POJ1850——Code

Code

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9492 Accepted: 4542

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

解：

字典序问题。

详解请看：字典序问题

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;long long c(int n,int m){long long num=1;for(int i=1;i<=m;i++){num=num*(n-i+1);num/=i;}return num;}int main(){char a[11];while(cin>>a){long long sum=1;int len=strlen(a);for(int i=1;i<len;i++)          {              if(a[i]<=a[i-1])              {                  printf("0\n");return 0;              }          }  for(int i=1;i<len;i++)sum+=c(26,i);int start = 0;for(int i=0;i<len;i++){for(int z=start;z<a[i]-'a';z++){sum+=c(26-z-1,len-i-1);}start=a[i]-'a'+1;}printf("%I64d\n",sum);memset(a,0,sizeof(a));}return 0;}