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
題意：給你兩組序列a和b。讓你從中選出兩個(gè)序列中找出最多n/2+1個(gè)下標(biāo)相同的數(shù)字。使得從兩個(gè)數(shù)組中選出來(lái)的這些數(shù)字和比原數(shù)組和一半要大。
思路：最多選取n/2+1個(gè)數(shù)字，貪心的思想肯定要選最多的數(shù)字。而且選的越大越好。所以我們對(duì)于a數(shù)組按著降序排序。根據(jù)a數(shù)組降序的位置去找b數(shù)組中較大的那個(gè)數(shù)。相當(dāng)于把原來(lái)的數(shù)組分成了n/2的集合。
代碼如下：

#include<bits/stdc++.h> #define ll long long using 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)<<endl;printf("%d",p[1].pos);//最大的數(shù)字最好選上for(int i=2;i<=n;i+=2){if(i==n) printf(" %d",p[i].pos);else printf(" %d",b[p[i].pos]>b[p[i+1].pos]?p[i].pos:p[i+1].pos);}cout<<endl; }

努力加油a啊，(^o)/~

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