B. Codeforces Subsequences

B. Codeforces S

B. Codeforces Subsequences

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least kk subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.

Formally, a codeforces subsequence of a string ss is a subset of ten characters of ss that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.

Help Karl find any shortest string that contains at least kk codeforces subsequences.

Input

The only line contains a single integer kk (1≤k≤1016)1≤k≤1016).

Output

Print a shortest string of lowercase English letters that contains at least kk codeforces subsequences. If there are several such strings, print any of them.

Examples

input

Copy

1

output

Copy

codeforces

input

Copy

3

output

Copy

codeforcesss

ubsequences

=========================================================================

首先不在字母本位置上进行增加是不会获得最大效果的，最终的答案就是各个字母个数的乘积。

原来是

1 1 1 1 1 1 1 1 1 1

加1的话，放在那个位置都行

2 1 1 1 1 1 1 1 1 1

再加1的话，可以加到2，也可以加到1，但是发现加到1带来的增效更高

2 2 1 1 1  1 1 1 1 1

再增加1也是同理，即尽量让10个字母数量相同能够使得最终个数最多

那么我们就一直这样暴力循环，每个最多不到100次，10个也就不到1000次循环就能超过k

#include
#include
#include
# include
 
#include
#define  mo 998244353;
using namespace std;
 
typedef long long int ll;
 
ll cnt[20];
 
char ans(int i)
{
    string s=" codeforces";
 
    return s[i];
}
int main()
{
 
  ll n;
 
  cin>>n;
 
  for(int i=1;i<=10;i++)
  {
      cnt[i]=1;
  }
 
  ll now=1;
 
  int t=1;
  while(now
  {
 
      now=now/cnt[t]*(cnt[t]+1);
 
      cnt[t]++;
 
      t++;
 
      if(t==11)
        t=1;
 
 
  }
 
  for(int i=1;i<=10;i++)
  {
      for(int j=1;j<=cnt[i];j++)
      {
          cout<<ans(i);
      }
  }
 
 
    return 0;
}