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Saleh127

UVA 10182

Jul 18th, 2021
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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. #define ll long long
  4. #define test int tt; cin>>tt; for(int cs=1;cs<=tt;cs++)
  5. pair<ll,ll>x[100005];
  6. int main()
  7. {
  8. ios_base::sync_with_stdio(0);
  9. cin.tie(0);cout.tie(0);
  10.  
  11. ll n,m,i,j,k,l;
  12.  
  13. x[1]={0,0},x[2]={0,1};
  14. n=3;
  15. for(i=1;i<100001;i++)
  16. {
  17. for(j=0;j<i;j++)
  18. {
  19. x[n]={x[n-1].first-1,x[n-1].second};
  20. n++;
  21. if(n>100000) break;
  22. }
  23. for(j=0;j<i;j++)
  24. {
  25. x[n]={x[n-1].first,x[n-1].second-1};
  26. n++;
  27. if(n>100000) break;
  28. }
  29. for(j=0;j<i;j++)
  30. {
  31. x[n]={x[n-1].first+1,x[n-1].second-1};
  32. n++;
  33. if(n>100000) break;
  34. }
  35. for(j=0;j<i;j++)
  36. {
  37. x[n]={x[n-1].first+1,x[n-1].second};
  38. n++;
  39. if(n>100000) break;
  40. }
  41. for(j=0;j<i+1;j++)
  42. {
  43. x[n]={x[n-1].first,x[n-1].second+1};
  44. n++;
  45. if(n>100000) break;
  46. }
  47. for(j=0;j<i;j++)
  48. {
  49. x[n]={x[n-1].first-1,x[n-1].second+1};
  50. n++;
  51. if(n>100000) break;
  52. }
  53. if(n>100000) break;
  54. }
  55.  
  56.  
  57. while(cin>>n)
  58. {
  59. cout<<x[n].first<<" "<<x[n].second<<endl;
  60. }
  61.  
  62.  
  63. return 0;
  64. }
  65.  
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