ACM POJ

Code

Time Limit: 1000MS

Memory Limit: 30000KB

64bit IO Format: %I64d & %I64u

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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

Source

Romania OI 2002

 

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int c[27][27];
void table()
{
    int i,j;
    for(i=0; i<=26; i++)
    {
        for(j=0; j<=i; j++)
        {
            if(!j||i==j)
                c[i][j]=1;
            else
                c[i][j]=c[i-1][j-1]+c[i-1][j];
        }
    }
    return;
}
int main()
{
    table();

    char s[17];

    int i,j;

    scanf("%s",s);

    int len=strlen(s);

    for(i=1; i<len; i++)
    {
        if(s[i-1]>=s[i])
        {
            printf("0
");
            return 0;
        }
    }

    int sum=0;

    for(i=1; i<len; i++)
    {
        sum+=c[26][i];
    }

    for(i=0; i<len; i++)
    {
        char ch=(i==0?'a':s[i-1]+1);
        while(ch<=s[i]-1)
        {
            sum+=c['z'-ch][len-1-i];
            ch++;
        }
    }

    printf("%d
",sum+1);

    return 0;
}