POJ 1850 Code/POJ 1496 Word Index（组合数学-字母串序号）_a - 1 b - 2 ... z - 26 ab - 27 ... az - 51 bc

本文介绍了一种基于字母顺序和长度的特殊编码系统，并提供了一个算法实现，用于确定给定单词在该系统中的编码值。文章包括了问题描述、解题思路及C++代码实现。

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Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9645		Accepted: 4617

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

Source

Romania OI 2002

题目意思：

有一个使用26个小写字母a~z的编码系统的工作原理如下：
•单词按其长度的递增顺序排列。
•具有相同长度的词按词典顺序排列（来自字典的顺序）。
•我们通过编号将这些词编码，从a开始，如下：
a-1
b-2
... ...
z-26
ab-27
... ...
az-51
bc-52
... ...
vwxyz - 83681
... ...

如果可以根据该编码系统编码，则输出给定字及其代码，否则输出0。

解题思路：

规律是：n位字符串的所有情况是C(26,n)。

对于给出的n位字符串，先利用规律求出其n-1位的编码，然后遍历字符串每一位累加计算编码。

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 30
int vis[MAXN][MAXN];//用于记忆化搜索，vis[i][j]表示组合数C(i,j)的值
string s;//输入串
int len;//串的长度
int c(int x,int y)//计算组合数C(x,y)
{
    int s=1,i,j;
    for (i=y+1, j=1; i<=x; i++,j++)
        s=s*i/j;
    return s;
}
void solve()
{
    memset(vis,-1,sizeof(vis));
    int ans=0;
    for(int i=1; i<=len-1; ++i)//计算len-1位的大小
    {
        if(vis[26][i]!=-1)//已经计算过
            ans+=vis[26][i];
        else//未计算
        {
            vis[26][i]=c(26,i);
            ans+=vis[26][i];
        }
    }
    for(int i=0; i<len; ++i)//计算len位中当前串的大小
    {
        char t;
        if(i==0) t='a';//首字符从a开始计算
        else t=s[i-1]+1;//后面的字符从前一个字符的后一位开始计算
        while(t<s[i])//一直计算到当前字符
        {
            if(vis['z'-t][len-1-i]!=-1)
                ans+=vis['z'-t][len-1-i];
            else
            {
                vis['z'-t][len-1-i]=c('z'-t,len-1-i);
                ans+=vis['z'-t][len-1-i];
            }
            ++t;//后一个字母
        }
    }
    cout<<++ans<<endl;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(cin>>s)
    {
        bool flag=false;//是否升序
        len=s.size();
        for(int i=1; i<len; ++i)
            if(s[i-1]>s[i])
            {
                flag=true;//存在降序
                break;
            }
        if(flag) cout<<"0"<<endl;//字符串非法
        else solve();//字符串合法
    }
    return 0;
}