CodeForces - 554B Ohana Cleans Up （模拟）水

CodeForces - 554B

Ohana Cleans Up

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Description

Ohana Matsumae is trying to clean a room, which is divided up into an n by n

Return the maximum number of rows that she can make completely clean.

Input

The first line of input will be a single integer n (1 ≤ n ≤ 100).

The next n lines will describe the state of the room. The i-th line will contain a binary string with n characters denoting the state of the i-th row of the room. The j-th character on this line is '1' if the j-th square in the i-th row is clean, and '0' if it is dirty.

Output

The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.

Sample Input

Input

40101100011110101

Output

2

Input

3111111111

Output

3

Hint

In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.

In the second sample, everything is already clean, so Ohana doesn't need to do anything.

Source

Codeforces Round #309 (Div. 2)

//题意：

每次可以将一列的数取反（0变1,1变0），问操作完后最多有几行是全0的？

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char map[200][200];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int maxx=0;
		memset(map,0,sizeof(map));
		for(int i=0;i<n;i++)
		scanf("%s",map[i]);
		for(int i=0;i<n;i++)
		{
			int cnt=1;
			for(int j=i+1;j<n;j++)
			{
				if(strcmp(map[i],map[j])==0)
				cnt++;
			}
			maxx=max(maxx,cnt);
		}
		printf("%d\n",maxx);
	}
	return 0;
}