poj 1850 code

Code
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 10060 Accepted: 4817

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

计数问题可转化为数位dp，与p进制不同，后面的位不能比前面的位大，于是可从长度1到当前str的长度从a开始计数。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define ll long longchar str[15];long long dp[11][30];int num[15];void init(){    for(int j = 1;j <= 26;++j){        dp[1][j] = 1;    }    for(int i = 2;i <= 10;++i){        for(int j = 1;j <= 26;++j){            for(int k = j + 1;k <= 26;++k){                dp[i][j] += dp[i - 1][k];            }        }    }}long long dfs(int i,int j,bool flag){    if(i == 0)        return 1;    if(flag){        int limit = num[i];        long long sum = 0;        for(int k = j + 1;k <= limit;++k){            sum += dfs(i - 1,k,k == limit);        }        return sum;    }    else{        return dp[i + 1][j];    }}long long solve(){    int k = 1,len = strlen(str);    for(int i = len - 1;i >= 0;--i){        int t = str[i] - 'a' + 1;        num[k++] = t;    }    long long sum = 0;    for(int i = 1;i < k;++i){        bool flag = i == k - 1 ? true : false;        int limit = flag ? num[i] : 26;        for(int j = 1;j <= limit;++j){            sum += dfs(i - 1,j,j == limit);        }    }    return sum;}int main(){    init();    while(~scanf("%s",str)){        bool flag = true;        for(int i = 1;str[i] != '\0';++i){            if(str[i] <= str[i - 1]){                flag = false;                break;            }        }        if(flag){            cout << solve() << endl;        }        else{            cout << 0 << endl;        }    }    return 0;}