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POJ 1496 POJ 1850 组合计数

2024-07-26 19:32:20   222人阅读

Code
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 8256 Accepted: 3906

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


/**************************************
     author     :  Grant Yuan
     time        :  2014/10/12 17:06
     algortihm: 组合计数
     source     : POJ 1496 POJ 1850
***************************************/

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
char a[27];
int num[27][27];
long long ans;
bool check()
{
    int l=strlen(a);
    if(l==1) return 1;
    for(int i=0;i<l-1;i++)
        if(a[i]>=a[i+1]) return 1;
    return 0;
}
void Get_num()
{
    int l=strlen(a);
    for(int i=0;i<=26;i++)
        for(int j=0;j<=i;j++)
    {
        num[i][j]=0;
        if(i==0||j==0) num[i][j]=1;
        else num[i][j]=num[i-1][j]+num[i-1][j-1];
    }
}
void  Get_sum1()
{
    int l=strlen(a);
    ans=0;
    for(int i=0;i<l;i++)
        ans+=num[26][i];
}
void Get_sum2()
{
    int l=strlen(a);
   for(int i=0;i<l;i++)
   {
       char j;
       if(i==0) j='a';
       else j=a[i-1]+1;
       for(;j<a[i];j++)
       {
           ans+=num['z'-j][l-i-1];
       }
   }
}
int main()
{
    Get_num();
    while(~scanf("%s",a)){
        int l;
        l=strlen(a);
        if(l==1) {printf("%d\n",a[0]-'a'+1);continue;}
        if(check()) {printf("0\n");continue;}
        Get_sum1();
        Get_sum2();
        printf("%lld\n",ans);
    }
    return 0;
}

POJ 1496 POJ 1850 组合计数


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