Codeforces 798D 构造

Mike and distribution

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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) isgreater 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

58 7 4 8 34 2 5 3 7

output

31 4 5

题意：给你两个序列  从中选出n/2+1个相同下标的数  使得两个序列的和  大于各自序列和的一半  题目保证有解

题解：先按A从大到小排序

如果n是奇数  就先把第一个选进去  然后两个两个的for  对于相邻的两个  选取B大的那个

然后就能保证a的和大于序列和的一半  因为a1>max(a2,a3)  min(a2,a3)>max(a4,a5)...

b选的也是大的  所以也能保证

如果n是偶数  先这样做一次  然后把1或者2再扔进去即可

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;struct node{int a,b,lab;bool operator <(const node& c)const{if(a==c.a)return b>c.b;return a>c.a;}}e[100005];vector<int>sp;int main(){    int n,i,j;    scanf("%d",&n);    for(i=1;i<=n;i++){    scanf("%d",&e[i].a);    e[i].lab=i;    }    for(i=1;i<=n;i++)scanf("%d",&e[i].b);    sort(e+1,e+1+n);    if(n%2){    sp.push_back(e[1].lab);    i=2;    }    else i=1;for(;i<=n;i+=2){    if(e[i].b<e[i+1].b)sp.push_back(e[i+1].lab);    else sp.push_back(e[i].lab);    }    if(n%2==0){    if(sp[0]!=e[1].lab)sp.push_back(e[1].lab);    else sp.push_back(e[2].lab);    }    printf("%d\n",sp.size());    for(i=0;i<sp.size();i++)printf("%d ",sp[i]);    printf("\n");    return 0;}