POJ1850——Code(组合数学)

Code

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

题目大意：

　　　　判断一个字符串在字典中的位置。

　　　　字符串必须保证为升序字符串才能合法。

解题思路：

　　　　组合数学问题。

　　　　首先通过dp打印出来com[][]杨辉三角形。

　　　　假设字符串的长度为len。

　　　　详情见备注。

Code：

/*************************************************************************
    > File Name: poj1850.cpp
    > Author: Enumz
    > Mail: 369372123@qq.com
    > Created Time: 2014年10月28日 星期二 01时32分11秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<list>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<bitset>
#include<climits>
#define MAXN 100000
using namespace std;
long long com[30][30];
void init()   /*打印杨辉三角*/
{
    for (int i=0; i<=26; i++)
        for (int j=0; j<=i; j++)
            if (!j||i==j)
                com[i][j]=1;
            else
                com[i][j]=com[i-1][j-1]+com[i-1][j];
    com[0][0]=0;
}
int main()
{
    init();
    char num[20];
    scanf("%s",num);
    int len=strlen(num),k;
    for (k=1;k<=len-1;k++)  /*判断是否为递增字符串*/
        if (num[k]<=num[k-1]) break;
    if (k!=len)
    {
        cout<<0<<endl;
        return 0;
    }
    long long ans=0;
    /*任意长度小于等于len的数都在其前面*/
    for (int i=1; i<=len-1; i++)
        ans+=com[26][i];
    /*计算最高位为[a,num[0])的个数*/
    for (char i='a'; i<num[0]; i++)
        ans+=com['z'-i][len-1];
    /*计算前j位相同的个数*/
    for (int j=1;j<=len-1;j++)
        /*第j位可能的数字为[num[j-1]+1,num[j]) */
        for (char i=num[j-1]+1;i<num[j];i++)
            ans+=com['z'-i][len-1-j];
    cout<<ans+1<<endl;
    return 0;
}