POJ 1850 Code/POJ 1496 Word Index（组合数学-字母串序号）

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

55

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;
}