B. Ohana Cleans Up
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/554/problem/B
Description
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
4
0101
1000
1111
0101
Sample Output
2
HINT
题意
有一个n*n的房间，干净为1，脏为0
然后每次可以打扫一列，使得干净的变成肮脏，肮脏的变干净
问你最多使多少行全部变干净
题解：
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
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** int n;
string s[];
map<string,int>H;
int main()
{
n=read();
int ans=;
for(int i=;i<n;i++)
{
cin>>s[i];
H[s[i]]++;
ans=max(ans,H[s[i]]);
}
cout<<ans<<endl; }