An Easy Problem

Description

Now we define that:

A(j)={i|0<i<j&&a[i]<a[j]}

Here a is an array with n element(1<=n<=1000000);

And then we define a function f that:

int f(int j){

if(A(j) is not empty) return max( A(j) );

else return -1;

}

Input

The first line contain an integer n(1<=n<=1000000),the size of the array a.

The second line contain n integers describe the array a.

The third line is an integer m(1<=m<=1000000),the number of the query.

And then follow m lines,each line is an integer j(1<=j<=n).

Output

For each query j,output a[f(j)] if f(j)>=1,Otherwise output -1 instead.

Sample Input

3

1 2 3

3

1

2

3

Sample Output

-1

1

2

单调DP，用另一数组记录下a[f(i)]，可用stack实现。

#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;
#define N 1000005
int a
,b
;
int main()
{
int n,m,i,temp,x;
stack<int> S;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(S.empty())
{
b[i]=-1;
S.push(a[i]);
}
else
{
temp=S.top();
if(temp<a[i])
{
b[i]=temp;
S.push(a[i]);
}
else
{
S.pop();
while(!S.empty())
{
temp=S.top();
if(temp<a[i])
{
b[i]=temp;
S.push(a[i]);
break;
}
else
S.pop();
}
if(S.empty())
{
b[i]=-1;
S.push(a[i]);
}
}
}
}
scanf("%d",&m);
while(m--)
{
scanf("%d",&x);
printf("%d\n",b[x-1]);
}
return 0;
}