标签：des style blog io color os ar for sp
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
Sample Output
55
题目大意：
判断一个字符串在字典中的位置。
字符串必须保证为升序字符串才能合法。
解题思路：
组合数学问题。
首先通过dp打印出来com[][]杨辉三角形。
假设字符串的长度为len。
详情见备注。
Code：
1 /************************************************************************* 2 > File Name: poj1850.cpp 3 > Author: Enumz 4 > Mail: 369372123@qq.com 5 > Created Time: 2014年10月28日 星期二 01时32分11秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<cstdio> 10 #include<cstdlib> 11 #include<string> 12 #include<cstring> 13 #include<list> 14 #include<queue> 15 #include<stack> 16 #include<map> 17 #include<set> 18 #include<algorithm> 19 #include<cmath> 20 #include<bitset> 21 #include<climits> 22 #define MAXN 100000 23 using namespace std; 24 long long com[30][30]; 25 void init() /*打印杨辉三角*/ 26 { 27 for (int i=0; i<=26; i++) 28 for (int j=0; j<=i; j++) 29 if (!j||i==j) 30 com[i][j]=1; 31 else 32 com[i][j]=com[i-1][j-1]+com[i-1][j]; 33 com[0][0]=0; 34 } 35 int main() 36 { 37 init(); 38 char num[20]; 39 scanf("%s",num); 40 int len=strlen(num),k; 41 for (k=1;k<=len-1;k++) /*判断是否为递增字符串*/ 42 if (num[k]<=num[k-1]) break; 43 if (k!=len) 44 { 45 cout<<0<<endl; 46 return 0; 47 } 48 long long ans=0; 49 /*任意长度小于等于len的数都在其前面*/ 50 for (int i=1; i<=len-1; i++) 51 ans+=com[26][i]; 52 /*计算最高位为[a,num[0])的个数*/ 53 for (char i=‘a‘; i<num[0]; i++) 54 ans+=com[‘z‘-i][len-1]; 55 /*计算前j位相同的个数*/ 56 for (int j=1;j<=len-1;j++) 57 /*第j位可能的数字为[num[j-1]+1,num[j]) */ 58 for (char i=num[j-1]+1;i<num[j];i++) 59 ans+=com[‘z‘-i][len-1-j]; 60 cout<<ans+1<<endl; 61 return 0; 62 }
标签：des style blog io color os ar for sp
原文地址：http://www.cnblogs.com/Enumz/p/4060391.html
