Educational Codeforces Round 75 (Rated for Div. 2) B题

B. Binary Palindromes

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
A palindrome is a string t which reads the same backward as forward (formally, t[i]=t[|t|+1−i] for all i∈[1,|t|]). Here |t| denotes the length of a string t. For example, the strings 010, 1001 and 0 are palindromes.

You have n binary strings s1,s2,…,sn (each si consists of zeroes and/or ones). You can swap any pair of characters any number of times (possibly, zero). Characters can be either from the same string or from different strings — there are no restrictions.

Formally, in one move you:

choose four integer numbers x,a,y,b such that 1≤x,y≤n and 1≤a≤|sx| and 1≤b≤|sy| (where x and y are string indices and a and b are positions in strings sx and sy respectively),
swap (exchange) the characters sx[a] and sy[b].
What is the maximum number of strings you can make palindromic simultaneously?

Input
The first line contains single integer Q (1≤Q≤50) — the number of test cases.

The first line on each test case contains single integer n (1≤n≤50) — the number of binary strings you have.

Next n lines contains binary strings s1,s2,…,sn — one per line. It’s guaranteed that 1≤|si|≤50 and all strings constist of zeroes and/or ones.

Output
Print Q integers — one per test case. The i-th integer should be the maximum number of palindromic strings you can achieve simultaneously performing zero or more swaps on strings from the i-th test case.

一道傻逼题，马的codeforces用自己的热点连不上，一直交不了题；好不容易上的分一次掉没了；

这道题比较明显的规律就是当s长度为奇数时必然可以回文；而且可以贡献0和1给s长度为偶数的字符串进行换位；那么长度为偶数的字符串想要回文的话必然0和1出现的次数要为偶数，按这个规律模拟就行；

代码：

#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define lson k<<1
#define rson k<<1|1
//ios::sync_with_stdio(false);
using namespace std;
const int N=200100;
const int M=500100;
const LL mod=1e9+7;
int main(){
	ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		string s[100];
		int sum=0;
		for(int i=1;i<=n;i++){
			cin>>s[i];
		}
		int ans=0;
		int c[2];
		memset(c,0,sizeof(c));
		for(int i=1;i<=n;i++){
			int len=s[i].length();
			if(len%2==1){
				sum++;
				for(int j=0;j<len;j++){
					c[s[i][j]-'0']++;
				}
			}
			else{
				int a[2];
				memset(a,0,sizeof(a));
				for(int j=0;j<len;j++){
					a[s[i][j]-'0']++;
				}
				if(a[0]%2){
					ans++;
				}
				else{
					sum++;
				}
			}
		}	
		cout<<sum+(ans/2)*2+min(c[0]+c[1],ans%2)<<endl;
	}
	return 0;
}