POJ 题目1850 Code（组合数学）

Code

Time Limit: 1000MS Memory Limit: 30000KTotal 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. 

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

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.csdn.net/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';elsech=s[i-1]+1;while(ch<s[i]){sum+=c['z'-ch][len-i-1];ch++;}}printf("%I64d\n",++sum);}}