POJ1850:Code(组合计数) java程序员

Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6374		Accepted: 2987

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

MYCode:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long LL;
LL c[27][27];
char str[15];
void make()
{
    c[1][0]=1;
    c[1][1]=1;
    int i,j;
    for(i=2;i<=26;i++)
    {
        c[i][0]=c[i][i]=1;
        for(j=1;j<i;j++)
        {
            c[i][j]=c[i-1][j-1]+c[i-1][j];
        }
    }
}
int main()
{
    make();
    while(scanf("%s",&str)!=EOF)
    {
        int i,j;
        bool flag=true;
        int len=strlen(str);
        for(i=1;i<len;i++)
        {
            if(str[i]<=str[i-1])
            {
                flag=false;
                break;
            }
        }
        if(flag==false)
        {
            printf("0\n");
            continue;
        }
        LL sum=0;
        for(i=1;i<len;i++)
        {
            sum+=c[26][i];
        }
        //cout<<"sum="<<sum<<endl;
        for(i=0;i<len;i++)
        {
            int lt,rt;
            if(i!=0)
            lt=str[i-1]-'a'+2;
            else
            lt=1;
            rt=str[i]-'a';
            for(j=lt;j<=rt;j++)
            {
                sum+=c[26-j][len-1-i];
            }
            //cout<<"sum="<<sum<<endl;
        }
        //printf("%lld\n",sum+1);
        cout<<sum+1<<endl;
    }
}
//

组合计数问题,注意利用杨辉三角求组合数,sum以及c[][]数组用LL类型.