B. Parity Alternated Deletions

B. Parity Alternated Deletions

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

I

nput

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

思路：

计量偶数与奇数分别有多少个，并且单独储存，当奇数偶数数量相同时，输出0，不相等让把多的那个的数加到sum里（加数量差并减去第一次任意删除的那一个）。

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
int main(){
	int n,sum=0;
	int x=0,y=0;
	while(cin>>n){
		sum=0;
		int a[3000],b[3000],c[3000];
		x=0,y=0;
		for(int i=0;i<n;i++){
			cin>>a[i];
			if(a[i]%2==0){
				b[x]=a[i];
				x++;
			}//把偶数计入
			else if(a[i]%2!=0){
				c[y]=a[i];
				y++;
			}//把奇数计入
		}
		int z,j,k;
		sort(b,b+x);sort(c,c+y);
		
		if(x==y){
			sum=0;
		}
		else if(x>y){
			z=x-y-1;
			for(j=0;j<=z-1;j++){
				sum+=b[j];
			}
		}
		else if(y>x){
			z=y-x-1;
			for(k=0;k<=z-1;k++){
				sum+=c[k];
			}
		}
		cout<<sum<<endl;
	}
	return 0;
}