poj 1496 Word Index

本文提供了一种将英语字母组成的五字母以下的单词编码为整数的简单编码方案,并通过C++代码实现了该算法。输入为一系列单词,长度从1到5不等,仅包含小写字母。输出为每个有效单词在字母排序列表中的位置索引,无效单词输出0。

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Word Index

Time Limit: 1000MS
Memory Limit:10000K

Total Submissions: 3586
Accepted:2030

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
 
  1: #include <iostream>
  2: using namespace std;
  3: 
  4: __int64 dp[16][32], sum[16], ans;
  5: 
  6: int main()
  7: {
  8:     int i, j, len;
  9:     char str[16];
 10: 
 11:     for (i = 0; i < 26; i++)
 12:         dp[0][i] = 1;
 13:     for (i = 1; i < 10; i++) {
 14:         for (j = 25 - i; j >= 0; j--) {
 15:             dp[i][j] += dp[i][j + 1];
 16:             dp[i][j] += dp[i - 1][j + 1];
 17:         }
 18:     }
 19:     for (i = 0; i < 10; i++) 
 20:         for (j = 0; j < 26; j++)
 21:             sum[i] += dp[i][j];
 22: 
 23: 	while(scanf("%s", str)!=EOF)
 24: 	{
 25: 		ans=0;
 26: 	    for (len = 0; str[len]; len++)
 27: 	        ans += sum[len];
 28: 	    ans -= sum[len - 1];
 29: 	    j = 0;
 30: 	    for (i = len - 1; i >= 0; i--) {
 31: 	        while (j < str[len - 1 - i] - 'a') {
 32: 	            ans += dp[i][j];
 33: 	            j++;
 34: 	        }
 35: 	        j++;
 36: 	    }
 37: 	    for (i = 0; i < len - 1; i++)
 38: 	        if (str[i] >= str[i + 1])
 39: 	            ans = -1;
 40: 	    printf("%I64d\n", ans + 1);
 41: 	}
 42:     return 0;
 43: }
 44: 
 45: 

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/21/2257575.html

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