[poj3250]单调栈 Bad Hair Day

解题关键：将每头牛看到的牛头数总和转化为每头牛被看到的次数，然后用单调栈求解，其实做这道题的目的只是熟悉下单调栈

此题为递减栈

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cstdlib>
 #include<stack>
 #include<iostream>
 using namespace std;
 typedef long long ll;
 ll a[];
 stack<ll>ss;
 int main(){
     ll n,ans;
     while(cin>>n){
         ans=;
         for(int i=;i<n;i++) cin>>a[i];
         for(int i=;i<n;i++){
             while(!ss.empty()&&a[i]>=ss.top()){
                 ss.pop();
             }
             ans+=ss.size();
             ss.push(a[i]);
         }
         cout<<ans<<endl;
         while(!ss.empty())    ss.pop();
     }
     return ;
 } 

数组实现：

 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 typedef long long ll;
 ll ss[],top;
 int main(){
     ios::sync_with_stdio();
     ll n,t,ans=;
     while(cin>>n){
         top=-,ans=;
         for(int i=;i<n;i++){
             cin>>t;
             while(top!=-&&ss[top]<=t) top--;
             ans+=top+;
             ss[++top]=t;
         }
         cout<<ans<<endl;
     }
     return ;
 }