diff --git a/C++/Dynamic Programming/interval_queries.cpp b/C++/Dynamic Programming/interval_queries.cpp
new file mode 100644
index 0000000..3f0d643
--- /dev/null
+++ b/C++/Dynamic Programming/interval_queries.cpp	
@@ -0,0 +1,49 @@
+#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;
+}
diff --git a/C++/Math/particular_sum.cpp b/C++/Math/particular_sum.cpp
new file mode 100644
index 0000000..3753bd8
--- /dev/null
+++ b/C++/Math/particular_sum.cpp
@@ -0,0 +1,32 @@
+//This is a detailed solution to provide intuition on solving CP problems based on finding the reduced max size of a given array whose elements are divisible by given number. (Elements of the array can be replaced by sum of elements present in the array).
+//Have a look into at the question: https://codeforces.com/problemset/problem/1176/B
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int t;
+    cin>>t; // Input to run the program for 't' test cases
+    while(t--){
+        int n,f=0,r1=0,r2=0;
+        cin>>n; // Input to know the total number of elements in the array
+        vector<int>a; // array initialization (a vector can also be used as an array)
+        for(int i=0;i<n;i++){
+            int x;
+            cin>>x;
+            if(x%3==0)
+            f++; // counting the number of elements which are divisible by 3
+            if(x%3==1)
+            r1++; // counting the number of elements which leave reminder 1 when divided by 3
+            if(x%3==2)
+            r2++; // counting the number of elements which leave reminder 2 when divided by 3
+        }
+        int ans=f;
+        int m=min(r1,r2); // We can add-up the elements whose reminder is 1 with those whose reminder is 2 such that the newly formed number is divisible by 3
+        ans+=m;
+        m=(r1+r2-2*m)/3; // Consider r1 as a set and r2 as a set. The remaining elements of a particular set, the one which has max elements and are not pared with the other remainder set but can still yield a sum which is divisible by 3.
+        ans+=m; 
+        cout<<ans<<endl;
+    }
+    return 0;
+}
diff --git a/C++/README.md b/C++/README.md
index 52cdcbe..192958c 100644
--- a/C++/README.md
+++ b/C++/README.md
@@ -22,6 +22,8 @@
 
 - [Longest Common Subsequence](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Dynamic%20Programming/lcs.cpp)
 
+- [Interval_queries](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Dynamic%20Programming/interval_queries)
+
 <a name="graph"></a>
 
 ## Graph Algorithms
@@ -66,6 +68,8 @@
 
 - [Sieve of eratosthenes](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Math/sieve_of_eratosthenes.cpp)
 
+-[particular_sum](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Math/Particular_sum)
+
 <a name="searching"></a>
 
 ## Searching
diff --git a/README.md b/README.md
index 5b70072..e9a97a7 100644
--- a/README.md
+++ b/README.md
@@ -55,6 +55,7 @@ You can also create an issue or [contact us](https://github.com/aniketsharma0041
 
 - [Coin change](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Dynamic%20Programming/dp_coin_change.cpp)
 - [Longest Common Subsequence](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Dynamic%20Programming/lcs.cpp)
+- [Interval_queries](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Dynamic%20Programming/interval_queries)
 
 <a name="graph"></a>
 
@@ -83,6 +84,7 @@ You can also create an issue or [contact us](https://github.com/aniketsharma0041
 - [Matrix exponentiation](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Math/matrix_exponentiation.cpp)
 - [Pascal triangle](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Math/pascal_triangle.cpp)
 - [Sieve of eratosthenes](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Math/sieve_of_eratosthenes.cpp)
+-[particular_sum](https://github.com/aniketsharma00411/algorithmsUse/blob/master/C%2B%2B/Math/Particular_sum)
 
 <a name="searching"></a>