Code(组合数学)

Code

Time Limit: 1000MS Memory Limit: 30000K

Total Submissions: 8766 Accepted: 4168

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

大神博客

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

typedef long long LL;

const int MAX = 1e5+10;

int Bin[35];

int c[33][33];

char str[20];

char ch;

void ComBinnations()//打表,计算组合数
{
for(int i=0; i<=32; i++)
{
for(int j=0; j<=i; j++)
{
if(!j||i==j)
{
c[i][j]=1;
}
else
{
c[i][j]=c[i-1][j]+c[i-1][j-1];
}
}
}
c[0][0]=0;
}
int main()
{
int sum;
bool flag;
ComBinnations();
scanf("%s",str);
int len=strlen(str);
flag=false;
for(int i=1; i<len; i++)//判断是否是合法字符
{
if(str[i-1]>=str[i])
{
printf("0\n");
flag=true;
break;
}
}
if(!flag)
{
sum=0;
for(int i=1; i<len; i++)//计算比它小的字母排列
{
sum+=c[26][i];
}
for(int i=0; i<len; i++)//计算长度相等是字符的个数
{
ch=!i?'a':str[i-1]+1;
while(ch<str[i])
{
sum+=c['z'-ch][len-i-1];
ch++;
}
}
printf("%d\n",sum+1);//加上本身
}
return 0;
}