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.

The only line contains a word. There are some constraints: 
• The word is maximum 10 letters length 
• The English alphabet has 26 characters. 

The output will contain the code of the given word, or 0 if the word can not be codified.

bf

55

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

typedef long long LL;

LL C[27][27];

char S[15];

void getC()

{

    memset(C, 0, sizeof(C));

    for(int i = 0; i <= 26; i++)

    {

        for(int j = 0; j <= i; j++)

        {

            if(!j || i == j)

                C[i][j] = 1;

            else

                C[i][j] = C[i-1][j] + C[i-1][j-1];

        }

    }

    C[0][0] = 1;

}

int main()

{

    getC();

    while(~scanf("%s", S))

    {

        LL ans = 0;

        int len = strlen(S);

        //判断升序否

        for(int i = 1; i < len; i++)

        {

            if(S[i] <= S[i-1])

            {

                printf("0\n");

                return 0;

            }

        }

        //先把长度比S小的数目都统计加起来

        for(int i = 1; i < len; i++)

        {

            ans += C[26][i];

        }

        //统计长度相同的，枚举相加

        for(int i = 0; i < len; i++)

        {

            int ch = i==0?'a':S[i-1]+1; //考虑到S升序

            for(; ch < S[i]; ch++)

            {

                ans += C['z' - ch][len - i - 1];  //从能选的字母中选出len-i+1个进行组合

            }

        }

        ans++;  //加自己

        printf("%lld\n", ans);

    }

    return 0;

}