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.(1≤∣S∣≤10⁵,1≤k≤100)(1 ≤ |S| ≤ 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 

输出

flo
low
owe
wer

利用map储存长为k的字符串，暴力计算即可。

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <string>
 #include <math.h>
 #include <algorithm>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <sstream>
 #include <ctime>
 const int INF=0x3f3f3f3f;
 typedef long long LL;
 const int mod=1e9+;
 const LL MOD=;
 const double PI = acos(-);
 const double eps =1e-;
 #define Bug cout<<"---------------------"<<endl
 const int maxn=1e5+;
 using namespace std;

 set<string> st;
 map<string,int> mp;

 int main()
 {
     #ifdef DEBUG
     freopen("sample.txt","r",stdin);
     #endif
 //    ios_base::sync_with_stdio(false);
 //    cin.tie(NULL);

     string str;
     int k;
     cin>>str>>k;

     if(k<=str.size())
     {
         for(int i=;i<=str.size()-k;i++)
         {
             string now=str.substr(i,k);
             mp[now]++;
             if(mp[now]==) st.insert(now);
         }
     }

     cout<<st.size()<<endl;
     for(set<string>::iterator it=st.begin();it!=st.end();it++)
         cout<<*it<<endl;

     return ;
 }

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