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  1.  
  2.     #include <bits/stdc++.h>
  3.  
  4.     using namespace std;
  5.  
  6.     using ll = long long;
  7.     const int MOD = 1000000007; 
  8.     const int MOD2 =  998244353; 
  9.     const ll INF = 1e18;
  10.     const int MX = 1000001; //check the limits, dummy
  11.  
  12.  
  13.     ll modExp(ll base, ll power) {
  14.         if (power == 0) {
  15.             return 1;
  16.         } else {
  17.             ll cur = modExp(base, power / 2); cur = cur * cur; cur = cur % MOD;
  18.             if (power % 2 == 1) cur = cur * base;
  19.             cur = cur % MOD;
  20.             return cur;
  21.         }
  22.     }
  23.  
  24.     ll inv(ll base) {
  25.         return modExp(base, MOD-2);
  26.     }
  27.  
  28.  
  29.     ll mul(ll A, ll B) {
  30.         return (A*B)%MOD;
  31.     }
  32.  
  33.     ll add(ll A, ll B) {
  34.         return (A+B)%MOD;
  35.     }
  36.  
  37.     ll dvd(ll A, ll B) {
  38.         return mul(A, inv(B));
  39.     }
  40.  
  41.     ll sub(ll A, ll B) {
  42.         return (A-B+MOD)%MOD;
  43.     }
  44.  
  45.     ll* facs = new ll[MX];
  46.     ll* facInvs = new ll[MX];
  47.  
  48.     ll choose(ll a, ll b) {
  49.         if (b > a) return 0;
  50.         if (a < 0) return 0;
  51.         if (b < 0) return 0;
  52.         ll cur = facs[a];
  53.         cur = mul(cur, facInvs[b]);
  54.         cur = mul(cur, facInvs[a-b]);
  55.         return cur;
  56.     }
  57.  
  58.     void initFacs() {
  59.         facs[0] = 1; 
  60.         facInvs[0] = 1;
  61.         for (int i = 1 ; i < MX ; i ++ ) {
  62.             facs[i] = (facs[i-1] * i) % MOD;
  63.             facInvs[i] = inv(facs[i]);
  64.         }
  65.     }
  66.     int main()  {
  67.         ios_base::sync_with_stdio(0); cin.tie(0);  
  68.         int t ; 
  69.         cin >> t;
  70.         while (t --) {
  71.             ll k; cin >> k; 
  72.             ll left = 1; 
  73.             ll right = 2e18;
  74.  
  75.             while (left < right) {
  76.                 ll mid =  left + (right - left)/2;
  77.  
  78.                 // find the smallest 
  79.                 ll n = mid - int(sqrtl(mid)); 
  80.                 if (n >= k) {
  81.                     right = mid; 
  82.                 } else {
  83.                     left = mid + 1;
  84.                 }
  85.             }
  86.  
  87.             cout << left << endl; 
  88.         }
  89.  
  90.         // for some n , dterminr   w
  91.         // light bulb k is on dependdent o the parity weof thenu mb ero fidbiosr we
  92.         return 0;
  93.     }
  94.  
Success #stdin #stdout 0s 5316KB
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https://ideone.com/ix3z4c
language:
C++ (gcc 8.3)
created:
3 days ago
visibility:
public

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