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最新推荐文章于 2019-02-14 20:40:58 发布

翻译最新推荐文章于 2019-02-14 20:40:58 发布 · 549 阅读

文章标签：

#string #character #constraints #dictionary #output #system

数论同时被 2 个专栏收录

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Code

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

组合数学。

代码如下：

#include<iostream>
#include<string.h>
using namespace std;

int c[27][27];
//杨辉三角：

void yanghui_table(void) {
    for (int i = 0; i < 27; i++)
        for (int j = 0; j <= i; j++)
            if (i == j || j == 0)
                c[i][j] = 1;
            else
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    c[0][0] = 0;
}

int main(void) {
    yanghui_table();
    char string[11];
    while (cin >> string) {
        int sum = 0;
        int len = strlen(string);
        int flag = 0;
        for (int i = 0; i < len - 1; i++)
            if (string[i] >= string[i + 1]) {//检查string字符串是否符合升序排列
                cout << 0 << endl;
                flag = 1;
                break;
            }
        if (flag)
            break;
        for (int i = 1; i < len; i++)
            sum += c[26][i]; //比string字符串长度小的长度为i的符合升序排列的字符串有26ci个
        char ch;
        for (int i = 0; i < len; i++) {
            if (i == 0)//当前位置的字符至少要比前一位置的字符大1
                ch = 'a';
            else
                ch = string[i - 1] + 1;
            while (ch < string[i]) {//当前位置的字符至多能比后一位置的字符串小1
                sum += c['z' - ch][len - i - 1]; //剩下的字符串中可供选择的有'z'-ch个，每次选择len-i-1个按照升序排列
                ch++;
            }
        }
        if (!flag)
            cout << ++sum << endl; //输出结果
    }
    return 0;
}