poj 1850(组合数学)

Code

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 8050 Accepted: 3772

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

参考详解(看了一定会懂得)：http://hi.baidu.com/you289065406/blog/item/d0900ff85e4b0b71024f563a.html#0

AC代码：

#include<iostream>#include<cstring>#include<algorithm>using namespace std;int C[27][27];void init(){    for(int i=0;i<=26;i++)    for(int j=0;j<=i;j++){        if(i==j || j==0)            C[i][j]=1;        else            C[i][j]=C[i-1][j]+C[i-1][j-1];    }}int main(){    init();    char str[12];    while(cin>>str){        int i,j,k;        int sucess=1;        int len=strlen(str);        long long sum=0;        for(i=1;i<len;i++){            if(str[i-1] >= str[i]){                sucess=0;                break;            }        }        if(!sucess){            cout<<'0'<<endl;            continue;        }        for(i=1;i<len;i++)            sum+=C[26][i];        for(i=0;i<len;i++){            char ch;            if(i==0)                ch='a';            else                ch=str[i-1]+1;            while(ch<=str[i]-1){                sum+=C['z'-ch][len-1-i];                ch++;            }        }        sum++;        cout<<sum<<endl;    }    return 0;}