POJ - 1850 Code 组合数

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9600		Accepted: 4592

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

Source

Romania OI 2002

也是找规律嘛

算出它之前的字符有多少个，再加上一

先把长度小于的算出来，C26,k
,k是字符串的长度

然后把每位字符类似成数位

从最高位向下找，找比这位小的字符串总数（不包括等于）

例如求ceg

先找长度为1，2的字符串个数

在从最高位第一个c字符开始，找最高位为a，b的字符个数（a：C26-(‘a'-’a‘+1) , len-i ; b :
C26-(‘b'-’a‘+1),len-i ,其中i代表第几位）

第二位就找比e小的，注意 你已经找了第一位是a或b，假如第二位是a,b,c，第一位肯定是a或b，所以第二位要从d开始，依次向下类推，最后前面的字符串个数加一就是要求的编号

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
typedef long long ll;
using namespace std;
int myc(int n,int r)
{
int sum=1;
for(int i=1;i<=r;i++)
{
sum=sum*(n+1-i)/i;
}
return sum;
}
int mya(int n,int r)
{
int sum=1;
for(int i=0;i<r;i++)
sum=sum*(n-i);
return sum;
}
int ans(char ch[],int len)
{
int ans=0,i,j,t;
for(i=1;i<len;i++)
{
ans+=myc(26,i);
}//
t=0;
// printf("ha%d\n",ans);
for(i=0;i<len;i++)
{
for(j=t;j<ch[i]-'a';j++)
{
ans+=myc(26-j-1,len-i-1);
}
t=ch[i]-'a'+1;
}
return ans;
}
int main()
{
char ch[20],maxx;
int len,i;
scanf("%s",ch);
len=strlen(ch);
maxx=ch[0];
for(i=1;i<len;i++)
{
if(ch[i]<=maxx)
{
printf("0\n");
return 0;
}
else
maxx=ch[i];
}
printf("%d\n",ans(ch,len)+1);
return 0;
}