Code--组合数学

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9320		Accepted: 4454

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

题目链接：http://poj.org/problem?id=1850

并不晓得为什么这道题会归到组合数学中去，可能是因为组合数吧，从网上跟大神学了一招，可以用杨辉三角来预处理组合数，当然，你也可以一个一个的计算。

这个题的数学的符号不好打，我从网上找了个大神的博客，他讲的很清楚，最重要的就是两个式子，组合数的转换，大神的博客上有，我打不出来。。。。。尴尬中。。。。。

大神的博客、

http://blog.csdn.net/lyy289065406/article/details/6648492

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[50][50];
void yanghui()//杨辉三角预处理组合数
{
int i,j;
for(i=0;i<=26;i++)
{
for(j=0;j<=i;j++)
{
if(!j||j==i)
a[i][j]=1;
else
a[i][j]=a[i-1][j]+a[i-1][j-1];
}
}
a[0][0]=0;
return ;
}
int main()
{
char s[100];
yanghui();
int sum=0;
int i;
while(~scanf("%s",s))
{
int len=strlen(s);
for(i=1;i<len;i++)//不符合条件
{
if(s[i-1]>=s[i])
{
break;
}
}
if(i!=len)
{
cout<<"0"<<endl;
continue;
}
for(i=1;i<len;i++)//计算比s字符串长度小的值的个数
{
sum+=a[26][i];//a[26][i]表示 长度为i的字符串的个数
}
char ch;
for(i=0;i<len;i++)
{
ch=(i==0)?'a':s[i-1]+1;// 根据升序规则，当前位置的ch至少要比s前一位置的字符大1
while(ch<=s[i]-1)//根据升序规则，当前位置的ch最多只能比 s这个位置实际上的字符 小1
{
sum+=a[(int)('z'-ch)][len-i-1];// 小于等于ch的字符不允许再被选择，所以当前能够选择的字符总数为'z'-ch
ch++;// ch位置后面（不包括ch）剩下的位数，就是从'z'-ch选择len-1-i个字符
}
}
printf("%d\n",sum+1);//别忘了sum要+1；
}
return 0;
}