Adding new binary search problem

Searching Algos & Problems/finding_maximum_divisors_in_a_range.cpp

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+/*
+QUESTION:
+Maximum divisors in a Range
+-----------------------------
+You are given two integers a and b. After that, you will be given q queries each of which contains
+two integers low and high. Let�s define S as the set of common divisors of a and b which lie in the range
+low to high i.e. low=d=high. Find the maximum element in this set S or report -1 if no such element is possible.
+
+
+
+INPUT:
+First line contains two integers a and b.
+q lines follow: Each line contains 2 space separated integers denoting low and high
+
+
+OUTPUT:
+Output q lines. Each line denotes maximum divisor which lies in range 
+[low,high]. Report -1 if no such divisor exists.	
+
+
+Constraints:
+1<=q<=10^5
+0<=low,high,a,b<=10^9
+
+
+EXAMPLE:
+input
+8 12
+4
+2 10
+3 7
+2 2 
+5 6
+output
+4
+4
+2
+-1
+
+Sample Test Case Explanation
+For a=8 and b=12, common divisors are [1,2,4].
+1. Between [2,10] and [3,7] maximum common divisor is 4.
+2. Between [2,2] maximum common divisor is 2.
+3. Between [5,6] there is no common divisor, so -1.
+*/
+
+
+//SOLUTION:
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int gcd(int a,int b){ //function for finding Greatest Common Divisor
+    if(b==0)
+        return a;
+    gcd(b,a%b);
+}
+
+void find_divisor(int arr[],int n){ //function for finding divisor
+    for(int i=1;i<=n;i++){
+        if(n%i==0){
+            arr[i-1] = i;
+        }
+    }
+    sort(arr,arr+n);
+    reverse(arr,arr+n);
+}
+
+int find_max_divisor(int arr[],int n,int i,int low,int high){
+    //loop for finding maximum divisor
+    //the array arr[] was reversed in the find_divisor(int,int) function
+    for(int i=0;i<n;i++){
+        if(arr[i]!=0){
+            if(arr[i]>=low && arr[i]<=high)
+                return arr[i];
+        }
+        else
+            break;
+    }
+    return -1;
+
+}
+
+int main(){
+    int a,b,q,low,high;
+    cin>>a>>b;
+    cin>>q;
+    //finding gcd
+    int n = gcd(a,b);
+
+    //taking array to store the divisors
+    int arr[n] = {0};
+
+    //function for finding divisors of the gcd
+    find_divisor(arr,n);
+
+    while(q--){
+        cin>>low>>high;
+        //finding the maximum divisor
+        int res = find_max_divisor(arr,n,0,low,high);
+        cout<<res<<endl;
+    }
+    return 0;
+}