poj Code(组合数)

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9918		Accepted: 4749

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

 

/*
组合数
题意是查一个字串的字典序排名
先求出长度比它小的，显然是ΣC 26 i(i<len)
然后求出长度等于它却比它小的。具体看代码。 
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 27

using namespace std;
int c[N][N],ans;
char str[N];

inline void combination()
{
    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-1]+c[i-1][j];  
    c[0][0]=0;  
    return;  
}

int main()
{
    combination();
    while(cin>>str)
    {
        ans=0;
        int len=strlen(str);
        for(int i=1;i<len;i++)
          if(str[i]<=str[i-1])
            {
                cout<<"0"<<endl;
                return 0;
            }
        for(int i=1;i<len;i++) ans+=c[26][i];//长度小于它的所有方案 
        for(int i=0;i<len;i++)
        {
            char ch=(!i)?'a':str[i-1]+1;//比上一个大
            while(ch<str[i])//比当前这个小 
            {
                ans+=c['z'-ch][len-i-1];//长度等于它且排在它前面的所有方案 
                ch++;
            }
        }
        cout<<++ans<<endl;
    }
    return 0;
}