[POJ1496]Word Index

Word Index
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 3200
Accepted: 1831

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

我的解法:
先生成所有代码的字典,然后读入,再二分查找
const max=83681;
var s,s0:string;
s1:
array[1..5] of char;
len,i,j:integer;
index,k:longint;
a:
array[1..max] of string[5];

function valid(s:string):boolean;
var i:integer;
begin
for i:=2 to len do
if s[i]<=s[i-1] then exit(false);
exit(true);
end;
//生成字典
procedure find(t,l:integer);
var i:char;
begin
if t>l then
begin
inc(index);
s0:
='';
for j:=1 to t-1 do
s0:
=s0+s1[j];
a[index]:
=s0;
exit;
end;
for i:='a' to 'z' do
if (t=1) or (i>s1[t-1]) then
begin
s1[t]:
=i;
find(t
+1,l);
end;
end;
//判定字符串s1不小于s2
function notsmall(s1,s2:string):boolean;
var i:integer;
begin
if (length(s1)>length(s2)) or ((length(s1)=length(s2)) and (s1>=s2)) then exit(true)
else exit(false);
end;

//二分法获取字符串index
function getIndex(s:string;l,r:longint):longint;
var i,k:longint;
begin
if l=r then exit(l)
else
begin
k:
=(l+r) div 2;
if notsmall(a[k],s) then getIndex:=getIndex(s,l,k)
else getIndex:=getIndex(s,k+1,r);
end;
end;

begin
for i:=1 to 5 do
find(
1,i);
while not EOF do
begin
readln(s);
len:
=length(s);
if valid(s) then writeln(getIndex(s,1,max))
else writeln(0);
end;
end.
posted @ 2011-04-14 08:52  Mose  阅读(239)  评论(0编辑  收藏  举报