https://codeforces.com/contest/1144/problem/E 

题意：给定两个字符串，定义一个序列是第一个字符串按字典序到第二个字符串的所有字符串。求这个序列的中间的字符串是什么（保证这个序列个数肯定为奇数)

题解：模拟

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
int b[N];
int c[N];
char str1[N],str2[N],str3[N];
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    cin>>str1>>str2;
    ll pos=0;
    for(int i=0;i<n;i++){
        a[i]=str1[i]-'a';
        b[i]=str2[i]-'a';
        c[i]=a[i]+b[i];
        temp=(c[i]+pos)%2;
        c[i]=(c[i]+pos)/2;
        pos=temp*26;
    }
    for(int i=n-1;i>=0;i--){
        if(c[i]>=26){
            c[i-1]+=c[i]/26;
            c[i]%=26;
        }
    }
    for(int i=0;i<n;i++){
        printf("%c",c[i]+'a');
    }

    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}