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;
}

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