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  1. #include <stdio.h>
  2.  
  3. int main(void)
  4. {
  5. 	//数列a_nは常にa_n>a_(n-1)を満たす
  6. 	//a_2 = 8
  7. 	int a_l, a_m, a_n, count;
  8. 	a_l = 1;
  9. 	a_m = 2;
  10. 	count = 0;
  11. 	for(a_n = 8;a_n<=1000;a_n = 3*a_m + 2*a_l)
  12. 	{
  13. 		if(a_n>=100)
  14. 		{
  15. 			count++;
  16. 		}
  17. 		a_l = a_m;
  18. 		a_m = a_n;
  19. 	}
  20. 	printf("100 <= a_n <= 1000 を与えるnは%d個ある",count);
  21. 	return 0;
  22. }
  23.  
Success #stdin #stdout 0s 5272KB
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100 <= a_n <= 1000 を与えるnは2個ある
https://ideone.com/0o48Qr
language:
C (gcc 8.3)
created:
3 days ago
visibility:
public

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