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 3

输出

4
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+7;
const LL MOD=20010905;
const double PI = acos(-1);
const double eps =1e-8;
#define Bug cout<<"---------------------"<<endl
const int maxn=1e5+10;
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=0;i<=str.size()-k;i++)
        {
            string now=str.substr(i,k);
            mp[now]++;
            if(mp[now]==3) st.insert(now);
        }
    }

    cout<<st.size()<<endl;
    for(set<string>::iterator it=st.begin();it!=st.end();it++)
        cout<<*it<<endl;
    
    return 0;
}

 

 

 

 

 

 

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