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Bad Hair Day
| Time Limit: 2000MS |
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Memory Limit: 65536K |
| Total Submissions: 14989 |
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Accepted: 4977 |
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Description
Some of Farmer John's N cows (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 cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.
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!
Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.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 of
c
1 through
cN.
Sample Input
6
10
3
7
4
12
2
Sample Output
5
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告诉一个序列，求每个数后面比他小的个数和
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#include
#include
#include
#include
#include
#include
#include
#define N 100009 typedef long 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];//严格单调，不去等号 b[i]=tt; } // for(int i=1;i<=n;i++) // cout<
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