Problem Description
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer Ai is written.
He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.
Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
Constraints
3≦N≦105
N is odd.
1≦Ai≦105
Ai is an integer.
Input
The input is given from Standard Input in the following format:
N
A1 A2 A3 … AN
Output
Print the answer.
Example
Sample Input 1
5
1 2 1 3 7
Sample Output 1
3
One optimal solution is to perform the operation once, taking out two cards with 1 and one card with 2. One card with 1 and another with 2 will be eaten, and the remaining card with 1 will be returned to deck. Then, the values written on the remaining cards in the deck will be pairwise distinct: 1, 3 and 7.
Sample Input 2
15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1
Sample Output 2
7
題意： 有一堆紙牌，牌上有對應的數字。每一次可以拿出其中三張，再將最大的和最小的丟掉，把中間的放回。要求最後剩的紙牌上的數字互不相同，問最多能剩幾張牌。
思路： 當有兩張相同的時，就可以選這兩張和其他一種，然後僅留下一張。所以當種類數爲偶數時，就會有一種牌被多選出來扔掉。
代碼：
#include<bits/stdc++.h>
#define cutele main
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1e6+7;
int st[N],n;
void solve(){
int cnt=0;
for(int i=0;i<N;i++) if(st[i]) cnt++;
if(cnt%2) cout<<cnt<<endl;
else cout<<cnt-1<<endl;
}
//思維
int cutele(){
cin>>n;
for(int i=1;i<=n;i++){
int x;
cin>>x;
st[x]++;
}
solve();
return 0;
}