POJ 1850 Code

最新推荐文章于 2019-04-10 20:22:00 发布

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本文介绍了一种基于字符序列的编码系统，该系统用于传输和记忆信息。文章详细解释了编码规则，并通过示例展示了如何根据这些规则为字符序列分配唯一编号。

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9141		Accepted: 4363

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

利用排列与组合公式，注意输入不合法时直接退出循环

#include <stdio.h>
#include <string.h>
char s[15];
int len;
int sum;
int com(int n, int m)
{
	if(n < m)
		return 0;
	if(m == 0 || m == n)
		return 1;
	return com(n-1, m-1) + com(n-1, m);
}
int main()
{
	int i;
	char j;
	s[0] = 'a' - 1;
	while(scanf("%s", &s[1]) != EOF)
	{
		sum = 0;
		len = strlen(&s[1]);
		for(i = 1; i <= len-1; i++)
		{
			if(s[i+1] <= s[i])
			{
				printf("0\n");
				return 0;
			}
		}
		for(i = 1; i <= len-1; i++)
			sum += com(26, i);
		for(i = 1; i <= len; i++)
		{
			for(j = s[i-1] + 1; j < s[i]; j++)
			{
				sum += com(25 - (j - 'a'), len - i);
			}
		}
		printf("%d\n", ++sum);
	}
	return 0;
}