AtCoder Beginner Contest 171 C - One Quadrillion and One Dalmatians

C - One Quadrillion and One Dalmatians

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 300300 points

Problem Statement

10000000000000011000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 11 through 10000000000000011000000000000001, but he gave them new names, as follows:

• the dogs numbered 1,2,⋯,261,2,⋯,26 were respectively given the names a, b, ..., z;
• the dogs numbered 27,28,29,⋯,701,70227,28,29,⋯,701,702 were respectively given the names aa, ab, ac, ..., zy, zz;
• the dogs numbered 703,704,705,⋯,18277,18278703,704,705,⋯,18277,18278 were respectively given the names aaa, aab, aac, ..., zzy, zzz;
• the dogs numbered 18279,18280,18281,⋯,475253,47525418279,18280,18281,⋯,475253,475254 were respectively given the names aaaa, aaab, aaac, ..., zzzy, zzzz;
• the dogs numbered 475255,475256,⋯475255,475256,⋯ were respectively given the names aaaaa, aaaab, ...;
• and so on.

To sum it up, the dogs numbered 1,2,⋯1,2,⋯ 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 NN?"

Constraints

• NN is an integer.
• 1≤N≤10000000000000011≤N≤1000000000000001

Input

Input is given from Standard Input in the following format:

NN

Output

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

Sample Input 1 Copy

Copy

2

Sample Output 1 Copy

Copy

b

Sample Input 2 Copy

Copy

27

Sample Output 2 Copy

Copy

aa

Sample Input 3 Copy

Copy

123456789

Sample Output 3 Copy

Copy

jjddja题意：给你一个数字，然后用字母表示题解：思路很简单，其实就是把数字转换为26进制，不过是用字母表示的我们通过n对26求余能得到每一个位的大小,然后把余数转化为相应的字符压入栈，最后再对n/26，继续一直到n=0，当然这里有个例外，就是当n是26的倍数的时候需要特判一下，当余数为0的时候将'z'压入栈并且n需要-1.代码如下：

#include
#include
using namespace std;
long long n;
void slove()
{
    stackp;
    while(n)
    {
        int t=n%26;
        
        if(n)
        {
        if(t==0)//特判是否为整除
        p.push('z');
        else
        p.push((char)('a'+t-1));
        }
        n/=26;
        if(!t)//特判是否为整除
        n--;
    }
    while(!p.empty())
    putchar(p.top()),p.pop();
    putchar('\n');
}
int main(void)
{
    
    while(~scanf("%lld",&n))
    {
        slove();
    }
    return 0;
}