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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 14989 | Accepted: 4977 |
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
Output
Sample Input
6 10 3 7 4 12 2
Sample Output
5
告诉一个序列，求每个数后面比他小的个数和
#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#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<n&&a[i]>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<<b[i]<<" ";
for(int i=1;i<=n;i++)
{
ans+=b[i]-i;
}
printf("%I64d\n",ans);
}
return 0;
}
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原文地址：http://blog.csdn.net/wust_zjx/article/details/46778367
