AtCoder Beginner Contest 171 C

C - One Quadrillion and One Dalmatians

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

To sum it up, the dogs numbered $\mathrm{1,2,\cdots 1,2,\cdots }$ 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 $\mathrm{NN}$?"

Constraints

• $\mathrm{NN}$ is an integer.
• $\mathrm{1\le N\le 10000000000000011\le N\le 1000000000000001}$

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$

Output

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

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Sample Input 1 Copy

Copy

2

Sample Output 1 Copy

Copy

b

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Sample Input 2 Copy

Copy

27

Sample Output 2 Copy

Copy

aa

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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<cstdio>
#include<stack>
using namespace std;
long long n;
void slove()
{
    stack<char>p;
    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('
');
}
int main(void)
{
    
    while(~scanf("%lld",&n))
    {
        slove();
    }
    return 0;
}