POJ 1850 Code（排列数组）

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6093		Accepted: 2865

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

 

 

 

 

就是排列组合。。

先算长度小的，再算长度相同的。

打表求出组合数

 

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;

int C[27][27];
void init()
{
    C[0][0]=1;
    C[1][0]=C[1][1]=1;
    for(int i=2;i<27;i++)
    {
        C[i][0]=1;
        for(int j=1;j<i;j++)
           C[i][j]=C[i-1][j]+C[i-1][j-1];
        C[i][i]=1;
    }
}

char str[20];

int main()
{
    init();

    while(scanf("%s",&str)!=EOF)
    {
         bool flag=true;
         int len=strlen(str);
         for(int i=1;i<len;i++)
           if(str[i]<=str[i-1])
           {
               flag=false;
               break;
           }
        if(!flag)
        {
            printf("0\n");
            continue;
        }
        int ans=0;

        int t=len-1;
        while(t>0)
        {
            ans+=C[26][t];
            t--;
        }
        for(int i=0;i<len;i++)
        {
            t=str[i]-'a';
            int t1;
            if(i==0)t1=0;
            else t1=str[i-1]-'a'+1;
            while(t>t1)
            {
                ans+=C[26-t][len-1-i];
                t--;
            }

        }
        printf("%d\n",ans+1);
    }
    return 0;
}