Suppose you compile and run the following micro-benchmark program in an attempt to see how quickly floating-point multiplications are computed. main() {        float f = 1.0;        for (int i=0; i<1000000; ++i) f = f * f; } That code ran very fast, so you change the loop termination condition from i<1000000 to i<1000000000 to make it take longer to run, and recompile the program. Oddly, the new version takes exactly the same amount of processor time to execute. What probably happened?

Evaluate the performance per line

a. for (int i = 0; true; i++) System.out.println (a[i]); b. for (int i = 0; true; i++) ; System.out.println (a[i]); c. for (int i = 0; i <= a.length; i++) System.out.println (a[i]); d. for (int i = 1; i <= a.length ; System.out.println (a[i-1]); 1) Which loop statement outputs the elements of array 'a'? 2/7 e. None of the above ... ...

Please make a Python code which can generate a multiplication table given input n.

Answer (JAVA LANGUAGE) 1) a. for (int i = 0; true; i++) System.out.println(a[i]); b. for (int i = 0; true; i++) ; %3D Which loop statement System.out.println (a[i]); outputs the elements of c. for (int i = 0; i <= a.length; i++) %3D array 'a'? System.out.println (a[i]); d. for (int i = 1; i <= a.length ; i+t) System.out.println (a[i-1]); e. None of the above for (int i = 0; i < a.length; i++) 2) Which if statement on the %3D right should be used inside a. if (a[i] $ 2 = 1) !! the for loop to check for b. if (ali] % 2 != 1) the odd elements in array c. if(i$2 1) 'a'? d. if (a[i] % 2 == 1) e. None of the above a. if (a [i-n] 8 2 0) 3) To check if the last element in an integer array b. if (a[n-1] % 2 0) == 'a' is even we write : f. if (a[a.length-i] % 2 g. if (a [a.length-1] % 2 0) 0) h. None of the above