习题2-6 排列 permutation

//把频数数组换成标记数组，某个数字i出现过则置a[i]=1。如果不符合条件，必然至少有一个数组元素没有被标记。
#include 
int main() {
	int abc,def,ghi,i;
	for(abc = 100; abc <= 333; abc++) {
		int a[10]= {0};
		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;
		for( i = 1; i <= 9; i++)
			if(a[i]==0) break;
		if(i==10) printf("%d %d %d\n", abc, def, ghi);
	}
	return 0;
}