Painting fence in cpp (Algo-Phantoms#1883)

* Painting fence in cpp

* Update paint_fence.cpp

* Update paint_fence.cpp

Author: aaadddiii

Algorithm/DP/readme.md

@@ -46,3 +46,4 @@ be avoided by constructinng a temporary array dp[] and memoizing the computed va
 - Painting Fence Algorithm ----> [Python](/Code/Python/PaintFenceAlgo.py) | [Java](/Code/Java/Paint_fence_algo.java)
 - Shortest Common Supersequence ---->[C++](/Code/C++/print_shortest_supersequence.cpp)
 - Word Wrap Problem ----> [C++](/Code/C++/word_wrap.cpp)
+- Painting Fence algorithm ----> [C++](/Code/C++/paint_fence.cpp)

Code/C++/paint_fence.cpp

@@ -0,0 +1,65 @@
+/*
+Problem statement:
+Given a fence with n posts and k colors , find out the number of ways of painting the fence
+such that atmost 2 adjacent posts have the same color.
+Suppose we have a single post, The number of ways of painting the fence with k colors is k
+Now we consider 2 posts, Then the number of ways of painting the second fence with the same
+color as the first post is k, and the number of ways of painting the second post with a different
+color is k*(k-1) as we have k-1 colors that are different from the first post's color.
+So the total number of ways of painting 2 fences is k*(k-1) + k. Now we consider 3 posts,
+Then the number of ways of painting the third fence same as the second fence is the number of
+ways of painting the second fence with a different color that is k*(k-1). The number of ways of
+painting the third fence using a different color is equal to the total number of ways of
+painting the second fence * (k-1). Hence the total number of ways of painting 3 fences is k*(k-1) + (k+k*(k-1))*(k-1).
+This way we can find the total number of ways of painting n fences
+*/
+
+#include <bits/stdc++.h>
+
+using namespace std;
+unsigned long long MOD = 1000000007;
+//paintfence function uses permutations to find the number of ways of painting fences
+unsigned long long paintfence(int n, int k)
+{
+    unsigned long long total, diff, same;
+    same = k;
+    diff = k * (k - 1);
+    total = k * k;
+    for (int i = 1; i < n - 1; i++)
+    {
+        same = diff;
+        diff = total * (k - 1);
+        total = same + diff;
+    }
+    return total%MOD;
+}
+
+int main()
+{
+    int n, k;
+    cout << "Enter the number of posts" << endl;
+    cin >> n;
+    cout << "Enter the number of colors" << endl;
+    cin >> k;
+    cout << "The number of ways of painting the fence is : " << paintfence(n, k) << endl;
+    return 0;
+}
+
+/*
+Sample I/O:
+Enter the number of posts
+2
+Enter the number of colors
+4
+The number of ways of painting the fence is : 16
+
+Enter the number of posts
+3
+Enter the number of colors
+2
+The number of ways of painting the fence is : 6
+
+Complexity Analysis
+Time complexity - O(n)
+Space complexity - O(1), where n is the number of posts
+*/