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  1. #include <stdio.h>
  2. int kaijoh(int x){
  3. 	int i,s=1;
  4. 	for(i=1;i<=x;i++){
  5. 		s=s*i;
  6. 	}
  7. 	return s;
  8. }
  9. int main(void) {
  10. 	int i;
  11. 	for(i=1;i<=10;i++){
  12. 	printf("%d\n",kaijoh(i));
  13. 	}
  14. 	return 0;
  15. }
  16.  
Success #stdin #stdout 0.01s 5324KB
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https://ideone.com/4liKyK
2Q-2 1-2
language:
C99 (gcc 8.3)
created:
9 days ago
visibility:
public

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