ssu 488. Dales and Hills

488. Dales and Hills

Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes

input: standard
output: standard

Let's consider a number sequence a₁, ·s, a_N. We call the continuous subsequence a_i, ·s, a_j, ·s, a_k (1 ≤ i < j < k ≤ N) of the sequence a if a_t < a_t+1 for any i ≤ t < j and a_t > a_t+1 for any j ≤ t < k. In this case we call the of the hill. Similarly, we call the continuous subsequence a if a_t > a_t+1 for any i ≤ t < j and a_t < a_t+1 for any j ≤ t < k. In this case we call the of the dale.

Compute the height of the highest hill and the depth of the deepest dale in the given sequence.

Input

The first line of the input file contains T (), the number of test cases. The test cases follow, occupying two lines each. The first of the two lines contains N (), the second the members of the sequence, separated by spaces. The sum of values of N over all test cases in the file does not exceed . The absolute values of the members of the sequences do not exceed .

Output

The output file should consist of T lines and each line should contain two integers, the height of the highest hill and the depth of the deepest dale. If there are no hills or no dales, output 0 in the corresponding position.

Example(s)

sample input	sample output
2 10 4 4 1 6 3 2 1 2 5 7 10 2 3 4 5 6 7 8 9 10 9	1 3 1 0

 

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
#include <cmath>
#include <queue>
#include <cstring>
#include <set>
#include <stack>
#include <string>
#define LL long long
#define maxn 1000010
#define mod 1000000007
#define INF 2000000
#define MAX 16000010
#define eps 1e-6
using namespace std;

int a[maxn] ;
int main()
{
    int i ,m , tt , j , n ;
    int x , y , ii ,jj ,T ;
     //freopen("in.txt","r",stdin) ;
    // 直接枚举了，不过记得到了n要把 j--
    cin >> T ;
    while( T-- )
    {
        scanf("%d",&n) ;
        for( i = 0 ; i < n ;i++)
          scanf("%d",&a[i]) ;
          a[n] = -9999999 ;
          x = y = 0 ;
        for( i = 0 ; i < n ;i++ ) if(a[i] < a[i+1])
        {
            for( j = i+1 ; j < n ;j++)
                if(a[j] >= a[j+1]) break ;
                if(j==n)j-- ;
                jj = j ;
            for( ; j < n ;j++ )
                if(a[j] <= a[j+1]) break ;
                if(j==n)j-- ;
            x = max(x,min(j-jj,jj-i)) ;
            i = j-1 ;// 这个记得要
        }
        a[n] = 9999999;
        for( i = 0 ; i < n ;i++ ) if(a[i] > a[i+1])
        {
            for( j = i+1 ; j < n ;j++)
                if(a[j] <= a[j+1]) break ;
                if(j==n)j-- ;
                jj = j ;
            for( ; j < n ;j++ )
                if(a[j] >= a[j+1]) break ;
                if(j==n)j-- ;
            y = max(y,min(j-jj,jj-i)) ;
            i = j-1 ;
        }
        printf("%d %d\n",x,y) ;
    }
    return 0 ;
}