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!
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.
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.
5 8 7 4 8 3 4 2 5 3 7
3 1 4 5
題解：先按A從大到小排序
如果n是奇數 就先把第一個選進去 然後兩個兩個的for 對於相鄰的兩個 選取B大的那個
然後就能保證a的和大於序列和的一半 因爲a1>max(a2,a3) min(a2,a3)>max(a4,a5)...
b選的也是大的 所以也能保證
如果n是偶數 先這樣做一次 然後把1或者2再扔進去即可
#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;
}