hdu Code 组合数学

Problem 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

Sample Output

***************************************************************************************************************************************************************************************************************

组合数学

详细解释

数学推理

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 int c[27][27];
 6 void init()//初始化组合数
 7 {
 8    int i,j;
 9    for(i=0;i<=26;i++)
10    {
11        for(j=0;j<=i;j++)
12        {
13            if(!j||i==j)
14            c[i][j]=1;
15            else
16            c[i][j]=c[i-1][j-1]+c[i-1][j];
17        }
18    }
19    return;
20 }
21 int main()
22 {
23     init();
24     char s[17];
25     int i,j;
26     scanf("%s",s);
27     int len=strlen(s);
28     for(i=1;i<len;i++)
29     {
30          if(s[i-1]>=s[i])//不满足条件
31          {
32              printf("0
");
33              return 0;
34          }
35     }
36     int sum=0;
37     for(i=1;i<len;i++)//把前面不等长的加起来
38     {
39          sum+=c[26][i];
40     }
41    for(i=0;i<len;i++)//这时等长
42    {
43         char ch=(i==0?'a':s[i-1]+1);
44         while(ch<=s[i]-1)//需要枚举相加
45         {
46              sum+=c['z'-ch][len-1-i];
47             ch++;
48         }
49    }
50       printf("%d
",sum+1);
51     return 0;
52 }

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