Code
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 9588 | Accepted: 4584 |
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.
• 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.
• 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
题意
输出某str在字典中的位置。
思路
规定输入的字符串必须是升序序列，否则是非法字符串，如果出现非法字符串时，应该输出0，并且结束程序。
对于长度为1的字符串，有a,b,c,…z，共 C[26][1]=26 个。
对于长度为2的字符串：
以a开头的，有ab,ac,…az，一共 C[25][1]=25 个。
以b开头的，有bc,bd,…bz，一共 C[24][1]=24 个。
……
以x开头的，有xy,xz，一共 C[2][1]=2 个。
以y开头的，有yz，一共 C[1][1]=1 个。
于是，长度为2的字符串个数有 C[25][1]+C[24][1]+…+C[2][1]+C[1][1]=C[26][2] 种。
因为有公式：
C[n][m]=∑{i=1->n−1}C[i][m-1]
同理
长度为3的字符串个数为 C[26][3] 个。
长度为4的字符串个数为 C[26][4] 个。
所以我们可以用一个循环计算所有小于str长度的字符串个数，剩下的，就是求长度等于str，但是在str字典序前面的字符串个数了。
对于字符串第一位，起始最小为’a’，其他位最小只能是它的前一位+1，因此，对于当前位所能进行变换的只有 [c[i−1]+1,c[i]−1] ，我们所要计算的是当前位的后几位，组合数需要在’z’-ch中选择剩余位数个，最终求得的和+1便是答案，+1的原因是因为我们所计算的是str前面的字符串个数。
AC 代码
#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
#include<stack>
#include<string.h>
#define PI (acos(-1))
using namespace std;
int C[27][27];
void init() //组合数打表
{
C[1][0]=C[1][1]=1;
for(int i=2; i<27; i++)
for(int j=0; j<=i; j++)
{
if(j==0)C[i][j]=1;
else C[i][j]=C[i-1][j-1]+C[i-1][j];
}
}
int main()
{
init();
string c;
while(cin>>c)
{
int len=c.length(),sum=0;
for(int i=1; i<len; i++)
if(c[i]<=c[i-1])
{
printf("0\n");
return 0;
}
for(int i=1; i<len; i++) //长度小于len的字符串
sum+=C[26][i];
for(int i=0; i<len; i++) //长度等于len的字符串
{
char ch=i!=0?c[i-1]+1:'a'; //对于第一位，起始最小为'a'，其余为当前已有位+1
while(ch<c[i])
sum+=C['z'-ch++][len-1-i];
}
printf("%d\n",sum+1);
}
return 0;
}
