贪心 Codeforces Round #309 (Div. 2) B. Ohana Cleans Up

 

题目传送门

/*
    题意：某几列的数字翻转，使得某些行全为1，求出最多能有几行
    想了好久都没有思路，看了代码才知道不用蠢办法，匹配初始相同的行最多能有几对就好了，不必翻转
*/
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <iostream>
using namespace std;

const int MAXN = 1e2 + 10;
const int INF = 0x3f3f3f3f;
string s[MAXN];

int main(void)        //Codeforces Round #309 (Div. 2) B. Ohana Cleans Up
{
//    freopen ("B.in", "r", stdin);

    int n;
    while (scanf ("%d", &n) == 1)
    {
        for (int i=1; i<=n; ++i)    cin >> s[i];
        int ans = 0;
        for (int i=1; i<=n; ++i)
        {
            int m = 0;
            for (int j=1; j<=n; ++j)
            {
                if (s[i] == s[j])    m++;
            }
            ans = max (ans, m);
        }

        printf ("%d\n", ans);
    }

    return 0;
}