B. Array Sharpening---------------------思维

You’re given an array a1,…,an of n non-negative integers.

Let’s call it sharpened if and only if there exists an integer 1≤k≤n such that a1<a2<…ak+1>…>an. In particular, any strictly increasing or strictly decreasing array is sharpened. For example:

The arrays [4], [0,1], [12,10,8] and [3,11,15,9,7,4] are sharpened;
The arrays [2,8,2,8,6,5], [0,1,1,0] and [2,5,6,9,8,8] are not sharpened.
You can do the following operation as many times as you want: choose any strictly positive element of the array, and decrease it by one. Formally, you can choose any i (1≤i≤n) such that ai>0 and assign ai:=ai−1.

Tell if it’s possible to make the given array sharpened using some number (possibly zero) of these operations.

Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤15 000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤n≤3⋅105).

The second line of each test case contains a sequence of n non-negative integers a1,…,an (0≤ai≤109).

It is guaranteed that the sum of n over all test cases does not exceed 3⋅105.

Output
For each test case, output a single line containing “Yes” (without quotes) if it’s possible to make the given array sharpened using the described operations, or “No” (without quotes) otherwise.

Example
inputCopy
10
1
248618
3
12 10 8
6
100 11 15 9 7 8
4
0 1 1 0
2
0 0
2
0 1
2
1 0
2
1 1
3
0 1 0
3
1 0 1
outputCopy
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No
Note
In the first and the second test case of the first test, the given array is already sharpened.

In the third test case of the first test, we can transform the array into [3,11,15,9,7,4] (decrease the first element 97 times and decrease the last element 4 times). It is sharpened because 3<11<15 and 15>9>7>4.

In the fourth test case of the first test, it’s impossible to make the given array sharpened.

题意:
给你n个数，每个数都可以无限次操作-1，使得这n个数满足
a1<a2<a3< ak > ak+1>ak+2> an

解析:
考虑极端情况
0,1,2,3,4…n,n-1,n-2,n-3…0

所以我们只需遍历一遍，到峰点之前要递增的。所以只要判断a[i]>=i-1.达到峰点后，我们需要判断一种特殊情况 0,1,1,0。然后峰点之后的一部分必须递减，只要满足a[i]>=n-i

#include<bits/stdc++.h>
using namespace std;
const int N=3e5+1000;
int n;
int t;
int a[N];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		int f=1;
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		int i;
		for(i=1;i<=n;i++)
		{
			if(a[i]>=i-1) continue;
			break;
		}
		if(a[i]==a[i-1]&&!(a[i]>=n-i+1)) f=0; //判断0,1,1,0,这种情况
		for(;i<=n;i++)
		{
			if(a[i]>=n-i) continue;
			f=0;
			break;
		}
		if(f==1) puts("YES");
		else puts("NO");
	}
}