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CodeForces - 1420A
For god’s sake, you’re boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that’s funny, is it? Oh it’s funny? Because we’ve been at this for twelve hours and you haven’t solved it either, so I don’t know why you’re laughing. You’ve got one hour! Solve it!
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
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is “NO”.
1. Question meaning
Given you an array, you can swap two adjacent numbers each time so that the array does not decrease monotonously. If you swap more than n*(n-1)/2 -1, output YES, otherwise output NO.
2. Thinking
This problem is to solve the number of bubble sort exchanges, but if you directly use bubble sort to solve, the time complexity of an example is O(n), and 1000 examples will definitely time out. When an array declines strictly monotonously, that is, in the worst case of bubble sorting, the number of exchanges is n-1+n-2+n-3+…+ 2+1, which is n*(n-1)/2, we can It is found that the number of times n*(n-1)/2 -1 required by the question is exactly one difference. Therefore, as long as the array is not strictly monotonically decreasing, the requirement can be met.
3. Code
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 5e4 + 10;
int a[N];
int main()
{
int T;
cin >> T;
while (T--)
{
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
int flag = 1;
for (int i = 0; i < (n-1); i++)
{
if (a[i] <= a[i + 1])
{
flag = 0;
break;
}
}
if(flag==0) puts("YES");
else puts("NO");
}
return 0;
}