https://codeforces.com/contest/1144/problem/B

题意：给定一个数组，进行操作：每次删除一个数，但是删除的数的奇偶性要与上一次相反。求删除后剩下的数的最小和

题解：两个队列，先删元素多的

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    priority_queue<int>odd;
    priority_queue<int>even;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        if(a[i]%2){
            odd.push(a[i]);
        }else{
            even.push(a[i]);
        }
        sum+=a[i];
    }
    if(odd.empty()){
        cout<<sum-even.top()<<endl;
        return 0;
    }
    if(even.empty()){
        cout<<sum-odd.top()<<endl;
        return 0;
    }
    if(even.size()>odd.size()){
        flag=0;
    }else{
        flag=1;
    }
    while(1){
        if(flag&&!odd.empty()){
            sum-=odd.top();
            odd.pop();
            flag=!flag;
        }else if(!flag&&!even.empty()){
            sum-=even.top();
            even.pop();
            flag=!flag;
        }else{
            break;
        }

    }
    cout<<sum<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}