poj 1850 Code

Code

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9322 Accepted: 4455

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

提示

题意：

按照a-1,b-2,c-3......的方式编码，给出字符串输出它的编码，如果输入不合法则输出0。

思路：

把问题转化为有几个题目所给字符串形式小于等于它，例如：abc

长度为1：c[26][1]（a,b,c...）=c[26][1]

长度为2：c[25][1]（ab,ac,ad...）+c[24][1]（bc,bd,be...）...=c[26][2]

长度为3：c[1][1](abc)

我们发现

长度为n：c[26][n]

我们先计算长度小于等于它本身的个数，之后减去之前求和中比它大的个数。

还是拿abc来看，

第一步：sum=c[26][1]+c[26][2]+c[26][3]

第二步：

比它大的第一位b~z，sum=sum-c[25][3]

比它大的第二位c~z，sum=sum-c[24][2]

比它大的第三位d~z，sum=sum-c[23][1]

示例程序

Source CodeProblem: 1850Code Length: 708BMemory: 392KTime: 16MSLanguage: GCCResult: Accepted#include <stdio.h>#include <string.h>int c[27][27];int main(){    int i,i1,len,num=0;    char ch[11];    memset(c,0,sizeof(c));    for(i=0;26>=i;i++)    {        c[i][0]=1;        for(i1=1;i>=i1;i1++)        {            c[i][i1]=c[i-1][i1-1]+c[i-1][i1];        }    }    scanf("%s",ch);    len=strlen(ch);    for(i=1;len>i;i++)    {        if(ch[i-1]>=ch[i])//是否合法        {            break;        }    }    if(i==len)    {        for(i=1;len>=i;i++)//计算长度为len的所有情况        {            num=num+c[26][i];        }        for(i=0;ch[i]!='\0';i++)        {            num=num-c['z'-ch[i]][len-i];//这里是计算出长度为len比输入字符串大的        }        printf("%d",num);    }    else    {        printf("0");    }    return 0;}