HDU 4342

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 1 #include<iostream>

 2 #include<set>

 3 #include<cstdio>

 4 #include<cmath>

 5 using namespace std;

 6 typedef long long int ll;

 7 set<ll> S;

 8 

 9 void prepare()

10 {

11      ll end = ll(sqrt(double((1<<31) - 1)));

12      for (ll i(2); i<=end; ++i) {

13          S.insert(i*i);    

14      }

15      end = (1<<31) - 1;

16      ++end;

17      //cout<<end<<endl;

18      S.insert(end);   

19 }

20 

21 int main()

22 {

23     int t;

24     prepare();

25     scanf("%d",&t);

26     while (t--) {

27           ll n;

28           scanf("%I64d",&n);

29           if (n<3) {

30              cout<<n+1<<" "<<n+1<<endl;

31              continue;         

32           }

33           ll ans1(0),cnt(3);

34           for (set<ll>::iterator i = S.begin(); i != S.end(); ++i) {

35               set<ll>::iterator j = i;

36               ++j;

37               if (cnt <= n && n <= cnt + (*j - *i - 2)) {

38                   ans1 = n - cnt + 1 + *i;

39                   break;      

40               } else {

41                   cnt += (*j - *i - 1);

42               }    

43           }

44           

45           ll ans2(3);

46           cnt = 2;

47           for (set<ll>::iterator i = S.begin(); i != S.end(); ++i) {

48               set<ll>::iterator j = i;

49               ++j;

50               if (*i <= ans1 && ans1 <= *j) {

51                  ans2 += cnt*(ans1 - *i + 1);

52                  break;       

53               } else {

54                  ans2 += cnt*(*j - *i);

55                  ++cnt;       

56               }   

57           }

58           cout<<ans1<<" "<<ans2<<endl;

59     }

60     return 0;    

61 }