Ian and Array Sorting

题目链接

题目描述:

To thank Ian, Mary gifted an array $a$ of length $n$ to Ian. To make himself look smart, he wants to make the array in non-decreasing order by doing the following finitely many times: he chooses two adjacent elements $a_i$ and $a_{i+1}$ $(1\le i\le n-1)$, and increases both of them by $1$ or decreases both of them by $1$. Note that, the elements of the array can become negative.

As a smart person, you notice that, there are some arrays such that Ian cannot make it become non-decreasing order! Therefore, you decide to write a program to determine if it is possible to make the array in non-decreasing order.

输入描述:

The first line contains a single integer $t(1\le t\le {10}^4)$ — the number of test cases. The description of test cases follows.

The first line of each test case consists of a single integer $n(2\le n\le 3\cdot {10}^5)$ — the number of elements in the array.

The second line of each test case contains $n$ integers $a_1,a_2,\dots ,a_n(1\le ai\le {10}^9)$ — the elements of the array $a$.

It is guaranteed that the sum of $n$over all test cases does not exceed $3\cdot {10}^5$.

输出描述:

For each test case, output "$YES$" if there exists a sequence of operations which make the array non-decreasing, else output "$NO$".

You may print each letter in any case (for example, "$YES$", "$Yes$", "$yes$", "$yEs$" will all be recognized as positive answer).

样例：

input:

5
3
1 3 2
2
2 1
4
1 3 5 7
4
2 1 4 3
5
5 4 3 2 1

output:

YES
NO
YES
NO
YES

AC代码：

思路:

这个题直接写没有思路，但是可以转换成$a_2-a_1,a_3-a_2...a_n-a_{n-1}$

设这个数组为$b_1,b_2,...b_{n-1}$

$a_1,a_2...a_n$是一个非递减序列当且仅当$b$数组为非负数

最后可以看到有两种改变的情况

• $i$在$1$~$n-3$时，$b_i$ 加或减 $1$，则 $b_{i+2}$ 减或加$1$

• 单独改变$b_2$和$b_{n-2}$且不影响其他数

当在$n$为奇数时，$n-2$为奇数。所以对于$b_{n-2}$我们可以给它加上足够大的数

那么$i$为奇数时，我们可以任意改变其他$i$为奇数的$b$的值使得它们为非负数

而$i$为偶数时同理，给$b_2$加上足够大的数，使得其他$i$为偶数的$b$的值为非负数

当$n$为偶数时，$n-2$为偶数，我们给$b_2$和$b_{n-2}$加上足够大的数，此时只能保证$i$为偶数时$b$的值都为非负数

$i$为奇数的$b$的值，我们可以随意分配，但是$b$的值的总和都是不变的

所以此时只要将$i$为奇数时的$b$的总和加起来，其值大于等于$0$则代表可以保证$b$的值为非负数

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long LL;
 
void solve()
{
	int n;
	cin >> n;
 
	vector<int> a(n + 1);
 
	for(int i = 1; i <= n; i ++)
		cin >> a[i];
 
	if(n % 2 == 1)
	{
		cout << "YES\n";
 
		return;
	}
 
	LL sum = 0;
 
	for(int i = 2; i <= n; i += 2)
		sum += a[i] - a[i - 1];
 
	if(sum >= 0)
	{
		cout << "YES\n";
 
		return;
	}
 
	cout << "NO\n";
}
 
int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
 
	int T;
	cin >> T;
 
	while(T --)
		solve();
 
	return 0;
}