POJ 1850 Code(找规律)

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 7913		Accepted: 3709

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
[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
[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