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Language:
Bad Hair Day
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 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：一群牛有高有低站成一队每个牛可以看到它右边比他个低的牛如果有一个牛比它高则此后的牛都看不到；求每个牛能看到的牛的数量和
思路：把每个牛能看到牛的个数转化为能看到当前牛的个数。用单调栈；
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=80010;
long long stack[maxn];
int main()
{int n,i,j,k;while(scanf("%d",&n)!=EOF){int top=0;long long ans=0,num; for(i=1;i<=n;++i){scanf("%d",&num);while(top>0&&stack[top-1]<=num)top--;ans+=top;stack[top++]=num;}printf("%lld\n",ans);}return 0;
}