POJ 1850 Code (组合数学)

POJ 1850 Code (组合数学)

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 lengthThe 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

输出某str在字典中的位置。

思路

规定输入的字符串必须是升序序列，否则是非法字符串，如果出现非法字符串时，应该输出0，并且结束程序。

对于长度为1的字符串，有a,b,c,…z，共 C126=26

对于长度为2的字符串：

以a开头的，有ab,ac,…az，一共 C125=25

以b开头的，有bc,bd,…bz，一共 C124=24

……

以x开头的，有xy,xz，一共 C12=2

以y开头的，有yz，一共 C11=1

于是，长度为2的字符串个数有 C125+C124+...+C12+C11=C226

因为有公式：

Cmn=∑i=1n−1Cm−1i

同理

长度为3的字符串个数为 C326

长度为4的字符串个数为 C426

所以我们可以用一个循环计算所有小于str长度的字符串个数，剩下的，就是求长度等于str，但是在str字典序前面的字符串个数了。

对于字符串第一位，起始最小为’a’，其他位最小只能是它的前一位+1，因此，对于当前位所能进行变换的只有 [c[i−1]+1,c[i]−1]

AC 代码

#include #include#include#include#include#include#include#define PI (acos(-1))using namespace std;int C[27][27];void init() //组合数打表{ C[1][0]=C[1][1]=1; for(int i=2; i<27; i++) for(int j=0; j<=i; j++) { if(j==0)C[i][j]=1; else C[i][j]=C[i-1][j-1]+C[i-1][j]; }}int main(){ init(); string c; while(cin>>c) { int len=c.length(),sum=0; for(int i=1; i

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