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[30][30];

void init()
{
    memset(c,0,sizeof(c));
    for(int i = 0; i <= 26; i++)
    {
        c[i][0] = 1;
        c[i][i] = 1;
    }

    for(int i = 2; i <= 26; i++)
    {
        for(int j = 1; j < i; j++)
        {
            c[i][j] = c[i-1][j-1] + c[i-1][j];
        }
    }
}


int main()
{
    init();
    int i,j,sum;
    char s[12];
    scanf("%s",s);
    int len = strlen(s);

    int flag = 1;
    for(i = 0; i < len; i++)
    {
        for(j = i+1; j < len; j++)
        {
            if(s[i] >= s[j])
            {
               flag = 0;
                break;
            }
        }
        if(flag == 0)
            break;
    }
    if(flag == 0)
        printf("0\n");
    else
    {
        sum = 0;
        for(i = 1; i <= len-1; i++)
            sum += c[26][i];

        for(j = 0; j <= s[0]-'a'-1; j++)
            sum += c[25-j][len-1];

        for(i = 1; i < len; i++)
        {
            for(j = s[i-1]-'a'+1; j <= s[i]-'a'-1; j++)
            {
                sum += c[25-j][len-1-i];
            }
        }

        printf("%d\n",sum+1);
    }

    return 0;
}