| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 7184 | Accepted: 3353 |
Description
Transmitting and memorizing information is a task that requires different coding systems for the best use of the available space. A well known system is that one where a number is associated to a character sequence. It is considered that the words are made only of small characters of the English alphabet a,b,c, ..., z (26 characters). From all these words we consider only those whose letters are in lexigraphical order (each character is smaller than the next character).
The coding system works like this:
• The words are arranged in the increasing order of their length.
• The words with the same length are arranged in lexicographical order (the order from the dictionary).
• We codify these words by their numbering, starting with a, as follows:
a - 1
b - 2
...
z - 26
ab - 27
...
az - 51
bc - 52
...
vwxyz - 83681
...
Specify for a given word if it can be codified according to this coding system. For the affirmative case specify its code.
Input
The only line contains a word. There are some constraints:
• The word is maximum 10 letters length
• The English alphabet has 26 characters.
• The word is maximum 10 letters length
• The English alphabet has 26 characters.
Output
The output will contain the code of the given word, or 0 if the word can not be codified.
Sample Input
bf
Sample Output
55 忽略了输出0的情况，wa了若干次。。。
#include<stdio.h>
#include<string.h>
int c[][]; void init()
{
memset(c,,sizeof(c));
for(int i = ; i <= ; i++)
{
c[i][] = ;
c[i][i] = ;
} for(int i = ; i <= ; i++)
{
for(int j = ; j < i; j++)
{
c[i][j] = c[i-][j-] + c[i-][j];
}
}
} int main()
{
init();
int i,j,sum;
char s[];
scanf("%s",s);
int len = strlen(s); int flag = ;
for(i = ; i < len; i++)
{
for(j = i+; j < len; j++)
{
if(s[i] >= s[j])
{
flag = ;
break;
}
}
if(flag == )
break;
}
if(flag == )
printf("0\n");
else
{
sum = ;
for(i = ; i <= len-; i++)
sum += c[][i]; for(j = ; j <= s[]-'a'-; j++)
sum += c[-j][len-]; for(i = ; i < len; i++)
{
for(j = s[i-]-'a'+; j <= s[i]-'a'-; j++)
{
sum += c[-j][len--i];
}
} printf("%d\n",sum+);
} return ;
}