poj 1850 Code

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5387		Accepted: 2487

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

#include <iostream>
using namespace std;

__int64 dp[16][32], sum[16], ans;

int main()
{
    int i, j, len;
    char str[16];

    for (i = 0; i < 26; i++)
        dp[0][i] = 1;
    for (i = 1; i < 10; i++) {
        for (j = 25 - i; j >= 0; j--) {
            dp[i][j] += dp[i][j + 1];
            dp[i][j] += dp[i - 1][j + 1];
        }
    }
    for (i = 0; i < 10; i++) 
        for (j = 0; j < 26; j++)
            sum[i] += dp[i][j];

    scanf("%s", str);
    for (len = 0; str[len]; len++)
        ans += sum[len];
    ans -= sum[len - 1];
    j = 0;
    for (i = len - 1; i >= 0; i--) {
        while (j < str[len - 1 - i] - 'a') {
            ans += dp[i][j];
            j++;
        }
        j++;
    }
    for (i = 0; i < len - 1; i++)
        if (str[i] >= str[i + 1])
            ans = -1;
    printf("%I64d\n", ans + 1);

    return 0;
}