Code
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 8404 | Accepted: 3976 |
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 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.
• 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
题目大意是舒输入个递增的字符串,问小于等于他的递增字符串的个数。如果不是合法的就输出0就好
思路详见:http://blog.youkuaiyun.com/lyy289065406/article/details/6648492
ac代码
#include<stdio.h>
#include<string.h>
#include<math.h>
int c[30][30];
void fun()
{
int i,j;
for(i=0;i<30;i++)
{
c[i][0]=1;
}
for(i=1;i<30;i++)
{
for(j=1;j<30;j++)
c[i][j]=c[i-1][j-1]+c[i-1][j];
}
}
int main()
{
char s[20];
fun();
while(scanf("%s",s)!=EOF)
{
int len=strlen(s);
__int64 sum=0;
int i,j,k;
for(i=0;i<len-1;i++)
{
if(s[i]>s[i+1])
{
printf("0\n");
return 0;
}
}
for(i=1;i<len;i++)
{
sum+=c[26][i];
}
for(i=0;i<len;i++)
{
int ch;
if(i==0)
ch='a';
else
ch=s[i-1]+1;
while(ch<s[i])
{
sum+=c['z'-ch][len-i-1];
ch++;
}
}
printf("%I64d\n",++sum);
}
}