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  1. #include <stdio.h>
  2. int n[3], t1, t2, t, i;
  3. int main(void) {
  4. 	scanf("%d%d",&t1,&t2); t=t2;
  5. 	for(i=0;i<3;i++)
  6. 	{
  7. 		n[i]=t1*(t2%10);
  8. 		t2/=10;
  9. 	}
  10. 	printf("%d\n%d\n%d\n%d",n[0],n[1],n[2],t1*t);
  11. 	return 0;
  12. }
  13.  
Success #stdin #stdout 0s 4388KB
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472
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stdout
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2360
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https://ideone.com/LAUqoY
language:
C (gcc 8.3)
created:
6 years ago
visibility:
public

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