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  1. #include <iostream>
  2. #include <cmath>
  3. using namespace std;
  4. typedef long long int ll ;
  5. ll mod = 1000000007;
  6. int main() {
  7. ll b ;
  8. ll a = 1;
  9. ll c ;
  10. ll d;
  11. ll n, k ;
  12. cin >> n >> k ;
  13.  
  14.  
  15. b = -((2*(n%mod))%mod + 3);
  16. //cout << b ;
  17. c = -(k - n*n - n );
  18. b = b%mod;
  19. d = (b*b)%mod - (4*1*(c)%mod)%mod ;
  20.  
  21. long double ans = (-b - sqrtl(d))/(2*1) ;
  22.  
  23.  
  24. if (ans < 0 )
  25. {
  26. ans = (-b + sqrtl(d))/(2*1) ;
  27. }
  28. cout << (ans) << endl ;
  29. return 0;
  30. }
  31.  
Success #stdin #stdout 0s 15232KB
stdin
9 26
stdout
3.69926