POJ 1850 Code

组合数学。。。。

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 7202		Accepted: 3361

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 

主要是这个公式。。。

  然后就是关键了，长度为2的字符串，根据开头字母不同，就有25种不同情况，编程去处理是很困难的。这里必须要用数学方法去处理。

所以用一个简单的循环就能计算出 比str长度少的所有字符串个数 了，这就是数学的威力，把受限的取法转换为不限制的取法

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; int c[ 30][ 30]; char str[ 30]; void Init() {      for( int i= 0;i< 30;i++)         c [0]=c=1;     for(int i=2;i<30;i++)         for(int j=1;j<i;j++)             c[j]=c[i-1][j-1]+c[i-1][j]; } int main() {     Init();     while(scanf("%s",str)!=EOF)     {         bool flag=true;         int len=strlen(str);         for(int i=0;i<len-1;i++)         {             if(str>=str[i+1])             {                 flag=false;                 break;             }         }         if(!flag) { printf("0\n");continue; }         int sum=0;         ///n-1...1         for(int i=1;i<len;i++)         {             sum+=c[26];         }         ///N         for(int i=0;i<len;i++)         {             char ch=(i==0)?'a':str[i-1]+1;             while(ch<str)             {                 sum+=c['z'-ch][len-1-i];                 ch++;             }         }         printf("%d\n",++sum);     }     return 0; }

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