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;

}
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