poj1850 Code

来源：互联网发布：php jenkins docker 编辑：程序博客网时间：2024/06/17 17:31

Code

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7638 Accepted: 3591

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

大致题意：给定一个字符串，问这个字符串按照给定规律的位置。

规律：字符串<=10,(s[i-1] < s[i]),如果存在s[i-1] >= s[i]，则输出0

解题思路：找出规律就容易解答了，当有一个字符时, ans = 26;

当有两个字符时a(b-z) = 25, b(c-z) = 24, c(d-z) = 23, ... ,yz = 1.

= c(25,1) + c(24,1) + ... + c(n, 1) + c(n-1，1） + c(1,1)

因为 c(n,m) = c(n-1,m-1) +c(n-1,m)；

c(n-1,m) = c(n-2, m-1) + c(n-2, m);

c(n-2,m) = c(n-3, m-1) + c(n-3, m);

...

c(m+1, m) = c(m, m-1) + c(m, m) = c(m, 1) + 1;

c(m, m) = 1;

所以 c(n, m) = c(n-1, m-1) + c(n-2, m-1) + ... + c(m, m-1) + 1;

c(n, m) =c(n-1, m-1) + c(n-2, m-1) + ... + c(2, m-1) + c(1, m-1);

(c(s,t) = 0 t>s);

所以两个字符为c(26,2), 同理 3个 c(26,3) ... 10个 c(26, 10);

例如给bfgi = 1.[len<4的所有] + 2.[len = 4（小于 bfgi）];

1. 将所有len<4的c（26，i）和计算；

2. 就要枚举<bfgi(长度相等的字符串), 如：a..., (ba.. to be..), (bfg. to bfgh)

也就是为a时 c（'z' - 'a', len-pos(a)-1）, 为什么是（'z' - 'a'），

因为a下一个，从b-z,len-pos(a)-1就是除了a余下的长度，同理可得余下的。

代码：

#include <stdio.h>#include <string.h>int sum;int c[27][27] = {0};char str[20];void init(){int i, j;for (i=0; i<27; ++i){for (j=0; j<=26; ++j){if (!j || i==j)c[i][j] = 1;elsec[i][j] = c[i-1][j-1] + c[i-1][j];}}}void solve(){int len = strlen(str);sum = 0;int i, j;for (i=1; i<len; ++i)sum += c[26][i];char ch;for (i=1; i<len; ++i){if (str[i-1] >= str[i]){puts("0");return;}}for (i=0; i<len; ++i){ch = i==0?'a':str[i-1]+1;while (ch < str[i]){sum += c['z' - ch][len-i-1];ch++;}}sum++;printf("%d\n", sum);}int main(){init();while (~scanf("%s", str)){solve();}return 0;}

0 0