有效地创造，从字符集顺序词表

问题描述

我需要找到一种有效的方式建立的从C / C给定的charset创建++顺序单词的列表。我给大家举一个例子：

如果字符集是ABC，算法应该输出：

  A
  b
  C
 AA
 AB
 AC
 巴
 BB
 公元前
 CA
 CB
 CC
AAA
AAB
...
 

我有一些想法，但都需要太多的数学，我真的需要一个快速的解决方案。谁是有一个想法？

解决方案

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆string.h中GT;字符* numToColumn（INT N，字符* outstr，为const char * baseset）{
    的char * p = outstr;
    INT LEN;
    LEN = strlen的（baseset）;
    而（N）{
        * P ++ = baseset [0 +（（N％LEN == 0）LEN：N％LEN） -  1];
        N =（N  -  1）/ LEN;
    }
    * P ='\\ 0';
    返回strrev（outstr）; // strrev不是ANSI C
}字符* incrWord（字符* outstr，为const char * baseset）{
    字符* P;
    INT大小，LEN;
    INT I，携带= 1;    大小= strlen的（baseset）;
    LEN = strlen的（outstr）;
    对于（I = LEN-1;开展和放大器;＆安培; I＆GT; = 0;  - 我）{
        INT POS;
        POS =和strchr（baseset，outstr [I]） -  baseset; // MUST NOT NULL
        POS + = 1; //增量
        如果（POS ==大小）{
            携带= 1;
            POS = 0;
        }其他{
            携带= 0;
        }
        outstr [I] = baseset [POS]
    }
    如果（进）{
        的memmove（安​​培; outstr [1]，＆放大器; outstr [0]，LEN + 1）;
        outstr [0] = baseset [0];
    }
    返回outstr;
}诠释主（）{
    为const char * CSET =ABC;
    字符的buff [16];
    INT I;    对于（i = 1; I＆LT; 16; ++ I）// 1原点
        的printf（％S \\ n，numToColumn（我，浅黄色，CSET））;    的strcpy（BUFF，CC）; //启动CC
    的printf（\\ n要重新启动\\ n％S \\ n，浅黄色）;
    的printf（％S \\ n，incrWord（BUFF，CSET））;
    的printf（％S \\ n，incrWord（BUFF，CSET））;
    返回0;
}
/ *结果：
一个
b
C
AA
AB
AC
巴
BB
公元前
CA
CB
CC
AAA
AAB
AAC重新开始
CC
AAA
AAB
* /
 

I need to find an efficient way to create a list of sequential words created from a given charset in C/C++. Let me give you an example:

If the charset is "abc", the algorithm should output:

  a
  b
  c
 aa
 ab
 ac
 ba
 bb
 bc
 ca
 cb
 cc
aaa
aab
...

I have some ideas, but all are require too much maths and I really need a fast solution. Who's got an idea?

解决方案

#include <stdio.h>
#include <string.h>

char* numToColumn(int n, char* outstr, const char* baseset){
    char* p = outstr;
    int len;
    len = strlen(baseset);
    while(n){
        *p++ = baseset[0 + ((n % len == 0)? len : n % len) - 1];
        n = (n - 1) / len;
    }
    *p = '\0';
    return strrev(outstr);//strrev isn't ANSI C
}

char* incrWord(char* outstr, const char* baseset){
    char *p;
    int size,len;
    int i,carry=1;

    size = strlen(baseset);
    len = strlen(outstr);
    for(i = len-1; carry && i>=0 ;--i){
        int pos;
        pos = strchr(baseset, outstr[i]) - baseset;//MUST NOT NULL
        pos += 1;//increment
        if(pos == size){
            carry=1;
            pos = 0;
        } else {
            carry=0;
        }
        outstr[i]=baseset[pos];
    }
    if(carry){
        memmove(&outstr[1], &outstr[0], len+1);
        outstr[0]=baseset[0];
    }
    return outstr;
}

int main(){
    const char *cset = "abc";
    char buff[16];
    int i;

    for(i=1;i<16;++i)//1 origin
        printf("%s\n", numToColumn(i, buff, cset));

    strcpy(buff, "cc");//start "cc"
    printf("\nrestart\n%s\n", buff);
    printf("%s\n", incrWord(buff, cset));
    printf("%s\n", incrWord(buff, cset));
    return 0;
}
/* RESULT:
a
b
c
aa
ab
ac
ba
bb
bc
ca
cb
cc
aaa
aab
aac

restart
cc
aaa
aab
*/

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