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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 n each 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 题意：给你两组序列a和b。让你从中选出两个序列中找出最多n/2+1个下标相同的数字。使得从两个数组中选出来的这些数字和比原数组和一半要大。 思路：最多选取n/2+1个数字，贪心的思想肯定要选最多的数字。而且选的越大越好。所以我们对于a数组按着降序排序。根据a数组降序的位置去找b数组中较大的那个数。相当于把原来的数组分成了n/2的集合。 代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=1e5+100;ll a[maxx];ll b[maxx];struct node{
   	int pos;	ll v;	bool operator<(const node &c)const{
   		return v>c.v;	}}p[maxx];int n;int main(){
   	scanf("%d",&n);	ll sum1=0,sum2=0;	for(int i=1;i<=n;i++) scanf("%I64d",&a[i]),p[i].pos=i,p[i].v=a[i];	for(int i=1;i<=n;i++) scanf("%I64d",&b[i]);	sort(p+1,p+1+n);	cout<<(n/2+1)<
    
     b[p[i+1].pos]?p[i].pos:p[i+1].pos);	}	cout<
    
   

努力加油a啊，(^o)/~

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