The flower(寻找出现m次以上,长度为k的子串)

链接：https://ac.nowcoder.com/acm/contest/3665/B
来源：牛客网

题目描述

Every problem maker has a flower in their heart out of love for ACM. As a famous problem maker, hery also has a flower.

Hery define string T as flower-string if T appears more than twice in string S and |T| = k. Hery wants you to find how many flower-string in S.

输入描述:

The first line contains a string S consist of ’f’, ’l’, ’o’, ’w’, ’e’, ’r’ and an integer k. $10^5, 1 ≤ k ≤ 100)(1≤∣S∣≤105,1≤k≤100)$

输出描述:

Output an integer m in the first line, the number of flower − string in S.
Next m lines, each line contains a flower-string and the lexicographical order of them should be from small to large

示例1

输入

flowerflowerflower 3

输出

4
flo
low
owe
wer

利用

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <string>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <sstream>
13 #include <ctime>
14 const int INF=0x3f3f3f3f;
15 typedef long long LL;
16 const int mod=1e9+7;
17 const LL MOD=20010905;
18 const double PI = acos(-1);
19 const double eps =1e-8;
20 #define Bug cout<<"---------------------"<<endl
21 const int maxn=1e5+10;
22 using namespace std;
23 
24 set<string> st;
25 map<string,int> mp;
26 
27 int main()
28 {
29     #ifdef DEBUG
30     freopen("sample.txt","r",stdin);
31     #endif
32 //    ios_base::sync_with_stdio(false);
33 //    cin.tie(NULL);
34     
35     string str;
36     int k;
37     cin>>str>>k;
38 
39     if(k<=str.size())
40     {
41         for(int i=0;i<=str.size()-k;i++)
42         {
43             string now=str.substr(i,k);
44             mp[now]++;
45             if(mp[now]==3) st.insert(now);
46         }
47     }
48 
49     cout<<st.size()<<endl;
50     for(set<string>::iterator it=st.begin();it!=st.end();it++)
51         cout<<*it<<endl;
52     
53     return 0;
54 }