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- There is more than one way to calculate the value of T. One way that this can be done is by generating random numbers. This works by recognizing that if you take a unit square that you can draw a quarter circle of unit radius inside the square. The area of the quarter circle is exactly π/4 and the area of the square is 1. So if you randomly generate a pair of uniform numbers between 0 and 1 they will be distributed uniformly across the square. If you count the total number of points generated and the number of points (x, y) where x² + y² = ² < 1 then the ratio of those two numbers will tend towards the ratios of the areas of the square and the circle as the number of points generated increases. The ratio of the areas is just π/4 so if you take that ratio and multiply it by 4 you get an estimate for . a) b Write code to estimate π using this method. You can generate random numbers in the range 0arrow_forwardGiven a positive integer 'n', find and return the minimum number of steps that 'n' has to take to get reduced to 1. You can perform any one of the following 3 steps:1.) Subtract 1 from it. (n = n - 1) ,2.) If its divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,3.) If its divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). Write brute-force recursive solution for this.Input format :The first and the only line of input contains an integer value, 'n'.Output format :Print the minimum number of steps.Constraints :1 <= n <= 200 Time Limit: 1 secSample Input 1 :4Sample Output 1 :2 Explanation of Sample Output 1 :For n = 4Step 1 : n = 4 / 2 = 2Step 2 : n = 2 / 2 = 1 Sample Input 2 :7Sample Output 2 :3Explanation of Sample Output 2 :For n = 7Step 1 : n = 7 - 1 = 6Step 2 : n = 6 / 3 = 2 Step 3 : n = 2 / 2 = 1 SolutionDp///.arrow_forwardbool isprime(long n) /* fixed from to https://www.geeksforgeeks.org/euclid-euler-theorem/?ref=lbp */{ // check whether a number is prime or not int i; for (i = 2; i * i <= n; i++) if (n % i == 0) return false; return true;}arrow_forwardGiven a positive integer 'n', find and return the minimum number of steps that 'n' has to take to get reduced to 1. You can perform any one of the following 3 steps:1.) Subtract 1 from it. (n = n - 1) ,2.) If its divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,3.) If its divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). Write brute-force recursive solution for this.Input format :The first and the only line of input contains an integer value, 'n'.Output format :Print the minimum number of steps.Constraints :1 <= n <= 200 Time Limit: 1 secSample Input 1 :4Sample Output 1 :2 Explanation of Sample Output 1 :For n = 4Step 1 : n = 4 / 2 = 2Step 2 : n = 2 / 2 = 1 Sample Input 2 :7Sample Output 2 :3Explanation of Sample Output 2 :For n = 7Step 1 : n = 7 - 1 = 6Step 2 : n = 6 / 3 = 2 Step 3 : n = 2 / 2 = 1 SolutionDp///.arrow_forwardBum 0; for (int i = 1; iarrow_forwardThe statement 2n< n! is valid for a.All positive integers b.All positive integers n ≥ 4 c.For only n=4d.All positive integers n ≤ 4arrow_forwardarrow_back_iosarrow_forward_ios