1. Title Description
It's a very simple question , Given $n$ A large integer , Find the maximum number of occurrences .

2. The basic idea
This problem can be done with hash ,ELFHash Wait for hash to pass .

3. Code

 /* 1800 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <bitset>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000") #define sti set<int>
#define stpii set<pair<int, int> >
#define mpii map<int,int>
#define vi vector<int>
#define pii pair<int,int>
#define vpii vector<pair<int,int> >
#define rep(i, a, n) for (int i=a;i<n;++i)
#define per(i, a, n) for (int i=n-1;i>=a;--i)
#define clr clear
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1 const int maxl = ;
const int MOD = ;
int H[MOD], cnt[MOD];
char s[maxl];
int n, ans; inline int ELFHash(char *s) {
unsigned int h = ;
unsigned int g;
int i = ; while (s[i]) {
h = (h<<) + s[i];
g = h & 0xf0000000L;
if (g) {
h ^= (g >> );
h &= ~g;
}
++i;
} return h & 0x7fffffff;
} inline void Hash(char *s) {
int i = ; while (s[i]=='') ++i;
int k = ELFHash(s+i);
int t = k % MOD; while (H[t]!=- && H[t]!=k)
t = (t + ) % MOD; if (H[t] == -) {
cnt[t] = ;
H[t] = k;
} else if (++cnt[t] > ans) {
ans = cnt[t];
}
} int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif while (scanf("%d",&n)!=EOF) {
memset(H, -, sizeof(H));
ans = ;
rep(i, , n) {
scanf("%s", s);
Hash(s);
}
printf("%d\n", ans);
} #ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif return ;
}

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