poj 1850(组合数学)

最新推荐文章于 2019-04-10 20:22:00 发布

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本文介绍了一种基于字符序列的编码系统，该系统用于传输和记忆信息。文章详细解释了编码规则，并提供了一个示例程序来展示如何根据这些规则对特定单词进行编码。

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8050		Accepted: 3772

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

参考详解(看了一定会懂得)：http://hi.baidu.com/you289065406/blog/item/d0900ff85e4b0b71024f563a.html#0

AC代码：

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int C[27][27];
void init(){
    for(int i=0;i<=26;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]+C[i-1][j-1];
    }
}
int main(){
    init();
    char str[12];
    while(cin>>str){
        int i,j,k;
        int sucess=1;
        int len=strlen(str);
        long long sum=0;
        for(i=1;i<len;i++){
            if(str[i-1] >= str[i]){
                sucess=0;
                break;
            }
        }
        if(!sucess){
            cout<<'0'<<endl;
            continue;
        }
        for(i=1;i<len;i++)
            sum+=C[26][i];
        for(i=0;i<len;i++){
            char ch;
            if(i==0)
                ch='a';
            else
                ch=str[i-1]+1;
            while(ch<=str[i]-1){
                sum+=C['z'-ch][len-1-i];
                ch++;
            }
        }
        sum++;
        cout<<sum<<endl;
    }
    return 0;
}