Almost Sorted Array HDU - 5532

We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.

We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array a1,a2,…,an
, is it almost sorted?
Input
The first line contains an integer T indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n integers a1,a2,…,an.

1≤T≤2000
2≤n≤105
1≤ai≤105
There are at most 20 test cases with n>1000
.
Output
For each test case, please output “YES” if it is almost sorted. Otherwise, output “NO” (both without quotes).
Sample Input

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

Sample Output

YES
YES
NO

Think:相邻的三个数就可以判断出来是否要删数,把升序和降序两种情况都跑一遍,之后判断一下就行了

#include 
#include 
#include 
#include 
using namespace std;
int a[100010];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for (int i=0;iscanf("%d",&a[i]);
        }
        int flag=0;
        int sum=0;
        if (a[0]>a[1]){
            flag=1;
        }
        for (int i=2;iif (a[i]1]){
                if (flag==1){
                    flag=2;
                    break;
                }
                else{
                    if (a[i]>=a[i-2]){
                        flag=1;
                    }
                    else if (i1]>=a[i-1]){
                        flag=1;
                        i++;
                    }
                    else if (i==n-1){
                        flag=1;
                    }
                    else{
                        flag=2;
                        break;
                    }
                }
            }
        }
        sum+=flag;
        flag=0;
        if (a[0]1]){
            flag=1;
        }
        for (int i=2;iif (a[i]>a[i-1]){
                if (flag==1){
                    flag=2;
                    break;
                }
                else{
                    if (a[i]<=a[i-2]){
                        flag=1;
                    }
                    else if (i1]<=a[i-1]){
                        flag=1;
                        i++;
                    }
                    else if (i==n-1){
                        flag=1;
                    }
                    else{
                        flag=2;
                        break;
                    }
                }
            }
        }
        sum+=flag;
        if (sum<4){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }
    return 0;
}

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