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 + ;
 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) == )
     {
         for (int i=; i<=n; ++i)    cin >> s[i];
         int ans = ;
         for (int i=; i<=n; ++i)
         {
             int m = ;
             for (int j=; j<=n; ++j)
             {
                 if (s[i] == s[j])    m++;
             }
             ans = max (ans, m);
         }

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

     return ;
 }