Codeforces Round #550 (Div. 3) E. Median Stringl

You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.

Let’s consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s=“az” and t=“bf” the list will be [“az”, “ba”, “bb”, “bc”, “bd”, “be”, “bf”].

Your task is to print the median (the middle element) of this list. For the example above this will be “bc”.

It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.

Input
The first line of the input contains one integer k (1≤k≤2*10^5) — the length of strings.

The second line of the input contains one string s consisting of exactly k lowercase Latin letters.

The third line of the input contains one string t consisting of exactly k lowercase Latin letters.

It is guaranteed that s is lexicographically less than t.

It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.

Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.

SampleInput 1
2
az
bf
SampleOutput 1
bc
SampleInput 2
5
afogk
asdji
SampleOutput 2
alvuw
SampleInput 3
6
nijfvj
tvqhwp
SampleOutput 3
qoztvz

题意就是输入一个长度n 然后输入两个长度为n的字符串 一个是s 一个是t s的字典序一定小于等于t
然后让求中间的那个串
其实这题就是模拟一下26进制 两个字符串 就是两个数 我们要求中间串 也就是求这两个数的和的二分之一 那么我们只要模拟一个26进制的加法就可以了

#include<bits/stdc++.h>
using namespace std;
int k;
int q[200005],w[200005];
char s[200005],t[200005];
int main(){

   cin>>k>>s>>t;
   for(int i=0;i<k;i++)
    q[k-1-i]=s[i]-'a',w[k-1-i]=t[i]-'a';

   for(int i=0;i<k;i++)
   {
       q[i]+=w[i];
       if(q[i]>=26)
       {
           q[i+1]+=1;
           q[i]%=26;
       }
   }
   if(q[k]) q[k-1]+=26;
   for(int i=k-1;i>=0;i--)
   {
       if(q[i]&1)
       {
           q[i-1]+=26;
       }
       q[i]/=2;
       printf("%c",'a'+q[i]);
   }
  return 0;
}
``