Ahahahahahahahaha(思维+奇偶性)

https://codeforces.com/contest/1407/problem/A

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题意：01串使得奇数位的和与偶数位的和相等，规定了输入的长度为偶数。

思路：考虑1的个数多还是0的个数多，取个数多的那一个。如果个数多的是奇数个，就减1，不然就是偶数个必然能满足条件。然后for一遍输出一定个数的1也是遵守题目的出现顺序。

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e3+100;
typedef long long LL;
LL a[maxn];
int main(void)
{
  cin.tie(0);std::ios::sync_with_stdio(false);
  LL t;cin>>t;
  while(t--){
  	LL n;cin>>n;
  	for(LL i=1;i<=n;i++) cin>>a[i];
  	LL num1=0;LL num0=0;
	for(LL i=1;i<=n;i++){
  		if(a[i]==1) num1++;
  		else if(a[i]==0) num0++;
	}
	if(num1==num0){
		cout<<n/2<<endl;
		LL sum=n/2;
		for(LL i=1;i<=n;i++){
			if(a[i]==0){
				cout<<0<<" ";
				sum--;
			}
			if(sum<=0) break;
		}
		cout<<endl;
	}
	if(num1>num0){
		if(num1&1) num1--;
		cout<<num1<<endl;
		LL sum=num1;
		for(LL i=1;i<=n;i++){
			if(a[i]==1&&sum>0){
				cout<<1<<" ";
				sum--;
			}
			if(sum<=0) break;
		}
		cout<<endl;
	}
	else if(num1<num0){
		if(num0&1) num0--;
	//	debug(num0) 
	   	cout<<num0<<endl;
	   	LL sum=num0;
	 //  	debug(sum);
	   	for(LL i=1;i<=n;i++){
	   		if(a[i]==0&&sum>0){
	   			cout<<0<<" ";
				sum--;	
			}	
			if(sum<=0) break;
		}
		cout<<endl;
	}
  }
return 0;
}