The sword refers to the source code of the offer series-string arrangement, and the sword refers to the offer

The sword refers to the source code of the offer series-string arrangement, and the sword refers to the offer

Question 1369: String sorting time limit: 1 second memory limit: 32 MB special question: No submitted: 2432 solution: 609 question Description: enter a string, print all characters in the string in Lexicographic Order. For example, if abc is input, all the strings abc, acb, bac, bca, cab, and CBA that can be arranged by characters a, B, and c are printed. Input: each test case contains one line. Enter a string of no more than 9 characters (repeated characters may exist), including only uppercase and lowercase letters. Output: corresponding to each group of data, all data is output in Lexicographic Order. Sample input: abcBCA sample output: abcacbbacbcacabcbaABCACBBACBCACABCBA

# Include <stdio. h> # include <stdlib. h> # include <string. h> void bubbleSort (char * arr, int begin, int length); void swap (char * a, char * B); void Permulation (char * arr, int k, int length); int main () {char arr [10]; int length; int I; while (gets (arr) {length = strlen (arr); bubbleSort (arr, 0, length); Permulation (arr, 0, length);} return 0;} void Permulation (char * arr, int k, int length) {int I; if (k = length) {for (I = 0; I <length; I ++) printf ("% c", arr [I]); printf ("\ n ");} else {for (I = k; I <length; I ++) {if (k! = I & arr [k] = arr [I]) continue; swap (& arr [k], & arr [I]); bubbleSort (arr, k + 1, length); Permulation (arr, k + 1, length); bubbleSort (arr, k + 1, length) ;}} void swap (char * a, char * B) {char c; c = * B; * B = * a; * a = c;} // bubble sort void bubbleSort (char * arr, int begin, int length) {int I, j; for (I = begin; I <length; I ++) {for (j = I + 1; j <length; j ++) {if (arr [I]> arr [j]) {swap (& arr [I], & arr [j]) ;}}}