Merge pull request #488 from ViMo018/patch-9

Add solution for Codeforces problem 1846

Author: Dheeraj-06

Problems/Binary-Search/Day-06/sol/Vishva/soln2.cpp

@@ -0,0 +1,97 @@
+#include<bits/stdc++.h>
+#include<queue>
+using namespace std;
+typedef unordered_map<int, int> umii;
+typedef unordered_map<long long, long long> umll;
+typedef unordered_map<char, long long> umci;
+typedef vector<pair<int, int>> vpi;
+typedef vector<int> vi;
+typedef long long ll;
+typedef vector<long long> vll;
+typedef unordered_map<int , bool> umib;
+#define sum(v) accumulate(v.begin(), v.end(), 0)
+#define endl '\n'
+#define f0(i, n) for(long long i = 0; i < n; i++)
+#define f1(i, n) for(long long i = 1; i < n; i++)
+#define as(v) sort(v.begin(), v.end())
+#define all(x) (x).begin(), (x).end()
+#define pb push_back
+template<class T> umll frequency(vector<T> &v) {umll freq;for(auto &x:v) freq[x]++; return freq;}
+template<class T> umci S_frequency(vector<T> &v) {umci freq;for(auto &x:v) freq[x]++; return freq;}
+template <class T> void input(vector<T> &v){for(auto &x:v)cin>>x;}
+ll power(ll x, ll y){ ll res = 1; while (y > 0){ if (y & 1) res = (ll)(res*x); y = y>>1; x = (ll)(x*x); } return res; }
+void pvll(const vector<long long> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+void pvi(const vector<int> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+
+
+// SUBMISSION LINK:- https://codeforces.com/contest/1846/submission/356198833
+
+
+//global set
+set<ll>st;
+
+void solve(){
+
+//-------------INPUT-------------
+    ll n; 
+    cin >> n;
+    if(n<3)
+    {
+        //not possible
+        cout << "NO" << endl;
+        return;
+    }
+    ll a = 4*n-3;
+    ll sqrt_a = sqrt(a);
+    bool flag = false;
+
+    for(ll i=max(0ll,sqrt_a-5);i<=sqrt_a+5;i++)
+    {
+        if(i*i==a)
+        {
+            if ((i-1)%2==0 && (i-1)/2 >1) {
+            flag=true;
+            break;
+            }
+        }
+    }
+    // if not found using the formula,check in set
+    if(flag || st.count(n))
+    {
+        cout << "YES" << endl;
+        return;
+    }
+    cout << "NO" << endl;
+    return;
+
+
+//-------------CODE--------------
+    
+// if 4n-3 is a prefect squaare and fint the form (2*k+1)^2
+
+}
+
+
+int main(){
+    //precompute this entire shit
+
+    for(ll k=2;k<=1e6;k++)
+    {
+        ll sum = k+1;
+        ll power = k*k;
+        for(ll i=3;i<=63;i++)
+        {
+            // till 63 only
+
+            sum+=power;
+            if(sum>1e18)break;
+            st.insert(sum);
+            if(power>1e18/ k)break;
+            power*=k;
+        }
+    }
+    int tt; cin >> tt; while(tt--)
+{solve();};
+}
+
+// time complexity  o(nlogn)