POJ 1850 Code/POJ 1496 Word Index（組合數學-字母串序號）

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. 

The only line contains a word. There are some constraints: 
• The word is maximum 10 letters length 
• The English alphabet has 26 characters. 

The output will contain the code of the given word, or 0 if the word can not be codified.

bf

55

Romania OI 2002

#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];//用於記憶化搜尋，vis[i][j]表示組合數C(i,j)的值
string s;//輸入串
int len;//串的長度
int c(int x,int y)//計算組合數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)//計算len-1位的大小
    {
        if(vis[26][i]!=-1)//已經計算過
            ans+=vis[26][i];
        else//未計算
        {
            vis[26][i]=c(26,i);
            ans+=vis[26][i];
        }
    }
    for(int i=0; i<len; ++i)//計算len位中當前串的大小
    {
        char t;
        if(i==0) t='a';//首字元從a開始計算
        else t=s[i-1]+1;//後面的字元從前一個字元的後一位開始計算
        while(t<s[i])//一直計算到當前字元
        {
            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;//後一個字母
        }
    }
    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;//字串非法
        else solve();//字串合法
    }
    return 0;
}