juinda

2的大數

Oct 4th, 2016
81
0
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
text 0.63 KB | None | 0 0
  1. #include "stdio.h"
  2. int main()
  3. {
  4. int n;
  5. int a[3015] = { 0 }, howlong = 0;
  6. while (scanf("%d", &n) != EOF)
  7. {
  8. for (int i = 0; i < 3015; i++)//將每位元預設為0
  9. {
  10. a[i] = 0;
  11. }
  12. howlong = 0;
  13. a[0] = 1;
  14. for (int i = 0; i < n; i++)
  15. {
  16. for (int j = 0; j < 3015; j++)//將每位元*2
  17. {
  18. a[j] *= 2;
  19. }
  20. for (int j = 0; j < 3015; j++)
  21. {
  22. if (a[j] >= 10)
  23. {
  24. a[j] -= 10;//過10-10
  25. a[j + 1]++;//下一位進1
  26. howlong = j + 1;//紀錄長度
  27. }
  28. }
  29. }
  30. int i = howlong;
  31. while (i >= 0)
  32. {
  33. printf("%d", a[i]);
  34. i--;
  35. }
  36. printf("\n");
  37. }
  38. return 0;
  39. }
Advertisement
Add Comment
Please, Sign In to add comment