Create interval queries

This is a detailed solution to provide intuition on solving CP problems based on answering queries about time interval intersection. Have a look into at the question: http://codeforces.com/gym/294377/problem/E

Author: kvv1618

C++/Algorithms/Greedy-Algorithms/interval queries

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+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+// your code goes here
+// this program will give an intuition behind the solution for CP questions based on interval quries. Refer to the question 
+//http://codeforces.com/gym/294377/problem/E
+
+//this following two lines are used for fast input-output
+ios_base::sync_with_stdio(false);
+ cin.tie(NULL);
+
+ int t;
+ cin>>t;
+ while(t--) //running the program for 't' test cases
+ {
+ int n;
+ cin>>n; //number of students in th college
+
+ //declare an array of max size (based on the input constraint) to indicate the timeline of all the students
+ int N=2*100000;
+ int a[N]={0};
+
+ for(int i=0;i<n;i++)
+ {
+ int x,y;
+ cin>>x>>y;
+ // while taking the entering and leaving time of each student, increase the (x-1)th index by one and decrease the (y-1)th index by 1. 
+ a[x-1]+=1;
+ a[y-1]-=1;// Here the student is outside the college during the yth sec, hence we subtract 1 from (y-1)th index. Else we would subtract the same from yth index 
+ }
+ for(int i=1;i<2*100000;i++)
+ {
+ // Summing the value in the current index by its previous index's value will give the total number of students at a perticular instance of time on the timeline a[N] 
+ a[i]+=a[i-1];
+ }
+ int q;
+ cin>>q;
+ while(q--)// running a loop to ans all the queries
+ {
+ int pp;
+ cin>>pp;// The required time to output number of students present at that particular instance 
+
+ // We can find out the number of students present at that perticular time by accessing the value stored in the array 'a' at (pp-1) index
+ cout<<a[pp-1]<<endl; 
+ }
+ }
+return 0;
+}