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 6Cow#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 【题意】n头奶牛站成一排，从左往右看，只要右边高度<=自己高度的都可以看到，但是一旦遇到比自己高的，则它后面的奶牛都看不到，即计数停止，要计算所有牛向右所能看到的数目和。（奶牛哪知道哪边是左哪边是右~~~汗，，，）【AC代码】
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; typedef long long ll; const int N=8e4+5; int main() { int m,n; ll a[N],vis[N]; while(~scanf("%d",&m)) { ll ans=0; //long long 可以通过 memset(a,0,sizeof(a)); memset(vis,0,sizeof(vis)); for(int i=1;i<=m;i++) { scanf("%lld",&a[i]); vis[i]=i; } for(int i=m-1;i>0;i--) { int s=i; while(a[s+1]<a[i]&&s<m)//相等也看不到 s=vis[s+1]; vis[i]=s; } for(int i=1;i<=m;i++) ans+=(vis[i]-i); printf("%lld\n",ans); } return 0; }
