poj1850 Code

Posted 2022-01-05 zbtrs

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10059		Accepted: 4816

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

Source

​​Romania OI 2002​​

大致题意：问你一个字符串是按照某种顺序排列的第几个.

分析：先求出长度小于当前字符串的字符串有多少种，可以利用组合数快速统计,没有位上的限制，只是后一位要比前一位大.如果当前统计位数为len的，那么方案数就是C(26,len).

          再来统计长度等于当前字符串的字符串有多少种.因为整个序列是单调上升的，如果第i位我们固定的字符是第j个，那么剩下的len-i位就只能用26-j个字符了，那么方案数就是C(26-j,len-i).每个位置的字符的取值是有一个范围限制的.即它必须大于上一个字符，小于当前给定的字符.

犯的一个错误：我处理字符是把字符-‘a’变成数字来处理的，结果有一处忘了-a.最好的解决方法是不要-a,直接用一个int来存.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int c[30][30], len, ans;
char s[15];

int main()

    c[0][0] = 1;
    for (int i = 1; i <= 26; i++)
    
        c[i][0] = 1;
        for (int j = 1; j <= 10; j++)
            c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    

    while (~scanf("%s", s + 1))
    
        ans = 0;
        bool can = true;
        len = strlen(s + 1);
        for (int i = 2; i <= len; i++)
            if (s[i] <= s[i - 1])
            
                can = false;
                break;
            
        if (!can)
        
            printf("0\\n");
            continue;
        
        for (int i = 1; i < len; i++)
            ans += c[26][i];
        for (int i = 1; i <= len; i++)
        
            int ch = s[i] - a, ch2;
            if (i == 1)
                ch2 = 0;
            else
                ch2 = s[i - 1] - a + 1;
            if (i == len)
                ans += ch - ch2;
            else
            
                while (ch2 < ch)
                
                    ans += c[26 - ch2 - 1][len - i];
                    ch2++;
                
            
        
        ans++;
        printf("%d\\n", ans);
    

    return 0;