题目描述
有NNN个由小写字母组成的模式串以及一个文本串TTT。每个模式串可能会在文本串中出现多次。你需要找出哪些模式串在文本串TTT中出现的次数最多。
输入输出格式
输入格式：输入含多组数据。
每组数据的第一行为一个正整数NNN，表示共有NNN个模式串，1≤N≤1501 leq N leq 1501≤N≤150。
接下去NNN行，每行一个长度小于等于707070的模式串。下一行是一个长度小于等于10610^6106的文本串TTT。
输入结束标志为N=0N=0N=0。
输出格式：对于每组数据，第一行输出模式串最多出现的次数，接下去若干行每行输出一个出现次数最多的模式串，按输入顺序排列。
输入输出样例
输入样例#1：
复制
2 aba bab ababababac 6 beta alpha haha delta dede tata dedeltalphahahahototatalpha 0
输出样例#1： 复制
4 aba 2 alpha haha
模板；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
string mode[maxn];
struct tree {
int fail;
int vis[30];
int num;
}ac[maxn];
int ans[maxn];
int n, siz;
void ins(string s, int v) {
int now = 0;
for (int i = 0; i < s.length(); i++) {
int o = s[i] - 'a';
if (!ac[now].vis[o])ac[now].vis[o] = ++siz;
now = ac[now].vis[o];
}
ac[now].num = v;
}
void getfail() {
int now = 0;
queue<int>q;
for (int i = 0; i < 26; i++) {
if (ac[0].vis[i]) {
q.push(ac[0].vis[i]);
ac[ac[0].vis[i]].fail = 0;
}
}
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
if (ac[u].vis[i]) {
ac[ac[u].vis[i]].fail = ac[ac[u].fail].vis[i];
q.push(ac[u].vis[i]);
}
else {
ac[u].vis[i] = ac[ac[u].fail].vis[i];
}
}
}
}
void query(string s) {
int now = 0;
for (int i = 0; i < s.length(); i++) {
now = ac[now].vis[s[i] - 'a'];
for (int j = now; j; j = ac[j].fail)ans[ac[j].num]++;
}
}
int main() {
ios::sync_with_stdio(0);
while (cin >> n && n) {
ms(ac); siz = 0; ms(ans);
for (int i = 1; i <= n; i++) {
cin >> mode[i];
ins(mode[i], i);
}
getfail();
string k; cin >> k;
query(k);
int tmp = 0;
for (int i = 1; i <= n; i++) {
tmp = max(tmp, ans[i]);
}
cout << tmp << endl;
for (int i = 1; i <= n; i++) {
if (ans[i] == tmp) {
cout << mode[i] << endl;
}
}
}
return 0;
}