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POJ - 1850 Code

2024-07-31 23:20:44   222人阅读

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

55

题意:按照题目描述给出的定义：

a->1，b->2……z->26，ab->27……vwxyz->83681.

合法的字符串序列是每一个小写字母比后一个小写字母ASCII码要大，不合法输出0。

思路：首先长度小于len 的话，那么就是计算C[26][{1,2...len-1}]，至于等于len 的情况就是讨论str[i-1]+1到str[i]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 50;

char str[maxn];
int C[maxn][maxn];

void init() {
	C[0][0] = 1;
	C[1][0] = C[1][1] = 1;
	for (int i = 2; i < 27; i++) {
		C[i][i] = C[i][0] = 1;
		for (int j = 1; j < i; j++) 
			C[i][j] = C[i-1][j] + C[i-1][j-1];
	}
}

int main() {
	init();
	while (scanf("%s", str) != EOF) {
		int flag = 1;
		int len = strlen(str);
		for (int i = 1; i < len; i++)
			if (str[i] <= str[i-1]) {
				flag = 0;
				break;
			}
		if (!flag) {
			printf("0\n");
			continue;
		}
		
		int ans = 0;
		for (int i = len-1; i > 0; i--)
			ans += C[26][i];
		for (int i = 0; i < len; i++) {
			char ch = (i == 0) ? 'a' : (str[i-1]+1);
			for (int j = ch; j < str[i]; j++)
				ans += C['z'-j][len-1-i];
		}
		
		printf("%d\n", ans+1);
	}	
	return 0;
}



POJ - 1850 Code


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