Parity Alternated Deletions

Polycarp has an array a consisting of n integers.

He wants to play a game with this array. The game consists of several moves. On the first move he chooses any element and deletes it (after the first move the array contains n−1 elements). For each of the next moves he chooses any element with the only restriction: its parity should differ from the parity of the element deleted on the previous move. In other words, he alternates parities (even-odd-even-odd-… or odd-even-odd-even-…) of the removed elements. Polycarp stops if he can’t make a move.

Formally:

If it is the first move, he chooses any element and deletes it;
If it is the second or any next move:
if the last deleted element was odd, Polycarp chooses any even element and deletes it;
if the last deleted element was even, Polycarp chooses any odd element and deletes it.
If after some move Polycarp cannot make a move, the game ends.
Polycarp’s goal is to minimize the sum of non-deleted elements of the array after end of the game. If Polycarp can delete the whole array, then the sum of non-deleted elements is zero.

Help Polycarp find this value.

Input
The first line of the input contains one integer n (1≤n≤2000) — the number of elements of a.

The second line of the input contains n integers a1,a2,…,an (0≤ai≤106), where ai is the i-th element of a.

Output
Print one integer — the minimum possible sum of non-deleted elements of the array after end of the game.

Examples

Input

5
1 5 7 8 2

Output

0

Input

6
5 1 2 4 6 3

Output

0

Input

2
1000000 1000000

Output

1000000

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define ll long long
#define mes(x,y) memset(x,y,sizeof(x))
using namespace std;
int main(){
	int n;
	while(cin>>n){
		int m;
		vector<int>a,b;
		while(n--){
			cin>>m;
			if(m%2==0){
				a.push_back(m);
			}
			else{
				b.push_back(m);
			}
		}
		long sum=0;
		long len_a=a.size(),len_b=b.size();
		if(len_a==len_b){
			sum=0;
		}
		else if(len_a>len_b){
			long len=len_a-len_b-1;
			sort(a.begin(),a.end());
			for(int i=0;i<len;i++){
				sum+=a[i];
			}
		}
		else{
			long len=len_b-len_a-1;
			sort(b.begin(),b.end());
			for(int j=0;j<len;j++){
				sum+=b[j];
			}
		}
		cout<<sum<<endl;
	}
	return 0;
}

标签：Deletions,Parity,Alternated,move,len,element,Polycarp,include,odd
来源： https://blog.csdn.net/weixin_44417851/article/details/89059365