Added a few codeforces Problems

codeforces/A/1420A.cpp

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+/*
+Codeforces Round #672 (Div. 2)
+Problem: Wheatley decided to try to make a test chamber. He made a nice test chamber, 
+but there was only one detail absent — cubes.
+For completing the chamber Wheatley needs n cubes. i-th cube has a volume ai.
+Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order 
+by their volume. Formally, for each i>1, ai−1≤ai must hold.
+
+To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1
+you can exchange cubes on positions i−1 and i.
+But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n⋅(n−1))/2−1
+exchange operations, he won't do this boring work.
+Wheatly wants to know: can cubes be sorted under this conditions?
+
+Input
+
+Each test contains multiple test cases.
+
+The first line contains one positive integer t
+(1≤t≤1000), denoting the number of test cases. Description of the test cases follows.
+
+The first line of each test case contains one positive integer n
+(2≤n≤5⋅104) — number of cubes.
+
+The second line contains n
+positive integers ai (1≤ai≤109) — volumes of cubes.
+
+It is guaranteed that the sum of nover all test cases does not exceed 105.
+
+Output
+
+For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can 
+be sorted and "NO" (without quotation marks) otherwise.
+*/
+/*
+Solution Explanation:
+We just need to find whether the array is strictly decreasing. If yes then to sort by swapping, (n*(n-1))/2 operations
+will be needed, which is greater than (n*(n-1))/2 -1.
+*/
+    #include <iostream>
+    #include <vector>
+    #include <cstring>
+    #include <algorithm>
+    #include<math.h>
+
+    using namespace std;
+
+
+    void solve()
+    {
+      int n,flag=0;
+      cin>>n;
+      int arr[n];
+      for(int i=0;i<n;i++)
+      {
+        cin>>arr[i];
+      }
+      for(int j=n-1;j>0;j--)
+      {
+        if(arr[j]>=arr[j-1])
+        {
+          flag=1;
+          break;
+        }
+      }
+      if(flag==1)
+      {
+        cout<<"YES\n";
+      }
+      else
+      {
+        cout<<"NO\n";
+      }
+
+      }
+
+    int main(){
+      int t;
+      cin>>t;
+      while(t--)
+      {
+         solve();
+      }
+      return 0;
+    }

codeforces/B/1420B.cpp

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+/*
+Codeforces Round #672 (Div. 2)
+Problem: Danik urgently needs rock and lever! Obviously, the easiest way to get these things is to ask 
+Hermit Lizard for them. Hermit Lizard agreed to give Danik the lever. But to get a stone, Danik needs 
+to solve the following task.
+You are given a positive integer n, and an array a of positive integers. The task is to calculate 
+the number of such pairs (i,j) that i<j and ai & aj≥ai⊕aj, where & denotes the bitwise AND operation, 
+and ⊕ denotes the bitwise XOR operation.
+Danik has solved this task. But can you solve it?
+
+Input
+Each test contains multiple test cases.
+The first line contains one positive integer t(1≤t≤10) denoting the number of test cases. Description 
+of the test cases follows.
+The first line of each test case contains one positive integer n(1≤n≤105) — length of the array.
+The second line contains n positive integers ai (1≤ai≤109) — elements of the array.
+It is guaranteed that the sum of n over all test cases does not exceed 105.
+
+Output
+For every test case print one non-negative integer — the answer to the problem.
+
+*/
+    #include <iostream>
+    #include <vector>
+    #include <cstring>
+    #include <algorithm>
+    #include<math.h>
+
+    using namespace std;
+
+
+    void solve()
+    {
+      int n;
+      cin>>n;
+      int arr[n];
+      for(int i=0;i<n;i++)
+      {
+        cin>>arr[i];
+      }
+      sort(arr,arr+n);
+      long long itr=2;
+      long long ans=0;
+      long long tem=0;
+      for(int i=0;i<n;i++)
+      {
+        if(arr[i]<itr)
+        {
+          tem++;
+        }
+        else
+        {
+          itr*=2;
+          if(tem>=2)
+          {
+            ans+=(((tem-1)*tem)/2);
+          }
+          tem=0;
+          i--;
+        }
+
+      }
+      if(tem>=2)
+          {
+            ans+=(((tem-1)*tem)/2);
+
+          }
+      cout<<ans<<"\n";
+      }
+
+    int main(){
+      int t;
+      cin>>t;
+      while(t--)
+      {
+         solve();
+      }
+      return 0;
+    }