poj1850——组合数学
Code
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 8492 | Accepted: 4020 |
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
题意:对递增的字符串,输出次序,非递增则输出0
思路:排列组合
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; const int maxn=1000100; const int INF=(1<<28); string str; int code; int C[30][30]; void creat_C() { memset(C,0,sizeof(C)); for(int i=0;i<=28;i++){ for(int j=0;j<=i;j++){ if(j==0||j==i) C[i][j]=1; else C[i][j]=C[i-1][j-1]+C[i-1][j]; } } } int main() { creat_C(); cin>>str; for(int i=0;i<str.length()-1;i++){ if(str[i]>=str[i+1]){ cout<<0<<endl; return 0; } } code=0; int len=str.length(); for(int i=1;i<=len-1;i++){ code+=C[26][i]; } for(int i=0;i<len;i++){ char ch=i?str[i-1]+1:'a';//ch至少要比前一个大 for(;ch<str[i];ch++){ code+=C['z'-ch][len-i-1]; } } cout<<code+1<<endl; return 0; }