长度为 N 的字母表中所有字母的排列
Permutation of all Letters of the Alphabet with length N
我最近开始了一个用 C 语言处理暴力密码破解和加密的个人项目。我一直在尝试开发一个输出长度为 N 的字母表的所有可能组合的函数。例如,如果 N = 4,则必须输出 aaaa - zzzz 的所有可能性。从逻辑上讲,我不明白我应该如何递归地处理这个问题。
char* alphabet[] = {"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
void passwords(int size){
char* password[size];
char* result;
//Determining static letter
for(int i = 0; i < size; i++){
for(int x = 0; x < size; x++){
password[x] = "a";
}
int index = i+1; //password index to modify
while(index < size){
for(int j = 0; j < 26; j++){
password[i] = alphabet[j];
printf("%s\n",password);
}
index++;
}
}
}
int main(int argc, char* argv[]){
passwords(3);
return 0;
}
目前该程序仅修改字母表中的一个字符并产生以下输出:
aaa
baa
caa
daa
...//e-z aa
aaa
aba
aca
ada
...//a e-z a
aaa
aab
aac
aad
...//aa e_z
任何建议都将非常感谢!
如果递归的使用不是强制要求,请尝试:
#include <stdio.h>
#include <stdlib.h>
#define N 26 // number of alphabets
void passwords(int size) {
int i;
char *password;
if (NULL == (password = malloc(size + 1))) {
perror("malloc");
exit(1);
}
for (i = 0; i < size; i++) {
password[i] = 'a'; // initialize the array
}
password[i] = '[=10=]'; // terminate the string
while (1) {
printf("%s\n", password);
password[size - 1]++; // increment rightmost character
for (i = size - 1; i >= 0; i--) {
if (password[i] >= 'a' + N) { // carry over
if (i == 0) { // end of permutation
free(password);
return;
} else {
password[i] = 'a';
password[i - 1]++;
}
}
}
}
}
int main()
{
passwords(3);
return 0;
}
我最近开始了一个用 C 语言处理暴力密码破解和加密的个人项目。我一直在尝试开发一个输出长度为 N 的字母表的所有可能组合的函数。例如,如果 N = 4,则必须输出 aaaa - zzzz 的所有可能性。从逻辑上讲,我不明白我应该如何递归地处理这个问题。
char* alphabet[] = {"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
void passwords(int size){
char* password[size];
char* result;
//Determining static letter
for(int i = 0; i < size; i++){
for(int x = 0; x < size; x++){
password[x] = "a";
}
int index = i+1; //password index to modify
while(index < size){
for(int j = 0; j < 26; j++){
password[i] = alphabet[j];
printf("%s\n",password);
}
index++;
}
}
}
int main(int argc, char* argv[]){
passwords(3);
return 0;
}
目前该程序仅修改字母表中的一个字符并产生以下输出:
aaa
baa
caa
daa
...//e-z aa
aaa
aba
aca
ada
...//a e-z a
aaa
aab
aac
aad
...//aa e_z
任何建议都将非常感谢!
如果递归的使用不是强制要求,请尝试:
#include <stdio.h>
#include <stdlib.h>
#define N 26 // number of alphabets
void passwords(int size) {
int i;
char *password;
if (NULL == (password = malloc(size + 1))) {
perror("malloc");
exit(1);
}
for (i = 0; i < size; i++) {
password[i] = 'a'; // initialize the array
}
password[i] = '[=10=]'; // terminate the string
while (1) {
printf("%s\n", password);
password[size - 1]++; // increment rightmost character
for (i = size - 1; i >= 0; i--) {
if (password[i] >= 'a' + N) { // carry over
if (i == 0) { // end of permutation
free(password);
return;
} else {
password[i] = 'a';
password[i - 1]++;
}
}
}
}
}
int main()
{
passwords(3);
return 0;
}