poj 1850 Code

poj 1850 Code

Code

Time Limit: 1000MS

Memory Limit: 30000KB

64bit IO Format: %I64d & %I64u

[Submit]
[Go
Back] [Status]

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

[Submit]
[Go
Back] [Status]

/*

//poj1850

题意：

输出某个str字符串在字典中的位置，由于字典是从a=1开始的，因此str的位置值就是在str前面所有字符串的个数+1

规定输入的字符串必须是升序排列。不降序列是非法字符串

解答太详细了
http://hi.baidu.com/you289065406/blog/item/d0900ff85e4b0b71024f563a.html#0
折服了

膜拜

竟然数学使此题变得如此简答

组合数学题，不知道为什么会被归类到递推数学，可能是因为杨辉三角和组合数之间的关系。。

*/

#include <iostream>

#include <string>

using namespace std;

char str[11];

int ta[27][27]={0};

void get_table()//利用了杨慧三角形

{

    int i,j;

    for(i=0;i<=26;i++)

    {

        for(j=0;j<=i;j++)

        {

            if(!j||i==j)ta[i][j]=1;

            else ta[i][j]=ta[i-1][j-1]+ta[i-1][j];

        }

    }

}

int main()

{

    int len,i,j,res;

    get_table();

    char ch;

    while(cin>>str)

    {

        len=strlen(str);

        for(i=1;i<len;i++)

        {

            if(str[i]<=str[i-1])

            {

                cout<<0<<endl;return 0;

            }

        }

        res=0;

        for(i=1;i<len;i++)//求nC1到nClen-1的和

        {

            res+=ta[26][i];

        }

        for(i=0;i<len;i++)

        {

            ch=(!i)?'a':str[i-1]+1;//要比前面的字符大1

            while(ch<=str[i]-1)   //ch<=str[i]-1根据升序规则，当前位置的ch最多只能比str这个位置实际上的字符小1

            {

                res+=ta['z'-ch][len-1-i];  //'z'-ch：小于等于ch的字符不允许再被选择，所以当前能够选择的字符总数为'z'-ch

                ch++;                      //len-1-i  ：ch位置后面（不包括ch）剩下的位数，就是从'z'-ch选择len-1-i个字符

            }

        }

        cout<<res+1<<endl;

    }

    return 0;

}