Coderforces Round #

Ohana Cleans Up
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

Description

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.

思路：

如果要让打扫后，有两行同时变干净的话，那么这两行是一样的才行

所以直接判每一行的字符串出现了多少次，然后输出最多次数就好了

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[110][110];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%s",map[i]);
}
int ans=0;
for(int i=1;i<=n;i++)
{
int s=0;
for(int j=1;j<=n;j++)
{
if(strcmp(map[i],map[j])==0)
{
s++;
}
}
ans=max(s,ans);
}
printf("%d\n",ans);
}
return 0;
}

如果要让打扫后，有两行同时变干净的话，那么这两行是一样的才行

所以直接判每一行的字符串出现了多少次，然后输出最多次数就好了