Codeforces554B:Ohana Cleans Up

B. Ohana Cleans Up

Time Limit: 2000ms

Memory Limit: 262144KB

64-bit integer IO format: %I64d      Java class name: (Any)

Submit Status

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.

 

Sample Input

Input

4
0101
1000
1111
0101

Output

2

Input

3
111
111
111

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.

题目大意：有一个广场   n*n大小，0表示不干净，1表示干净，有一个神奇厉害的大妈，每次扫地只扫一列，扫完之后，干净变成不干净，不干净变成干净（如此神奇）,问最多能使得几行变得干净。

思路：遍历一遍即可，看看最多有几行是相同的。用vi[i]保存遍历过的，防止重复搜索。

代码：

#include <stdio.h>
#include <iostream>
#include <string.h>

using namespace std;

int main(){
    char s[][];
    int vi[]={};
    int max=;
    int num=;
    int n;
    cin>>n;
    for(int i=;i<n;i++)
    cin>>s[i];
    for(int i=;i<n;i++){
        num=;
        if(vi[i]==)
            continue;
        vi[i]=;
        for(int j=;j<n ;j++){
                if(strcmp(s[i],s[j])==){num++;vi[j]=;}
        }
        if(max<num)
            max=num;
    }
    cout<<max;
}