows you the entire deck of nn cards. You see that the values of cards from the topmost to the bottommost are integers a1,a2,…,ana1,a2,…,an, and all values are different.

Then he asks you to shuffle the deck mm times. With the jj-th shuffle, you should take bjbj topmost cards and move them under the remaining (n−bj)(n−bj) cards without changing the order.

And then, using some magic, Monocarp tells you the topmost card of the deck. However, you are not really buying that magic. You tell him that you know the topmost card yourself. Can you surprise Monocarp and tell him the topmost card before he shows it?

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of testcases.

The first line of each testcase contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of cards in the deck.

The second line contains nn pairwise distinct integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n) — the values of the cards.

The third line contains a single integer mm (1≤m≤2⋅1051≤m≤2⋅105) — the number of shuffles.

The fourth line contains mm integers b1,b2,…,bmb1,b2,…,bm (1≤bj≤n−11≤bj≤n−1) — the amount of cards that are moved on the jj-th shuffle.

The sum of nn over all testcases doesn't exceed 2⋅1052⋅105. The sum of mm over all testcases doesn't exceed 2⋅1052⋅105.

Output

For each testcase, print a single integer — the value of the card on the top of the deck after the deck is shuffled mm times.

Example

input

Copy

output

Copy

2
3
3

Note

In the first testcase, each shuffle effectively swaps two cards. After three swaps, the deck will be [2,1][2,1].

In the second testcase, the second shuffle cancels what the first shuffle did. First, three topmost cards went underneath the last card, then that card went back below the remaining three cards. So the deck remained unchanged from the initial one — the topmost card has value 33.

无限循环数组，这个是一维的所以，把操作数对数组长度n取模就能随时提取操作后数组的第一个数

#include<iostream>
#include<algorithm>
using namespace std;
const int manx = 2e5 + 10;
int a[manx];
int b[manx];
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		int n, m;
		cin >> n;
		for (int i = 0; i < n; i++)
		{
			cin >> a[i];
		}
		cin >> m;
		long long sum = 0;
		for (int i = 0; i < m; i++)
		{
			cin >> b[i];
			sum += b[i];
		}
		sum = sum % n;
		cout << a[sum] << endl;
	}
}