HDU-2100-lovekey

Question Link

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2100

At the beginning of this question, I have been thinking about converting the 26th to the 10th, and converting the 26th to the 26th, and the 200 characters won't work,

I am stuck and won't do it. Later I learned that it is a 200-length string, so I used the idea of adding large numbers.

Reference the idea of http://blog.csdn.net/stm32f98233/article/details/4427447

Solution: it is also a problem of hexadecimal conversion. In fact, it is an inverse process to convert a letter into a 10-digit system and then convert it back to a 26-digit system. There is no need to convert it into a 10-digit system. Just add the 26-digit system. If the sum of the sum is greater than 25, carry the number. Don't be fooled by the question prompts. Because it is a string of 200 characters in length, it is the thought of adding large numbers.

Note: if the data is aaaaaaaaa *, the final result is * and * is any letter. This is a special case.

 

 

Sort the strings in reverse order. For ease of calculation, add zero to the strings with less length, and output them in reverse order.

 

My AC code

# Include <stdio. h>
# Include <string. h>
Char A [210], B [210], C [210], d [210];
Int main (void)
{
Int I, p, q, N, S;
While (scanf ("% S % s", c, d) = 2)
{
P = strlen (C );
Q = strlen (d );
For (I = p-1; I> = 0; I --)
A [p-1-i] = C [I];
A [p] = '\ 0 ';
For (I = q-1; I> = 0; I --)
B [q-1-i] = d [I];
B [Q] = '\ 0 ';
N = P;
If (P <q)
{
For (I = P; I <q; I ++)
A [I] = 'a ';
A [I] = '\ 0 ';
N = Q;
}
If (P> q)
{
For (I = Q; I <p; I ++)
B [I] = 'a ';
A [I] = '\ 0 ';
N = P;
}
S = 0;
For (I = 0; I <n; I ++)
{
A [I] = A [I] + B [I] + S-'A ';
If (A [I]> 'Z ')
{
S = 1;
A [I] = A [I]-26;
}
Else
S = 0;
}
If (S = 1)
{
A [I] = 'B ';
N ++;
}
For (I = n-1; I> = 0; I --)
{
If (A [I] = 'A ')
N --;
Else
Break;
}
For (I = n-1; I> = 0; I --)
Printf ("% C", a [I]);
Printf ("\ n ");
}
Return 0;
}

 

I think more about the questions to make progress.