codeforces 667C C. Reberland Linguistics(dp)

题目链接：

C. Reberland Linguistics

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.

For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word — some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction — it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).

Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.

Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.

Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: , where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.

 

Input

 

The only line contains a string s (5 ≤ |s| ≤ 104) consisting of lowercase English letters.

 

Output

 

On the first line print integer k — a number of distinct possible suffixes. On the next k lines print suffixes.

Print suffixes in lexicographical (alphabetical) order.

 

Examples

 

input

abacabaca

output

3
aca
ba
ca

input

abaca

output

0

Note

The first test was analysed in the problem statement.

In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.

 

 

题意：

给一个字符串,去掉一个长度至少为5的前缀,剩下的切成长度为2或者3的串,且相邻的串不相同,问,最后可以切成多少个串且分别是什么;

思路：

dp[i][2]==1表示i,i+1形成的串可以满足要求,dp[i][3]==1同理;再就是转移了;

考虑i位置,长度为2;

(1)如果<i,i+1>与<i+2,i+3>相同,那么只有dp[i+2][3]==1时dp[i][2]=1;

如果<i,i+1>与<i+2,i+3>不同,那么只要dp[i+2][2]==1或者dp[i+2][3]==1时dp[i][2]=1;

(2)如果<i,i+1,i+2>与<i+3,i+4,i+5>相同,只有dp[i+3][2]==1时dp[i][3]=1;

如果不同,那么只要dp[i+3][2]==1或者dp[i+3][3]==1时dp[i][3]=1;

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e4+10;
const LL mod=1e9+7;
const double PI=acos(-1.0);
char str[N];
int dp[N][4];
map<string,int>mp;
vector<string>ve;
int main()
{
    scanf("%s",str);
    int n=strlen(str);
    string x,y="";
    int ans=0;
    for(int i=n-2;i>4;i--)
    {
        if(n-i==2)
        {
            x=y+str[i]+str[i+1];
            mp[x]=1;
            ve.push_back(x);
            dp[i][2]=1;
        }
        else if(n-i==3)
        {
            x=y+str[i]+str[i+1]+str[i+2];
            mp[x]=1;
            ve.push_back(x);
            dp[i][3]=1;
        }
        else if(n-i==4)
        {
            if(str[i]!=str[i+2]||str[i+1]!=str[i+3])
            {
                x=y+str[i]+str[i+1];
                if(!mp[x])
                {
                    mp[x]=1;
                    ve.push_back(x);
                }
                dp[i][2]=1;
            }
            else
            {
                x=y+str[i]+str[i+1];
                dp[i][2]=0;
            }
        }
        else
        {

            if(str[i]==str[i+2]&&str[i+1]==str[i+3])
            {
                if(dp[i+2][3]){
                dp[i][2]=1;
                x=y+str[i]+str[i+1];
                if(!mp[x])
                {
                    mp[x]=1;
                    ve.push_back(x);
                }
                }
            }
            else
            {
                if(dp[i+2][2]||dp[i+2][3])
                {
                    dp[i][2]=1;
                    x=y+str[i]+str[i+1];
                    if(!mp[x])
                    {
                        mp[x]=1;
                        ve.push_back(x);
                    }
                }
            }
            if(str[i]==str[i+3]&&str[i+1]==str[i+4]&&str[i+2]==str[i+5])
            {
                if(dp[i+3][2])
                {
                    dp[i][3]=1;
                    x=y+str[i]+str[i+1]+str[i+2];
                    if(!mp[x])
                    {
                        mp[x]=1;
                        ve.push_back(x);
                    }
                }
            }
            else
            {
                if(dp[i+3][3]||dp[i+3][2])
                {
                    dp[i][3]=1;
                    x=y+str[i]+str[i+1]+str[i+2];
                    if(!mp[x])
                    {
                        mp[x]=1;
                        ve.push_back(x);
                    }
                }
            }
        }
    }
    ans=ve.size();
    sort(ve.begin(),ve.end());
    printf("%d\n",ans);
    for(int i=0;i<ans;i++)
    {
        cout<<ve[i]<<"\n";
    }

}