poj1850——组合数学

poj1850——组合数学

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8492		Accepted: 4020

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

题意：对递增的字符串，输出次序，非递增则输出0
思路：排列组合

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=1000100;
const int INF=(1<<28);

string str;
int code;
int C[30][30];

void creat_C()
{
    memset(C,0,sizeof(C));
    for(int i=0;i<=28;i++){
        for(int j=0;j<=i;j++){
            if(j==0||j==i) C[i][j]=1;
            else C[i][j]=C[i-1][j-1]+C[i-1][j];
        }
    }
}

int main()
{
    creat_C();
    cin>>str;
    for(int i=0;i<str.length()-1;i++){
        if(str[i]>=str[i+1]){
            cout<<0<<endl;
            return 0;
        }
    }
    code=0;
    int len=str.length();
    for(int i=1;i<=len-1;i++){
        code+=C[26][i];
    }
    for(int i=0;i<len;i++){
        char ch=i?str[i-1]+1:'a';//ch至少要比前一个大
        for(;ch<str[i];ch++){
            code+=C['z'-ch][len-i-1];
        }
    }
    cout<<code+1<<endl;
    return 0;
}