Codeforces Round #698 (Div. 2) C. Nezzar and Symmetric Array（思维）

题目链接：http://codeforces.com/contest/1478/problem/C

Long time ago there was a symmetric array a1,a2,…,a2n consisting of 2n distinct integers. Array a1,a2,…,a2n is called symmetric if for each integer 1≤i≤2n, there exists an integer 1≤j≤2n such that ai=−aj.

For each integer 1≤i≤2n, Nezzar wrote down an integer di equal to the sum of absolute differences from ai to all integers in a, i. e. di=∑2nj=1|ai−aj|.

Now a million years has passed and Nezzar can barely remember the array d and totally forget a. Nezzar wonders if there exists any symmetric array a consisting of 2n distinct integers that generates the array d.

Input
The first line contains a single integer t (1≤t≤105) — the number of test cases.

The first line of each test case contains a single integer n (1≤n≤105).

The second line of each test case contains 2n integers d1,d2,…,d2n (0≤di≤1012).

It is guaranteed that the sum of n over all test cases does not exceed 105.

Output
For each test case, print “YES” in a single line if there exists a possible array a. Otherwise, print “NO”.

You can print letters in any case (upper or lower).

Example

input

6
2
8 12 8 12
2
7 7 9 11
2
7 11 7 11
1
1 1
4
40 56 48 40 80 56 80 48
6
240 154 210 162 174 154 186 240 174 186 162 210
1
2
3
4
5
6
7
8
9
10
11
12
13

output

YES
NO
NO
NO
NO
YES
1
2
3
4
5
6

---

分析

存在满足要求的 a 的前提是：

1. 每个元素都有且仅有一个与其相同的元素
2. 每个元素都是偶数

我们可以直接算出绝对值最大的点，就是 d 数组中的最大值 max / (2 *n) ，然后也可以计算出每组点之间的距离的绝对值，如果都能被规定的数整除，并且和是大于 max / (2 *n) 的，那么答案就是 yes 。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int t,n;
ll d[200007];
map<ll,ll> mp;
ll idx[200007],cnt;

int main()
{
	scanf("%d",&t);
	while(t--)
	{
		cnt = 0;
		int flag = 1;
		mp.clear();
		scanf("%d",&n);
		for(int i=1;i<=n*2;i++)
		{
			scanf("%lld",&d[i]);
			if(d[i] % 2 != 0)
			{
				flag = 0;
			}
			if(mp[d[i]] == 0) idx[++cnt] = d[i];
			else if(mp[d[i]] == 2) flag = 0;
			mp[d[i]]++;
		}
		if(flag == 0)
		{
			printf("NO\n");
			continue;
		}
		for(int i=1;i<=n*2;i++)
		{
			//cout<<"----"<<mp[d[i]]<<endl;
			if(mp[d[i]] % 2 != 0)
			{
				flag = 0;
				break;
			}
		}
		if(flag == 0)
		{
			printf("NO\n");
			continue;
		}

		sort(idx + 1, idx + 1 + cnt);
		if(idx[cnt] % (1ll * n * 2) != 0) flag = 0;
		ll t = idx[cnt] / (n * 2);
		for(ll i=1;i<cnt;i++)
		{
			ll tmp = 1ll * i * 2;
			if((idx[i + 1] - idx[i]) % tmp != 0) flag = 0;
			t -= (idx[i + 1] - idx[i]) / tmp;
		}
		if(t <= 0 || flag == 0) printf("NO\n");
		else printf("YES\n");
	}
	return 0;
}
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