Codeforces Round #410 (Div. 2) D. Mike and distribution(贪心)

Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive integers A = [a1, a2, ..., an] and B = [b1, b2, ..., bn] of
length neach which he uses to ask people some quite peculiar questions.

To test you on how good are you at spotting inequality in life, he wants you to find an "unfair" subset of the original sequence. To be more precise, he wants you to select k numbers P = [p1, p2, ..., pk] such
that 1 ≤ pi ≤ n for 1 ≤ i ≤ k and
elements in P are distinct. Sequence P will
represent indices of elements that you'll select from both sequences. He calls such a subset P "unfair" if
and only if the following conditions are satisfied: 2·(ap1 + ... + apk) is greater than
the sum of all elements from sequence A, and 2·(bp1 + ... + bpk)is greater than
the sum of all elements from the sequence B. Also, k should
be smaller or equal to 

 because
it will be to easy to find sequence P if he allowed you to select too many elements!

Mike guarantees you that a solution will always exist given the conditions described above, so please help him satisfy his curiosity!

Input

The first line contains integer n (1 ≤ n ≤ 105)
— the number of elements in the sequences.

On the second line there are n space-separated integers a1, ..., an (1 ≤ ai ≤ 109)
— elements of sequence A.

On the third line there are also n space-separated integers b1, ..., bn (1 ≤ bi ≤ 109)
— elements of sequence B.

Output

On the first line output an integer k which represents the size of the found subset. k should
be less or equal to 

.

On the next line print k integers p1, p2, ..., pk (1 ≤ pi ≤ n)
— the elements of sequence P. You can print the numbers in any order you want. Elements in sequence P should
be distinct.

Example

input

5
8 7 4 8 3
4 2 5 3 7

output

3

1 4 5

分析：按第一组数排序后，先把第一个选上，然后后边按顺序两两配对，每对中选同编号中第二组数较大的那一个数的编号，很容易发现这样选下来两组数都是符合条件的。

#include <bits/stdc++.h>
#define N 100005
using namespace std;
struct thing
{
int v,n;
friend bool operator < (thing a,thing b)
{
return a.v > b.v;
}
}a
;
int n,cnt,b
,ans
;
int main()
{
scanf("%d",&n);
for(int i = 1;i <= n;i++)
{
scanf("%d",&a[i].v);
a[i].n = i;
}
sort(a+1,a+1+n);
for(int i = 1;i <= n;i++) scanf("%d",&b[i]);
ans[++cnt] = a[1].n;
for(int i = 2;i <= n;i+=2)
{
if(i == n) ans[++cnt] = a[i].n;
else
if(b[a[i].n] > b[a[i+1].n]) ans[++cnt] = a[i].n;
else ans[++cnt] = a[i+1].n;
}
cout<<cnt<<endl;
for(int i = 1;i <= cnt;i++) cout<<ans[i]<<" ";
}