Algorithmic competition Getting Started classic exercise 2-10 arrangement (permutation)

Exercise 2-10

The 1,2,3,....,9 consists of 3 three-digit abc,def and GHI, each using exactly once, requiring abc:def:ghi=1:2:3. Outputs all solutions.

#include <stdio.h> #include <stdlib.h> #include <string.h>int main (int argc, char *argv[]) {  int ABC, Def, Ghi;  int a[10], count = 0;    memset (A, 0, sizeof (a)); printf ("n\n");  for (ABC = 123; ABC <= 329; abc++)  {     def = 2*abc;     Ghi = 3*ABC;          A[ABC/100] = a[abc/10%10] = a[abc%10] = 1;     A[DEF/100] = a[def/10%10] = a[def%10] = 1;     A[GHI/100] = a[ghi/10%10] = a[ghi%10] = 1;     int i;     for (i = 1; I <= 9; i++)        count + = A[i];          if (count = = 9) printf ("%d%d%d\n", ABC, Def, Ghi);      Count = 0;     memset (A, 0, sizeof (a));  System ("PAUSE");  return 0;}

Summary: 1 will be all possible numbers as a one-dimensional array subscript, and finally determine whether the sum is 9, if less than 9, there will be coincident, whereas each number has only one

2 after judging, Count and array are zeroed.

Algorithmic competition Getting started classic exercise 2-10 arrangement (permutation)