poj1850Code

Code

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

之前写了几道数位dp的题目 改成了字母的就又不会了

其实思路也很简单，就是分成小于长度的和等于长度的两种情况

小于长度的情况直接排列组合就可以了 等于长度的情况从高位枚举

http://blog.csdn.net/lyy289065406/article/details/6648492 看了这篇博文 感觉很好理解

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>

using namespace std;

int c[30][30];
char str[12];

void Cmn()
{
    c[0][0] = 1;
    for(int i = 1; i < 30; i++){
        c[i][0] = c[i][i] = 1;
        for(int j = 1;j < i; j++){
            c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
        }
    }
}

void solve()
{
    int sum = 0;
    for(int i = 1; i < strlen(str); i++){
        sum += c[26][i];
    }

    for(int i = 0; i < strlen(str); i++){
        char ch = (!i)?'a':str[i - 1] + 1;
        while(ch <= str[i] - 1){
            sum += c['z' - ch][strlen(str) - i - 1];
            ch++;
        }
    }
    sum ++;
    cout<<sum <<endl;
}

int main()
{
    Cmn();
    cin>> str;
    for(int i = 1; i < strlen(str); i++){
        if(str[i] <= str[i - 1]){
            cout<< 0<<endl;
            return 0;
        }
    }
    solve();
	return 0;
}