POJ-1850 Code

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9132 Accepted: 4360

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

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int f[27][27],h[11],dp[11],ans;char s[11];int main(){f[0][0] = 1;for(int i = 1;i <= 26;i++){f[i][0]=1;for(int j = 1;j <= i;j++) f[i][j] = f[i-1][j-1] + f[i-1][j];}for(int i = 1;i <= 10;i++)     dp[i] = dp[i-1] + f[26][i];cin.sync_with_stdio(false);cin>>s;int n = strlen(s);for(int i = 1;i <= n;i++) h[i] = s[i-1] - 'a' + 1;int maxn = 0;for(int i = 1;i <= n;i++) if(h[i] <= h[i-1]) { cout<<0<<endl; return 0; }ans+=dp[n-1]+1;for(int i = 1;i <= n;i++){int now = h[i-1] + 1;if (now == h[i]) continue;for(int j = now;j < h[i];j++) ans+=f[26-j][n-i];} cout<<ans<<endl;}