There are n psychos standing in a line. Each psycho is assigned a unique integer from 1 to n. At each step every psycho who has an id greater than the psycho to his right (if exists) kills his right neighbor in the line. Note that a psycho might kill and get killed at the same step.
You're given the initial arrangement of the psychos in the line. Calculate how many steps are needed to the moment of time such, that nobody kills his neighbor after that moment. Look notes to understand the statement more precise.
Input
The first line of input contains integer n denoting the number of psychos, (1 ≤ n ≤ 105). In the second line there will be a list of n space separated distinct integers each in range 1 to n, inclusive — ids of the psychos in the line from left to right.
Output
Print the number of steps, so that the line remains the same afterward.
Examples
input
10 10 9 7 8 6 5 3 4 2 1
output
2
input
6 1 2 3 4 5 6
output
0
题意：给出1到n的一个序列a[1].a[2]...a[n]，每一轮，如果a[i]>a[i+1],则称a[i]杀掉s[i+1],则可以从数列中去掉a[i+1],并且，一个数可以同时将右边一个数杀掉并且被左边数杀掉，问：最少经过几轮，不会存在杀掉关系？
思路：一开始其实并没有想到用栈，想的是搜索？？
大体的思想是，从后往前遍历数组，ans数组记录杀到能杀的人需要的步数
ac代码：
#include<iostream>
#include<stack>
using namespace std;
int a[100005],ans[100005];
int main()
{
int n,i;
stack<int> s;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=n;i>=1;i--)
{
if(s.empty()) {
s.push(a[i]);
ans[i]=0;
continue;
}
if(a[i]<s.top())
{
s.push(a[i]);
ans[a[i]]=0;
}
else if(a[i]>s.top())
{
int temp=max(1,ans[s.top()]);//这里和下面好好想想，当前a[i]的ans和它之前的ans的关系
s.pop();
while(!s.empty()&&a[i]>s.top())
{
temp=max(temp+1,ans[s.top()]);//其实就是找它前面的人杀人所需最多的步数 ，但是至少也得每次加1
s.pop(); //杀完一个就弹出，也避免了上面的+1重复
}
ans[a[i]]=temp;
s.push(a[i]);
}
}
int maxx=0;
for(i=1;i<=n;i++)
maxx=max(maxx,ans[i]);
cout<<maxx<<endl;
return 0;
}