poj1850--Code(组合篇2)-优快云博客

本文链接：https://blog.youkuaiyun.com/winddreams/article/details/43017047

本文介绍了一种针对英语小写字母组成的字典序单词进行编码的方法，通过将单词按长度及字典序排列并赋予唯一编号实现。文章详细解释了编码流程，并提供了一个高效的DP算法实现方案。

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Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8414		Accepted: 3983

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

Source

Romania OI 2002

给出一个字符串的字典中的编号，编号是升序的，否则输出0

dp[i][j] i记录这是字符串中的第i位字符，j代表以a到z开头时的总数

num[i] 代表在第i位之前出现的个数。

按字符串从高位开始向后累加，注意最后加上字符串自身。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
LL dp[30][30] , num[30] ;
char str[30] ;
void init()
{
    int i , j , k ;
    memset(dp,0,sizeof(dp)) ;
    memset(num,0,sizeof(num)) ;
    for(j = 0 ; j < 26 ; j++)
        dp[1][j] = 1 ;
    for(i = 2 ; i <= 26 ; i++)
    {
        for(j = 0 ; j < 26 ; j++)
        {
            for(k = j+1 ; k < 26 ; k++)
                dp[i][j] += dp[i-1][k] ;
        }
        num[i] = num[i-1] ;
        for(j = 0 ; j < 26 ; j++)
            num[i] += dp[i-1][j] ;
    }
    return ;
}
int main()
{
    int i , j , l , k , last ;
    LL ans = 0 ;
    init() ;
    while( scanf("%s", str) != EOF )
    {
        l = strlen(str) ;
        for(i = 1 ; i < l ; i++)
            if( str[i-1] >= str[i] )
                break ;
        if( i < l )
        {
            printf("0\n") ;
            continue ;
        }
        ans = num[l] ; last = -1 ;
        for(i = l ; i >= 1 ; i--)
        {
            k = str[l-i] - 'a' ;
            for(j = last+1 ; j < k ; j++)
                ans += dp[i][j] ;
            last = k ;
        }
        ans++ ;
        printf("%I64d\n", ans) ;
    }
    return 0;
}