题干：
Eugeny has n cards, each of them has exactly one integer written on it. Eugeny wants to exchange some cards with Nikolay so that the number of even integers on his cards would equal the number of odd integers, and that all these numbers would be distinct.
Nikolay has m cards, distinct numbers from 1 to m are written on them, one per card. It means that Nikolay has exactly one card with number 1, exactly one card with number 2 and so on.
A single exchange is a process in which Eugeny gives one card to Nikolay and takes another one from those Nikolay has. Your task is to find the minimum number of card exchanges and determine which cards Eugeny should exchange.
Input
The first line contains two integers n and m (2 ≤ n ≤ 2·105, 1 ≤ m ≤ 109) — the number of cards Eugeny has and the number of cards Nikolay has. It is guaranteed that n is even.
The second line contains a sequence of n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers on Eugeny's cards.
Output
If there is no answer, print -1.
Otherwise, in the first line print the minimum number of exchanges. In the second line print n integers — Eugeny's cards after all the exchanges with Nikolay. The order of cards should coincide with the card's order in the input data. If the i-th card wasn't exchanged then the i-th number should coincide with the number from the input data. Otherwise, it is considered that this card was exchanged, and the i-th number should be equal to the number on the card it was exchanged to.
If there are multiple answers, it is allowed to print any of them.
Examples
Input
6 2 5 6 7 9 4 5
Output
1 5 6 7 9 4 2
Input
8 6 7 7 7 7 8 8 8 8
Output
6 7 2 4 6 8 1 3 5
Input
4 1 4 2 1 10
Output
-1
题目大意：
尤金有 n 张卡片。每张卡片上都有一个整数。 尤金希望与尼古拉交换一些卡片，使得他的偶数卡片上的数量等于奇数卡片的数量，并且所有这些数字不同的。
尼古拉有 m 张卡，上面的数字为 1 到 m ，也就是每种数字的卡各一张。
每次两人可以交换一张卡，问最少交换几次可以满足尤金的要求。
Input
第一行包含两个整数 n 和 m (2 ≤ n ≤ 2·105, 1 ≤ m ≤ 109) — 分别表示两人拥有的卡片数。 保证 n 是偶数。
第二行包含一个由 n 个正整数组成的数列 a1, a2, ..., an (1 ≤ ai ≤ 109) — 表示尤金的卡。
Output
如果没有答案，输出 -1.
否则，在第一行输出最小需要换的卡片数。在第二行输出 n 个整数 — 交换后尤金的卡牌。 卡牌的顺序要和读入时一样，如果有多种方案，输出任意一种。
解题报告：
先预处理出来在<=m的前提下有多少奇数和偶数可以使用，然后先保证能不变的先不变（打个ok标记），剩下的贪心填。注意一下贪心的顺序，并且时刻注意填的数要<=m这个限制就好。
注意一定要先保证能不变的就不变，不然的话你不确定后面已经有多少奇数偶数了，那么就不能确定这一位优先填奇数还是偶数，所有有可能操作次数不是最少的。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 6e5 + 5;
ll n,m,ji,ou,resji,resou,needji,needou;
ll a[MAX];
set<ll> sji,sou,ans;
bool vis[MAX],ok[MAX];
vector<int> o,j;
int main()
{cin>>n>>m;if(m&1) resou=m/2,resji=resou+1;else resou=resji=m/2;for(int i = 1; i<=n; i++) { scanf("%lld",a+i);if(a[i] < MAX) vis[a[i]]=1;if(a[i]&1) {ji++;if(a[i] <= m && sji.find(a[i]) == sji.end()) resji--;sji.insert(a[i]);}else {ou++;if(a[i] <= m && sou.find(a[i]) == sou.end()) resou--;sou.insert(a[i]);}}ll bij = n/2-sji.size();//必须要去掉的奇数，准确的说，是还需要换上的奇数，还缺少的奇数ll bio = n/2-sou.size();//必须要去掉的偶数if(resji < bij || resou < bio) return 0,puts("-1");ll tmp = max(bij,0LL)+max(bio,0LL);printf("%lld\n",tmp);for(int i = 2; i<=min(m,1LL*MAX-1); i+=2) {if(vis[i]) continue;o.pb(i);}for(int i = 1; i<=min(m,1LL*MAX-1); i+=2) {if(vis[i]) continue;j.pb(i);}ll curji=0,curou=0;for(int i = 1; i<=n; i++) {if(ans.count(a[i])) continue;ans.insert(a[i]);if((a[i]&1) && curji < n/2) curji++,ok[i]=1;else if((a[i]&1)==0 && curou<n/2)curou++,ok[i]=1;}for(int i = 1; i<=n; i++) {if(ok[i]) {continue;}if(a[i]&1) {//如果是奇数if(ans.count(a[i])) {if(curji < n/2) a[i]=j.back(),j.pop_back(),curji++;else if(curou < n/2) a[i] = o.back(),o.pop_back(),curou++;}else {if(curji < n/2) ans.insert(a[i]),curji++;else a[i] = o.back(),o.pop_back(),curou++;}}else {if(ans.count(a[i])) {if(curji < n/2) a[i]=j.back(),j.pop_back(),curji++;else if(curou < n/2) a[i] = o.back(),o.pop_back(),curou++;}else {if(curou < n/2) ans.insert(a[i]),curou++;else a[i] = j.back(),j.pop_back(),curji++;}}}for(int i = 1; i<=n; i++) printf("%lld ",a[i]);return 0 ;
}
