E. Two Small Strings

Portal ：http://codeforces.com/problemset/problem/1213/E

You are given two strings s and t both of length 2 and both consisting only of characters ‘a’, ‘b’ and ‘c’.
Possible examples of strings s and t: “ab”, “ca”, “bb”.
You have to find a string res consisting of 3n characters, n characters should be ‘a’, n characters should be ‘b’ and n characters should be ‘c’ and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings “ab”, “ac” and “cc” are substrings of the string “abacc”, but the strings “bc”, “aa” and “cb” are not substrings of the string “abacc”.
If there are multiple answers, you can print any of them.

Input

The first line of the input contains one integer n (1≤n≤105) — the number of characters ‘a’, ‘b’ and ‘c’ in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters ‘a’, ‘b’ and ‘c’.
The third line of the input contains one string t of length 2 consisting of characters ‘a’, ‘b’ and ‘c’.

Output

If it is impossible to find the suitable string, print “NO” on the first line.
Otherwise print “YES” on the first line and string res on the second line. res should consist of 3n characters, n characters should be ‘a’, n characters should be ‘b’ and n characters should be ‘c’ and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.

Examples

input
2
ab
bc
output
YES
acbbac

input
3
aa
bc
output
YES
cacbacbab

input
1
cb
ac
output
YES
abc

The question

   Ask string str Whether there is ,str from n individual ’a’,n individual ’b’,n individual ’c’ form , And the input length is n String st1,st2 No str The string of , If exist , Output any case .

Ideas

   Just began to think for a long time , No idea , Because of the comparison of dishes , The classification point cannot be found , So I just enumerate it directly .
   Enumerate a certain number of str, And then use string Of find function , If you can't find it , It outputs ,return 0; If you can't find it at last , Just NO.
   If you enumerate , because str By n individual “abc” form , Then we can next_permutation, find “abc” All situations of , then n Double expansion .

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<ctime>
#include<utility>
#include<map>
#define ll long long
#define ld long double
#define ull unsigned long long
using namespace std;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f3f;
const double eps = 1e-6;
const int maxn = 150010;
string abc = "abc";
string st1,st2;
vector<string> st;
int main(void)
{
    
    int n;
    cin>>n>>st1>>st2;
    
    do{
    
        string st3;
        for(int i=0;i<n;i++)
            st3 += abc;
        st.push_back(st3);
        st.push_back(string(n,abc[0])+string(n,abc[1])+string(n,abc[2]));
    }while(next_permutation(abc.begin(),abc.end()));
    
    vector<string>::iterator it = st.begin();
    while(it!=st.end()){
    
        string st4 = *it;
        if(st4.find(st1)==-1&&st4.find(st2)==-1){
    
            printf("YES\n");
            cout<<st4<<endl;
            return 0;
        }
        it++;
    }
    printf("NO\n");

    return 0;
}