POJ 1850：Code 组合数学

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8710		Accepted: 4141

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

题意是给了一个字符串，问其在给定字典中的位置，字符串本身要判断是否符合字典排序。

真的很讨厌做这种题目，感觉没什么分析，不会的话就怎么想都是不会了。但这个题目还算比较简单，找规律，要利用c[i][j]=c[i-1][j-1]+c[i-1][j];预处理出来比这个字符串长度少的所有数量，再从起始字符开始逐个找之前对应长度的字符串，数量都加起来就是答案。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int c[33][33],len,i,sum;
char temp[50];

void init()
{
int i,j;
memset(c,0,sizeof(c));

for(i=0;i<=32;i++)
{
for(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];
}
}
}

bool pend(string temp)
{
for(i=1;i<len;i++)
{
if(temp[i]<temp[i-1])
return false;
}
return true;
}

void dfs(int step)
{
if(step==len)
return;

}

int main()
{
while(scanf("%s",temp)!=EOF)
{
init();
sum=0;
len=strlen(temp);

if(pend(temp))
{
for(i=1;i<len;i++)
{
sum+=c[26][i];
}
for(i=0;i<len;i++)
{
char temp_c;
if(i!=len-1)
{
if(i==0)
temp_c='a';
else
temp_c=temp[i-1]+1;
for(;temp_c<temp[i];temp_c++)
{
sum += c[26-(temp_c-'a'+1)][len-(i+1)];
}
}
else
{
if(len==1)
sum+=temp[i]-'a'+1;
else
sum += temp[i]-temp[i-1];
}
}
cout<<sum<<endl;
}
else
{
cout<<0<<endl;
}
}
return 0;
}