[ACM] poj 1496 Word Index（组合计数）

Word Index

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4541		Accepted: 2567

Description

Encoding schemes are often used in situations requiring encryption or information storage/transmission economy. Here, we develop a simple encoding scheme that encodes particular types of words with five or fewer (lower case) letters
as integers.

Consider the English alphabet {a,b,c,...,z}. Using this alphabet, a set of valid words are to be formed that are in a strict lexicographic order. In this set of valid words, the successive letters of a word are in a strictly ascending order; that is, later
letters in a valid word are always after previous letters with respect to their positions in the alphabet list {a,b,c,...,z}. For example,

abc aep gwz

are all valid three-letter words, whereas

aab are cat

are not.

For each valid word associate an integer which gives the position of the word in the alphabetized list of words. That is:

a -> 1

b -> 2

.

.

z -> 26

ab -> 27

ac -> 28

.

.

az -> 51

bc -> 52

.

.

vwxyz -> 83681

Your program is to read a series of input lines. Each input line will have a single word on it, that will be from one to five letters long. For each word read, if the word is invalid give the number 0. If the word read is valid, give the word's position index
in the above alphabetical list.

Input

The input consists of a series of single words, one per line. The words are at least one letter long and no more that five letters. Only the lower case alphabetic {a,b,...,z} characters will be used as input. The first letter of
a word will appear as the first character on an input line.

The input will be terminated by end-of-file.
Output

The output is a single integer, greater than or equal to zero (0) and less than or equal 83681. The first digit of an output value should be the first character on a line. There is one line of output for each input line.

Sample Input

z
a
cat
vwxyz

Sample Output

26
1
0
83681

Source

East Central North America 1995

解题思路：
题目中所输入的串必须严格上升是很关键的一点，在26个字母中随便选5个字母，则这5个字母只有一个排列（满足严格上升），所以根据此来计数。

比如要求 c e h 这个字典序是多少。该串长3 ，所以先把串长1和2的情况加上，即sum+=c (26,1) + c(26,2) , 该串为3了，注意到第一个字母最初应该从'a'开始,a->b 剩下两位随便在25->24个数里选 ,即 sum+=c (2 5,2) + c( 24 ,2) ，注意第一个字母不能到达 c，否则获得的串可能会比c e h大。给出的串第二个字母最初不能从a开始，因为要严格上升，要从原串上一个字母的下一个字母开始，第一个字母为c，所以第二个字母要从d开始计数, sum+=c(
24 , 1) ,最后一个字母也一样。这样最后所得的sum为 ceg的字典序，再加上1就是ceh的字典序了。 对于ceh 有 s+=c[26][1]+c[26][2]+c[25][2]+c[24][2]+c[22][1]+2+1;

代码：

#include <iostream>
#include <string.h>
using namespace std;

int c[30][30];

void C()//求组合数
{
for(int i=0;i<=26;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];
}
}
}

int main()
{
string s;
C();
while(cin>>s)
{
int len=s.length();
int i;
for(i=1;i<len;i++)
{
if(s[i]<=s[i-1])
break;
}
if(i<len)
{
cout<<0<<endl;
continue;
}
int sum=0;
for(i=1;i<len;i++)
sum+=c[26][i];//长度比该串短的先加上

for(i=0;i<len;i++)//从高位进行处理对于每一位处理到该位的前一个，比如该位为‘d'，就处理到c
{
char ch=i==0?'a':(s[i-1]+1);
for(char j=ch;j<s[i];j++)
sum+=c['z'-j][len-1-i];
}
cout<<sum+1<<endl;//加上串本身
}
return 0;
}