Added a program to calculate power of a number using binary exponentiation in log n time complexity

Power_of_a_number_using_binary_exponentiation.cpp

@@ -0,0 +1,36 @@
+#include <bits/stdc++.h>
+typedef long long int ll;
+#define M 1000000007
+using namespace std;
+
+// Time Complexity-> O(Log(n))
+ll power(ll base, ll n, ll m)
+{
+    ll ans = 1;
+    base = base % m;
+    while (n != 0)
+    {
+        if (n % 2 == 1)
+        {
+            n = n - 1;
+            ans = (ans * base) % m;
+        }
+        else
+        {
+            n = n / 2;
+            base = (base * base) % m;
+        }
+    }
+    return ans;
+}
+
+int main()
+{
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    cout << power(2, 10, M) << endl;
+    //(2^10)%M this will be calculated
+    // 1024
+    return 0;
+}

dsu.cpp

@@ -0,0 +1,53 @@
+#include <bits/stdc++.h>
+typedef long long int ll;
+#define M 1000000007
+using namespace std;
+
+// Program for disjoint set union
+
+ll parent[100005];
+ll sz[100005];
+
+ll findParent(ll node)
+{
+    if (parent[node] == node)
+    {
+        return node;
+    }
+    return parent[node] = findParent(parent[node]);
+}
+
+void uni(ll u, ll v)
+{
+    ll paru = findParent(u);
+    ll parv = findParent(v);
+    if (sz[paru] < sz[parv])
+    {
+        parent[paru] = parv;
+        sz[parv] += sz[paru];
+    }
+    else
+    {
+        parent[parv] = paru;
+        sz[paru] += sz[parv];
+    }
+}
+
+void init(ll n)
+{
+    for (ll i = 1; i <= n; i++)
+    {
+        parent[i] = i;
+        sz[i] = 1;
+    }
+}
+
+int main()
+{
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    ll n;
+    cin >> n;
+    init(n);
+    return 0;
+}