CodeForces 554B Ohana Cleans Up

http://codeforces.com/problemset/problem/554/B

B. Ohana Cleans Up

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ohana Matsumae is trying to clean a room, which is divided up into an n by n grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.

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.

Examples

input

4
0101
1000
1111
0101

output

2

input

3
111
111
111

output

3

Note

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.

题意：

扫地只能扫同一列，并且扫帚在 0 扫过后变成 1，在 1 扫过后变成 0。

问最后都为 1 的行数最多是多少。

思路：

当时没怎么理解，再一看给定的是二进制的字符串，直接两个for遍历相同的行数个数就行了。

AC CODE：

#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;
const int MYDD=113;
/*:如果她扫干净的广场,它将成为脏,如果她扫过一个肮脏的广场,它将变得干净。
她想打扫房间的一些列的行数最大化完全干净。
不允许扫描在列的一部分,Ohana只能扫描整个列*/
int main() {
	int n;
	char Map[MYDD][MYDD];
	scanf("%d",&n);
	for(int j=0; j<n; j++) {
		scanf("%s",Map[j]);
	}

	int ans=0,now=0;
	for(int j=0; j<n; j++) {
		now=1;
		for(int k=j+1; k<n; k++) {
			if(strcmp(Map[j],Map[k])==0)	now++;
		}
		ans=max(now,ans);
	}
	printf("%d\n",ans);
	return 0;
}
/*
3
111
101
111
*/