POJ - 1850Code 组合数

Code Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9600 Accepted: 4592

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

也是找规律嘛

算出它之前的字符有多少个，再加上一

先把长度小于的算出来，C26,k ,k是字符串的长度

然后把每位字符类似成数位

从最高位向下找，找比这位小的字符串总数（不包括等于）

例如求ceg

先找长度为1，2的字符串个数

在从最高位第一个c字符开始，找最高位为a，b的字符个数（a：C26-(‘a'-’a‘+1) , len-i ; b : C26-(‘b'-’a‘+1),len-i ,其中i代表第几位）

第二位就找比e小的，注意 你已经找了第一位是a或b，假如第二位是a,b,c，第一位肯定是a或b，所以第二位要从d开始，依次向下类推，最后前面的字符串个数加一就是要求的编号

#include<stdio.h> #include<string.h> #include<math.h> #include<string> #include<algorithm> #include<queue> #include<vector> #include<map> #include<set> typedef long long ll; using namespace std; int myc(int n,int r) { int sum=1; for(int i=1;i<=r;i++) { sum=sum*(n+1-i)/i; } return sum; } int mya(int n,int r) { int sum=1; for(int i=0;i<r;i++) sum=sum*(n-i); return sum; } int ans(char ch[],int len) { int ans=0,i,j,t; for(i=1;i<len;i++) { ans+=myc(26,i); }// t=0; // printf("ha%d\n",ans); for(i=0;i<len;i++) { for(j=t;j<ch[i]-'a';j++) { ans+=myc(26-j-1,len-i-1); } t=ch[i]-'a'+1; } return ans; } int main() { char ch[20],maxx; int len,i; scanf("%s",ch); len=strlen(ch); maxx=ch[0]; for(i=1;i<len;i++) { if(ch[i]<=maxx) { printf("0\n"); return 0; } else maxx=ch[i]; } printf("%d\n",ans(ch,len)+1); return 0; }