描述

传送门：poj-1850

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.

Examples

intput
1
bf
output
1
55

思路

对于长度为$N$的字符串应该先求长度为$N-1$的字符串有多少种。很明显可以看出长度为i的字符串就是从26个字母中选择i个字符组合（因为有大小限制，所以不是排列）。所以长度为$N-1$的字符串有：$
sum_{i=1}^{Len-1}C_{26}^{i}$种。

对于剩下的，就从a[0]开始，每次计算这个字母开头的单词个数，遍历完就行了。

代码

#include<iostream>
#include<string.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repd(i,a,n) for(int i=n-1;i>=a;i--)

int C(int n,int r){return r? C(n,r-1)*(n-r+1)/r:1;}

int main()
{
    std::ios::sync_with_stdio(false);
    int sum=0;char a[15];
    while(cin>>a){
        int Len=strlen(a);sum=0;

        rep(i,1,Len)
            if(a[i]<=a[i-1]) sum=-1;

        if(sum==-1) {cout<<0<<endl;continue;}

        rep(i,1,Len) sum+=C(26,i);
        rep(i,0,Len){
            char tem= i? a[i-1]+1:'a';
            while(tem<a[i])
                sum+=C('z'-tem++,Len-i-1);
        }
        cout<<sum+1<<endl;
    }
    return 0;
}