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  1. #include <stdio.h>
  2.  
  3. int main(void) {
  4. 	int a,b,c,d,e,f,g,h;
  5. 	scanf("%d",&a);
  6. 	printf("式の値を入力してください; %d\n",a);
  7.  
  8. 	for(b=0;b<=a;b++){
  9. 	 for(c=0;c<=b;c++){
  10. 	 	d=b-c;
  11. 	 	for(e=0;e<=a;e++){
  12. 	 		f=d*e;
  13. 	 		for(f=0;f<=a;f++){
  14. 	 			h=f+g;
  15. 	 			if(h==a) printf("(%d-%d)*%d+%d\n",b,c,e,g);
  16. 	 		}
  17. 	 	}
  18.  
  19. 	 }
  20. 	}
  21. 	return 0;
  22. }
  23.  
Success #stdin #stdout 0.01s 5304KB
comments (?)
stdin 
3
stdout 
式の値を入力してください; 3
(0-0)*0+0
(0-0)*1+0
(0-0)*2+0
(0-0)*3+0
(1-0)*0+0
(1-0)*1+0
(1-0)*2+0
(1-0)*3+0
(1-1)*0+0
(1-1)*1+0
(1-1)*2+0
(1-1)*3+0
(2-0)*0+0
(2-0)*1+0
(2-0)*2+0
(2-0)*3+0
(2-1)*0+0
(2-1)*1+0
(2-1)*2+0
(2-1)*3+0
(2-2)*0+0
(2-2)*1+0
(2-2)*2+0
(2-2)*3+0
(3-0)*0+0
(3-0)*1+0
(3-0)*2+0
(3-0)*3+0
(3-1)*0+0
(3-1)*1+0
(3-1)*2+0
(3-1)*3+0
(3-2)*0+0
(3-2)*1+0
(3-2)*2+0
(3-2)*3+0
(3-3)*0+0
(3-3)*1+0
(3-3)*2+0
(3-3)*3+0
https://ideone.com/FcBenT
language:
C (gcc 8.3)
created:
10 days ago
visibility:
public

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