PTA 7-98 Number
_October has a question:

You are given n integers a1,a2,…,an. Find the maximum value of max(al,al+1,…,ar)⋅min(al,al+1,…,ar)over all pairs (l,r) of integers for which 1≤l

INPUT:
The first line contains a single integer n(2≤n≤10
3
) .

The second line contains n integers a1,a2,…,an(1≤ai≤10
6
).

OUTPUT:
Please print a single integer — the maximum possible value of the product from the statement.

SAMPLE INPUT1:
3
2 4 3
SAMPLE OUTPUT1:
12
Note:
Let f(l,r)=max(al,al+1,…,ar)⋅min(al,al+1,…,ar).
In the first test case,
f(1,2)=max(a1,a2)⋅min(a1,a2)=max(2,4)⋅min(2,4)=4⋅2=8.
f(1,3)=max(a1,a2,a3)⋅min(a1,a2,a3)=max(2,4,3)⋅min(2,4,3)=4⋅2=8.
f(2,3)=max(a2,a3)⋅min(a2,a3)=max(4,3)⋅min(4,3)=4⋅3=12.
So the maximum is f(2,3)=12.
SAMPLE INPUT2:
4
3 2 3 1
SAMPLE OUTPUT2:
6
Advanced challenge ：What if the maxinum of n is 3∗10
5
?
代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB
代码：

#include int main()
{long int n = 0, arr[1000], min = 0, max = 0, result = 0, num;scanf( "%ld" , &n );for( int i = 0; i < n; i++ )//循环输入到数组中{scanf( "%ld" , &arr[i] );}for( int i = 0; i < n; i++ ){for( int j = i + 1; j < n; j++ ){min = arr[i];//假设最小值是数组中的某一个max = arr[i];//假设最大值是数组中的某一个for( int k = i; k <= j; k++ ){if( arr[k] > max )//找出最大值{max = arr[k];}if( arr[k] < min )//找出最小值{min = arr[k];}}if( min * max > result ){result = min * max;//把最小值和最大值相乘}}}printf( "%ld", result );return 0;
}