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POJ 1496 POJ 1850 组合计数
2015-07-20 17:29:46 来源: 作者: 【 】 浏览:2
Tags:POJ 1496 1850 组合 计数
Code
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 8256 Accepted: 3906

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


/**************************************
     author     :  Grant Yuan
     time        :  2014/10/12 17:06
     algortihm: 组合计数
     source     : POJ 1496 POJ 1850
***************************************/

#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        using namespace std; char a[27]; int num[27][27]; long long ans; bool check() { int l=strlen(a); if(l==1) return 1; for(int i=0;i
       
        =a[i+1]) return 1; return 0; } void Get_num() { int l=strlen(a); for(int i=0;i<=26;i++) for(int j=0;j<=i;j++) { num[i][j]=0; if(i==0||j==0) num[i][j]=1; else num[i][j]=num[i-1][j]+num[i-1][j-1]; } } void Get_sum1() { int l=strlen(a); ans=0; for(int i=0;i
        
         

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