Create birthdayPradox.cpp

birthdayPradox.cpp

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+// C++ program to approximate number of people in Birthday Paradox 
+// problem 
+//Using the above approximate formula sqrt(2*365*log(1/(1-p)) , we can approximate number of people for a given probability.
+//The questiopn is to find the number of person we should have so that the probability that any 2 of them have their birthday on the same date is n.
+
+#include <math.h> 
+#include <iostream> 
+using namespace std; 
+int find(double p) 
+{ 
+	return ceil(sqrt(2*365*log(1/(1-p)))); 
+} 
+int main() 
+{ 
+int n;//probability for which we have to find
+cout << find(n)<<endl; 
+} 

max_inversions.cpp

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+// C++ program to Count 
+// Inversions in an array 
+// using Merge Sort 
+#include <bits/stdc++.h> 
+using namespace std; 
+
+int _mergeSort(int arr[], int temp[], int left, int right); 
+int merge(int arr[], int temp[], int left, int mid, int right); 
+int mergeSort(int arr[], int array_size) 
+{ 
+	int temp[array_size]; 
+	return _mergeSort(arr, temp, 0, array_size - 1); 
+} 
+int _mergeSort(int arr[], int temp[], int left, int right) 
+{ 
+	int mid, inv_count = 0; 
+	if (right > left) { 
+		mid = (right + left) / 2; 
+		inv_count += _mergeSort(arr, temp, left, mid); 
+		inv_count += _mergeSort(arr, temp, mid + 1, right); 
+		inv_count += merge(arr, temp, left, mid + 1, right); 
+	} 
+	return inv_count; 
+} 
+
+int merge(int arr[], int temp[], int left, 
+		int mid, int right) 
+{ 
+	int i, j, k; 
+	int inv_count = 0; 
+
+	i = left; /* i is index for left subarray*/
+	j = mid; /* j is index for right subarray*/
+	k = left; /* k is index for resultant merged subarray*/
+	while ((i <= mid - 1) && (j <= right)) { 
+		if (arr[i] <= arr[j]) { 
+			temp[k++] = arr[i++]; 
+		} 
+		else { 
+			temp[k++] = arr[j++]; 
+			inv_count = inv_count + (mid - i); 
+		} 
+	} 
+	while (i <= mid - 1) 
+		temp[k++] = arr[i++]; 
+	while (j <= right) 
+		temp[k++] = arr[j++]; 
+	for (i = left; i <= right; i++) 
+		arr[i] = temp[i]; 
+
+	return inv_count; 
+} 
+
+
+int main() 
+{ 
+	int arr[] = { 1, 20, 6, 4, 5 }; 
+	int n = sizeof(arr) / sizeof(arr[0]); 
+	int ans = mergeSort(arr, n); 
+	cout << " Number of inversions are " << ans; 
+	return 0; 
+} 
+
+