diff --git a/code/code/dynamic_programming/src/Adjacent Bit Counts.cpp b/code/code/dynamic_programming/src/Adjacent Bit Counts.cpp
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+++ b/code/code/dynamic_programming/src/Adjacent Bit Counts.cpp	
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+/*
+For a string of n bits x1,x2,x3,...,Xn the adjacent bit count of the string (AdjBC(x)) is given by
+X1*X2 + X2*X3 + X3*X4 + ... + Xn-1 * Xn
+which counts the number of times a 1 bit is adjacent to another 1 bit. For example:
+AdjBC(011101101) = 3
+AdjBC(111101101) = 4
+AdjBC(010101010) = 0
+Write a program which takes as input integers n and k and returns the number of bit strings x of n bits (out of 2ⁿ) that satisfy AdjBC(x) = k. For example, for 5 bit strings, there are 6 ways of getting AdjBC(x) = 2:
+11100, 01110, 00111, 10111, 11101, 11011
+Input
+The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by a decimal integer giving the number (n) of bits in the bit strings, followed by a single space, followed by a decimal integer (k) giving the desired adjacent bit count. The number of bits (n) will not be greater than 100.
+Output
+For each data set there is one line of output. It contains the data set number followed by a single space, followed by the number of n-bit strings with adjacent bit count equal to k. As answer can be very large print your answer modulo 10^9+7.
+Sample Input
+10
+1 5 2
+2 20 8
+3 30 17
+4 40 24
+5 50 37
+6 60 52
+7 70 59
+8 80 73
+9 90 84
+10 100 90
+Sample Output
+1 6
+2 63426
+3 1861225
+4 168212501
+5 44874764
+6 160916
+7 22937308
+8 99167
+9 15476
+10 23076518
+*/
+
+// C++ program to count number of binary strings 
+// with k times appearing consecutive 1's. 
+//http://shuklas664.blogspot.com/2015/01/gnyr09f-solution.html
+#include <bits/stdc++.h> 
+using namespace std; 
+#define ll long long int 
+#define M 1000000007
+
+int count(int n, int k) 
+{ 
+	// dp[i][j][0] stores count of binary 
+	// strings of length i with j consecutive 
+	// 1's and ending at 0. 
+	// dp[i][j][1] stores count of binary 
+	// strings of length i with j consecutive 
+	// 1's and ending at 1. 
+	int dp[n + 1][k + 1][2]; 
+	memset(dp, 0, sizeof(dp)); 
+
+	// If n = 1 and k = 0. 
+	dp[1][0][0] = 1; //ADJACENT KUCH NHI HAI TOH  TOH BAAR BAAR ZERO HI HO JAIGA TO EK STRING POSSIBLE HOGI ZERO KI
+	dp[1][0][1] = 1; 
+
+	for (int i = 2; i <= n; i++) { //I IS SIZE J IS ADJACENT BIT COUNT 3RD IS THE BIT TO BE ADDED
+		// number of adjacent 1's can not exceed i-1 
+		for (int j = 0; j <=k; j++) { 
+			dp[i][j][0] = ((dp[i - 1][j][0]%M) + (dp[i - 1][j][1]%M))%M; 
+			dp[i][j][1] = (dp[i - 1][j][0])%M; 
+			if (j - 1 >= 0) // AVOIDING OVERFLOW CONDITION WHEN J=0.
+				dp[i][j][1] += (dp[i - 1][j - 1][1])%M; 
+		} 
+	} 
+
+	return ((dp[n][k][0]%M) + (dp[n][k][1]%M))%M; 
+} 
+
+// Driver code 
+int main() 
+{ 
+	ll n,m,k,i,j,x,z,t;
+    cin>>t;
+    while(t--)
+    {
+        cin>>i>>j>>k;
+        cout<<i<<" "<<count(j,k)<<endl;
+    }
+	return 0; 
+}