POJ1850——Code(组合数学)

Code

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

题目大意：

　　　　判断一个字符串在字典中的位置。

　　　　字符串必须保证为升序字符串才能合法。

解题思路：

　　　　组合数学问题。

　　　　首先通过dp打印出来com[][]杨辉三角形。

　　　　假设字符串的长度为len。

　　　　详情见备注。

Code：

 /*************************************************************************
     > File Name: poj1850.cpp
     > Author: Enumz
     > Mail: [e-mail]
     > Created Time: 2014年10月28日 星期二 01时32分11秒
  ************************************************************************/

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<string>
 #include<cstring>
 #include<list>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #include<algorithm>
 #include<cmath>
 #include<bitset>
 #include<climits>
 #define MAXN 100000
 using namespace std;
 long long com[][];
 void init()   /*打印杨辉三角*/
 {
     for (int i=; i<=; i++)
         for (int j=; j<=i; j++)
             if (!j||i==j)
                 com[i][j]=;
             else
                 com[i][j]=com[i-][j-]+com[i-][j];
     com[][]=;
 }
 int main()
 {
     init();
     char num[];
     scanf("%s",num);
     int len=strlen(num),k;
     for (k=;k<=len-;k++)  /*判断是否为递增字符串*/
         if (num[k]<=num[k-]) break;
     if (k!=len)
     {
         cout<<<<endl;
         return ;
     }
     long long ans=;
     /*任意长度小于等于len的数都在其前面*/
     for (int i=; i<=len-; i++)
         ans+=com[][i];
     /*计算最高位为[a,num[0])的个数*/
     for (char i='a'; i<num[]; i++)
         ans+=com['z'-i][len-];
     /*计算前j位相同的个数*/
     for (int j=;j<=len-;j++)
         /*第j位可能的数字为[num[j-1]+1,num[j]) */
         for (char i=num[j-]+;i<num[j];i++)
             ans+=com['z'-i][len--j];
     cout<<ans+<<endl;
     return ;
 }