POJ 1496 Word Index组合数学（解题报告）

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本文介绍了一种将特定类型的五字母或更少的英文单词编码为整数的简单编码方案。编码基于英文字母表，形成严格升序的词汇集合，并为每个有效单词分配其在字母排序列表中的位置。程序读取一系列输入单词，处理并输出相应的编码值或错误标识。

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

解题思路请在本站搜索poj 1850

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
 
__int64 factorial(int n,int m)
{
	__int64 sum=1;
	int i;
	for(i=n;i>n-m;i--)
	{
		sum*=i;
	}
	return sum;
}
 
__int64 factorial2(int n)
{
	__int64 sum=1;
	int i;
	for(i=n;i>=2;i--)
	{
		sum*=i;
	}
	return sum;
}
 
__int64 factorial3(int n,int i)
{
	return factorial(n,i)/factorial2(i);
}
 
int main()
{
	char a[100];
	int str,i,j;
	__int64 sum;
	//cout<<factorial(26,1)/factorial2(1)<<endl;
	while(scanf("%s",a)!=EOF)
	{
	str=strlen(a);
	sum=0;
	int flag=0; 
	for(i=1;i<str;i++)
	{
		if(a[i]<a[i-1])
		{
			flag=1;
			break;
		}
	}
	if(flag)
	{
		printf("0\n");
	}
	else
	{
		for(i=1;i<=str;i++)
	   {
		   sum+=factorial3(26,i);
	   }
		//printf("%I64d\n",sum);
	   for(i=1,j=str-1;j>=0;j--)
	   {
			sum-=factorial3('z'-a[j],i);
			i++;
		}
			printf("%I64d\n",sum);
	}
	}
	return 0;
}