POJ 1850 Code(找规律)

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Code

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//首先观察规律

1+1.....1 +
(25+...+1)+
((24+..+1) + (23+..+1)+..+1) 

+

[(24+..+1) + (23+..+1)+..+1)]+[(23+..+1)+..+1)]+...[1]+....//所以我维护了一个num[10][27]的数组

for(i=26;i>n;i--)
    {
        sum+=num[n-1][i];
        num[n][i]=sum;
    }num[i]中的所有数字相加  就是 代表 第i层完成之后的编号：举例：bcd这说明 前面的a开头的肯定完整了 所以这时候就是需要num[0]+num[1]所有元素的和 +num[2][1](代表三个字符的以a开头的所有数量) =bcd=26+325+300=651  然后加上bcd自己 就是652

这只是第一层上面的字母对于a偏移了  ，如果  第n层对第n-1层偏移的话  如aef  e对b偏移了 那么这个怎么算了我们可以忽略a 因为这里a对其不产生影响，唯一的影响是对e的起始的偏移位置的影响 所以我们可以用num[2][b-‘a‘+数组中起始有值的位置]+num[2][c-‘a‘+数组中起始有值的为位置]+....

/*
26 +
(25+...+1)
(25*24/2 +24*23/2 +...) +
( (24*23/2 +...) +(23*22/2+.....) )
*/
#include<stdio.h>
#include<string.h>
__int64 num[10][27];
void dfs(__int64 n)
{
    __int64 i;
    if(n>10) return ;

    __int64 sum=0;
    for(i=26;i>n;i--)
    {
        sum+=num[n-1][i];
        num[n][i]=sum;
    }
    dfs(n+1);
}
int main(void)
{
    __int64 i,j;
    char str[15];
    for(i=0;i<27;i++) num[0][i]=1;
    dfs(1);
    while(scanf("%s",&str[1])!=EOF)
    {
        __int64 len=strlen(str)-1;
        __int64 tol=0;
        //除掉不满足情况的
        for(i=2;i<=len;i++)
        {
            if(str[i]<=str[i-1]){ printf("0\n");
            return 0;}
        }

        for(i=0;i<len-1;i++)//先算总层
        {
            for(j=26;j>=i;j--)
            {    tol+=num[i][j];
            }
        }
        str[0]=‘a‘-2;//起始的时候处理一下
        for(j=len;j>0;j--){
            for(i=0;i<str[len-j+1]-(str[len-j]+1);i++)
            {
                tol+=num[j-1][j+i+(str[len-j]+1-‘a‘)];
            }
        }

        printf("%I64d\n",tol);
    }
    return 0;
}

ps:woshi1993