[7-10] PTA string penultimate N

Title repeat

Given a string arithmetic entirely composed of lowercase letters increasing sequence, each string in the sequence is fixed to L, starting from the L a, in increments of 1. For example, when L is 3, the sequence {aaa, aab, aac, ..., aaz, aba, abb, ..., abz, ..., zzz}. The penultimate 27 string sequence is zyz. For any given L, this problem requires you to give the corresponding N penultimate sequence string.

Input formats:

In the given input row two positive integers L (2 ≤ L ≤ 6) and N (≦ 10
. 5
).

Output formats:

In the row corresponding to the output sequences of the penultimate N string. Topic ensure that this string exists.

Sample input:

3 7417

Sample output:

pat

answer

In fact, this is a 26-ary problem.

• Taking into account only from L 2-6 so he spent a half-day strength if else want to judge them, and found that this can be cumbersome and error-prone, in theory, this is the simplest method possible.

• We might as well sit down and think about it, a lot of trouble dealing with letters, numbers so we can deal with that for a deal of ascll code.

• First, L-digit determines the overall likely scenario is 26 ^L kinds of topics make a reciprocal N-th, so the number should be the first positive 26 ^L - N number of time being credited x

• L = the first digit (possibly consisting of the remaining number of bits) x / + 'a', to know here is that char ( 'a' + 1) = 'b'

• In order to not interfere with the foregoing, x = (remaining bits may be composed of) x modulo of
  the second bit = x / L number of bits (number of bits consisting of the remaining possible) + 'a'

…and many more

100 ++ AND

#include <iostream>

using namespace std;

int main()
{
    char str[15];

    int L,N;
    cin>>L>>N;
    int sum=1;
    //根据位数决定有多少种可能
    for(int i=0;i<L;i++)
    {
        sum*=26;
    }
    //倒数第N个,所以应该找的是第sum-N个
    sum-=N;

    //对于字符数组，从第一个开始访问，
    for(int i=L-1,j=0;i>=0;i--,j++)
    {
        int tmpsum=1;
        for(int k=0;k<i;k++)
        {
            tmpsum*=26;
        }

        //sum/剩下位数的可能 就是当前位应该填多少
        str[j]=sum/tmpsum+'a';
        //更新sum，以消除前面位数的影响
        sum=sum%tmpsum;

    }
    for(int i=0;i<L;i++)
    {
        cout<<str[i];
    }


    return 0;
}