Problem:
Score : 300 points

Problem Statement

1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:

the dogs numbered 1,2,\cdots,26 were respectively given the names a, b, ..., z;

the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names aa, ab, ac, ..., zy, zz;

the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names aaa, aab, aac, ..., zzy, zzz;

the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names aaaa, aaab, aaac, ..., zzzy, zzzz;

the dogs numbered 475255,475256,\cdots were respectively given the names aaaaa, aaaab, ...;

and so on.

To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:

a, b, ..., z, aa, ab, ..., az, ba, bb, ..., bz, ..., za, zb, ..., zz, aaa, aab, ..., aaz, aba, abb, ..., abz, ..., zzz, aaaa, ...

Now, Roger asks you:

"What is the name for the dog numbered N?"

Constraints

N is an integer.

1 \leq N \leq 1000000000000001

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer to Roger's question as a string consisting of lowercase English letters.

Sample Input 1

2

Sample Output 1

b

Sample Input 2

27

Sample Output 2

aa

Sample Input 3

123456789

Sample Output 3

jjddja

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>

int main(void){
  int N;
  scanf("%d",&N);
  typedef char string[1000];
  string p[1000000], name;
   
  /*int num=1;
  for(int i1=1;i1<=26;i1++){
    (p[num])[0]=96+i1;
    printf("%s",p[num]);
    num++;
  }*/
  
  int keta=0;
  //桁数の判別方法
  //N-26^(n-1)>0⇒n桁以上
  while(N-pow(26,keta)>=0){
  keta++;
  }
  
  for(int i=keta-1;i>=0;i--){
  int pls= N % (int)pow(26,i+1);
  name[i]=pls+96;
  printf("%c",name[i]);
    //だめ
  }
  
  
  //strcatで繋げる？
  //Nの値によってどこまで生成するか決めないとオーバーフロー
  //11回しないと
  //if11回重ねて開始桁数変えるか
  //累乗の数がstring内の桁と一致
  //if1桁：N%26 +96 =(1桁目のchar)
  
  //if2桁:N%26^2 +96 =(2桁目のchar), (N-26^2)%26 +96 =(1桁目のchar)
  
  //if3桁:N%26^3 +96 =(3桁目のcharr), (N-26^3)%26^2 +96 =(2.) ,..
  
  //:
  
  //ifn桁:N%26^n +96 =(n桁目のchar), (N-26^n)%26^(n-1) +96 =(n-1.),..
  
  
  
  return 0;
}