Codeforces Round #410 (Div. 2) D. Mike and distribution-优快云博客

本文介绍了一个关于从两个整数序列中选取特定子集的问题，以确保所选元素的两倍总和超过原序列总和。通过一种二维贪心算法解决了这一问题，并提供了实现代码。

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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 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. SequenceP 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 $\text{[math]}$ 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 $\text{[math]}$ .

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：给你两个长度为n的数组a,b，让你从中选择n/2+1个下标，使得对于每个数组，选择的元素之和的2倍大于整个数组的元素之和。

思路（二维贪心，先让一维有序）：先把下标按a数组的大小从大到小排列，把第一个选上，然后接下来两个两个为一组，选择b数组值较大的那个下标，最后如果n是偶数，则还要把最后一个也选上。

#include<cstdio>
#include<algorithm>
using namespace std;

int n,k;
struct node
{
	int num;
	int id;
} a[100005],b[100005];

int ans[100005];

int cmp(node a,node b)
{
	return a.num>b.num;
}

int main()
{
	scanf("%d",&n);
	for(int i=0; i<n; i++)
	{
		scanf("%d",&a[i].num);
		a[i].id=i;
	}
	for(int i=0; i<n; i++)
	{
		scanf("%d",&b[i].num);
		b[i].id=i;
	}	
	sort(a,a+n,cmp);
	k=n/2+1;
	ans[0]=a[0].id+1;
	int j=1;
	for(int i=1; i+1<n; i=i+2)
	{
		if(b[a[i].id].num>b[a[i+1].id].num)
			ans[j]=a[i].id+1;
		else
			ans[j]=a[i+1].id+1;
		j++;
	}
	printf("%d\n",k);
	if(!(n&1))
	  ans[j]=a[n-1].id+1;
	for(int i=0;i<k;i++)
	   printf("%d ",ans[i]);
}