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
",a[0]-'a'+1);continue;}
if(check()) {printf("0
");continue;}
Get_sum1();
Get_sum2();
printf("%lld
",ans);
}
return 0;
}