題干：

A sequence of??integers??is called a peak, if and only if there exists exactly one integer??such that?, and??for all?, and??for all?.

Given an integer sequence, please tell us if it's a peak or not.

Input

There are multiple test cases. The first line of the input contains an integer?, indicating the number of test cases. For each test case:

The first line contains an integer??(), indicating the length of the sequence.

The second line contains??integers??(), indicating the integer sequence.

It's guaranteed that the sum of??in all test cases won't exceed?.

Output

For each test case output one line. If the given integer sequence is a peak, output "Yes" (without quotes), otherwise output "No" (without quotes).

Sample Input

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

Sample Output

Yes No No No Yes No No

題目大意：

? 給你一組數，問你能否湊出一個是山峰的情況。即這組數先遞增然后再遞減。

解題報告：

? ? 注意判斷不能一直遞增或者一直遞減就可以了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int n; ll a[MAX]; int main() {int t;cin>>t;while(t--) {scanf("%d",&n);for(int i = 1; i<=n; i++) scanf("%lld",a+i);int tar,flag = 1;if(a[1] >= a[2] || a[n-1] <= a[n]) {printf("No\n");continue;}for(int i = 2; i<=n-1; i++) {if(a[i] < a[i+1]) continue;else {tar = i;break;}}for(int i = tar; i<=n-1; i++) {if(a[i] > a[i+1]) continue;else {flag = 0;break;}}if(flag) printf("Yes\n");else printf("No\n");}return 0 ; }

?

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