Codeforces Round #672 (Div. 2) Problem A

今日份的题目。（指9月24日，因为比赛从晚上10点半持续到12点半）
问题A是水题，题面如下（大半夜了，就不翻译了，赶着睡觉）（其他题目明天再发）
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.

For completing the chamber Wheatley needs n
cubes. i-th cube has a volume ai

.

Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1
, ai−1≤ai

must hold.

To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1
you can exchange cubes on positions i−1 and i

.

But there is a problem: Wheatley is very impatient. If Wheatley needs more than n⋅(n−1)2−1

exchange operations, he won’t do this boring work.

Wheatly wants to know: can cubes be sorted under this conditions?
Input

Each test contains multiple test cases.

The first line contains one positive integer t
(1≤t≤1000

), denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains one positive integer n
(2≤n≤5⋅104

) — number of cubes.

The second line contains n
positive integers ai (1≤ai≤109

) — volumes of cubes.

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

.
Output

For each test case, print a word in a single line: “YES” (without quotation marks) if the cubes can be sorted and “NO” (without quotation marks) otherwise.

Example
Input

3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1

Output

YES
YES
NO

实际上就是求逆序对。
主要运用的数学定理是，任意一个排列经过和逆序对数等同次数的相邻元素交换，可以得到有序的排列。
所以就是求逆序对的数量，再和n(n-1)/2-1做比较即可。
同时注意一下取值范围。
代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 500005
using namespace std;
int T;
int n;
long long aim;
int num[maxn];
int r_num[maxn],l_num[maxn];
long long merge(int l,int r){
	if(l==r){
		return 0;
	}
	int mid=(l+r)>>1;
	long long ans=0;
	ans=merge(l,mid)+merge(mid+1,r);
	int lind=0,rind=0;
	for(int i=l;i<=mid;i++){
		l_num[i-l]=num[i];
	}
	for(int i=mid+1;i<=r;i++){
		r_num[i-mid-1]=num[i];
	}
	for(int i=l;i<=r;i++){
		if(lind>(mid-l)){
			num[i]=r_num[rind++];
		}else if(rind>(r-mid-1)){
			num[i]=l_num[lind++];
			ans+=rind;
		}else if(l_num[lind]<=r_num[rind]){
			num[i]=l_num[lind++];
			ans+=rind;
		}else{
			num[i]=r_num[rind++];
		}
	}
	return ans;
}
int main(){
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		aim=1LL*n*(n-1)/2-1;
		for(int i=0;i<n;i++){
			scanf("%d",&num[i]);
		}
		long long step=merge(0,n-1);
//		printf("%d ",step);
		if(step<=aim){
			printf("YES\n");
		}else{
			printf("NO\n");
		}
	}
	return 0;
}