The 9 numbers from 1 to 9 consist of 3 3 digits, and three numbers are 1:2:3 in proportion, trying to find out all the combinations

A classmate is doing ACM, gave me a problem, the title is the same. Finally write the following

/* Use 1 to 9 to make up 3 3-digit, and three-number ratio of 1:2:3, to find out all the number of satisfying conditions */#include <stdio.h> #include <stdbool.h> int noc_1 (int n);	Returns the number of the hundreds of n int noc_2 (int n);	Returns the N of the 10-bit numeric int noc_3 (int n);		Returns the digit of the n digit bool check (int x,int y,int z); Verify that the number on each bit of x y x is the same if it is not the same return true, otherwise it returns false int main (void) {int a,b,c;//a b C is 3 three-digit number int i,j,k;//i J K is a Bashing bit of a	the number int n_1,n_2,n_3;
		Use n_1 n_2 n_3 to rotate the three digits of a for (n_1=1;n_1<4;n_1++) {i=n_1;
			for (n_2=1;n_2<10;n_2++) {if (n_2==n_1)//If there is the same number that discards the combination of this series continue;
			
			j=n_2; for (n_3=1;n_3<10;n_3++) {if (n_2==n_3| |
				N_1==n_3)//If there is the same number to discard the combination of this series continue;
				K=n_3;
				A=i*100+j*10+k;
				b=a*2;
				c=a*3;
				if (c>999)//If this combination is discarded for a four-digit number continue;
			if (check (a,b,c) ==true) printf ("%5d%5d%5d\n", a,b,c);
}}} return 0;

} int noc_1 (int n) {return n/100;}
	int noc_2 (int n) {N=N/10;
return n%10;

} int Noc_3 (int n) {return n%10;}
	BOOL Check (int x,int y,int z) {int a[9]; A[0]=noc_1 (x);
	a[1]=noc_2 (x);
	A[2]=noc_3 (x);
	A[3]=noc_1 (y);
	A[4]=noc_2 (y);
	A[5]=noc_3 (y);
	A[6]=noc_1 (z);
	A[7]=noc_2 (z);
	A[8]=noc_3 (z);
	int i,j;
		for (i=0;i<9;i++) {if (a[i]==0)//b C may have a bit on the number of 0 if 0 then return false to false;
			for (j=0;j<9;j++) {if (j==i) continue;
		if (A[i]==a[j]) return false;
}} return true; }

Compile, run the following as shown below

hyp@debian:~$ gcc-wall shu.c 
hyp@debian:~$./a.out 
  192  384  576
  219  438  657
  273  546  819
  327  654  981
hyp@debian:~$ 

This I think there should be a lot of algorithms, I have not yet seen how many algorithmic knowledge, now written as above.

------------------------------------------20121127-------------------------------

I didn't think of that. Indeed this is much shorter and more efficient, and I have done a lot of useless thinking. In fact, tests can be put to the end, and some judgments can be streamlined. Haha, long insight, everyone has a misunderstanding of thinking, considered too hasty.

#include <stdio.h>
int main ()
{
	int a,b,c,i,j,s[9];
	for (a=100;a<333;a++)
	{
		b=2*a;c=3*a;
		s[0]=a%10;s[1]=a%100/10;s[2]=a/100;
		s[3]=b%10;s[4]=b%100/10;s[5]=b/100;
		s[6]=c%10;s[7]=c%100/10;s[8]=c/100;
		for (i=0;i<8;i++)
			{for
				(j=i+1;j<9;j++)
					if (s[i]==s[j]) break;
				if (j<9) break;
			}
		if ((i==8) && (j==9))
			printf ("%d,%d,%d\n", a,b,c);
	}
	return 0;
}