POJ 3250 Bad Hair Day simulated monotonous Stack

POJ 3250 Bad Hair Day simulated monotonous Stack

 

Bad Hair Day

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:14989		Accepted:4977

 

Description

Some of Farmer John'sNCows (1 ≤N≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows 'heads.

Each cowIHas a specified heightHi(1 ≤Hi≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cowICan see the tops of the heads of cows in front of her (namely cowsI+ 1,I+ 2, and so on), for as long as these cows are strictly shorter than cowI.

Consider this example:

        ==       ==   -   =         Cows facing right -->=   =   == - = = == = = = = =1 2 3 4 5 6 

Cow #1 can see the hairstyle of cows #2, 3, 4
Cow #2 can see no cow's hairstyle
Cow #3 can see the hairstyle of cow #4
Cow #4 can see no cow's hairstyle
Cow #5 can see the hairstyle of cow 6
Cow #6 can see no cows at all!

LetCiDenote the number of cows whose hairstyle is visible from cowI; Please compute the sumC1 throughCN. For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

Line 1: The number of cows, N.
Lines 2. N + 1: Line I+ 1 contains a single integer that is the height of cow I.

Output

Line 1: A single integer that is the sum C1 through CN.

Sample Input

610374122

Sample Output

5

 

Tell a sequence to calculate the sum of numbers smaller than each other.

 

# Include
 
  
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        # Define N 100009 typedef long ll; using namespace std; ll a [N]; int main () {int n; while (~ Scanf (% d, & n) {for (int I = 1; I <= n; I ++) scanf (% I64d, & a [I]); ll ans = 0; int B [N]; B [n] = n; for (int I = n-1; I> = 1; I --) {int tt = I; while (tt
       
         A [tt + 1]) tt = B [tt + 1]; // strictly monotonous, not equal to B [I] = tt;} // for (int I = 1; I <= n; I ++) // cout <