Code (组合数)

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. 

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 ;
 }