CSci 1113, Fall 2018Lab Exercise 3 (Week 4): Repeat, Again and AgainIterationImperative programming languages such as C++ provide high-level constructs that support both conditionalselection and iteration. Conditional iteration or repetition refers to the ability to repeatedly executestatements, or groups of statements, until some dynamically tested criteria is met. Iteration is what gives themodern electronic digital computer much of its extraordinary power. Using computational methods such asexhaustive enumeration, we can search through a large number of possible problem solutions in a matter ofseconds to determine the correct or best one.C++ provides 3 basic repetition mechanisms: while-loops, for-loops and do-while loops, but the latter two arejust simple variants of the first. Learning to solve problems using repetition is often challenging at first. It isimportant to remember that, although you construct the loop as a static structure, it will be executeddynamically. You need to develop the ability to envision how the steps of the loop will unfold as it executes.(This lab is longer than the previous labs as we are now at the point in the class where we know enough C++to have longer, more complicated (but also more interesting) lab problems. Usually we will want you to getthrough all the way through the warm-up, stretch, and workout problems to get credit for the lab. However, if--- due to the length of this lab --- you do not get through all the workout problems, then do as much of themas you can.)Mystery-Box ChallengeWe are introducing a new feature to the Lab Exercises: the Mystery-Box Challenge. It is designed to helpyou develop a better understanding of how programs work. The Mystery-Box Challenge presents a shortcode fragment that you must interpret to determine what the computation it is intending to accomplish at ahigh level. For example, the following line of code,3.14159 * radius * radiusis intended to compute the area of a circle whose radius is contained in a variable named radius. Ananswer such as multiply 3.14159 by radius and then multiply it by radius again, although technicallyaccurate, would not be considered a correct answer!Here is your first mystery-box challenge: determine (and describe to one of your Lab TAs) what the followingC++ code fragment is intended to do:int sum = 0;for( int i=0; i{ if( i % 7 == 0 ) sum++;}cout 1Warm-up1) Fibonacci NumbersFibonacci numbers pop up all over the place and are related to the golden ratio. You can find the nextFibonacci number by adding together the previous two. The first two Fibonacci numbers are F0 = 0, F1 = 1.Thus:F2 = F1 + F0 = 1 + 0 = 1,F3 = F2 + F1 = 1 + 1 = 2,F4 = F3 + F2 = 2 + 1 = 3,...Write a program that asks how many Fibonacci numbers to compute and then show each number.2) Loop EquivalenceUsing the compiler or paper and pencil, rewrite the following while loop as an equivalent for loop. Youand your partner should first do this individually. When both of you are done, compare answers, and come upwith a common answer that both of you agree on.int i = 1;int total = 0;while( i { total += i; i++;}3) Nested LoopsWrite a short C++ program that will use nested for loops to print out all the values for a 10 x 10multiplication table as follows: 1 2 3 4 5 6 7 8 9 10 2 4 6 8 10 12 14 16 18 20 3 6 9 12 15 18 21 24 27 30 4 8 12 16 20 24 28 32 36 40 5 10 15 20 25 30 35 40 45 50 6 12 18 24 30 36 42 48 54 60 7 14 21 28 35 42 49 56 63 70 8 16 24 32 40 48 58 64 72 80 9 18 27 36 45 54 63 72 81 90 10 20 30 40 50 60 70 80 90 100Each entry in the table must be calculated as the product of two variables and output from within the loop (doNOT simply write out character strings!). Format your output so the columns line up as shown.2Stretch1) Square Root, Babylonian StyleThe ancient Babylonians produced a remarkably efficient process to compute the square root of any numbern. The method is referred to as a guess and check algorithm because it involves guessing the answer andthen checking to see if the guess produces the correct answer. The trick is how to (systematically) improvethe guess if it isnt close enough, in order to make a better guess the next time. The simple intuition that theBabylonians discovered is to make the next (improved) guess equal to the average of the old guess and theratio of n to the old guess:new_guess = (old_guess + ( n / old_guess ) ) / 2(see: 6:25 to 7:25 in https://www.youtube.com/watch?v=csInNn6pfT4 )This algorithm converges quickly, which means that if we begin with any arbitrary guess and update it asufficient number of times, we will obtain a very good approximation of the square root of the value. In fact,this method converges quadratically and will produce excellent square root approximations in a small numberof iterations.The algorithm consists of three simple steps:1. Make an initial guess at the answer (start with 1 or n/2)2. Revise the guess:new_guess = ( old_guess + n / old_guess ) / 23. Determine if the guess is sufficiently close to the square root of n. In this problem we will say theguess is sufficiently close if the new guess is within 1% of the old guess. If it is not, repeat Step 2 fora sufficient number of iterations until that criterion is satisfied. The more that Step 2 is repeated, thecloser the guess will become to the square root of n.Write a program that has the user input a positive integer for n, and then uses a while-loop to iterate throughthe Babylonian algorithm until the stopping criteria in Step 3 is met. Output the result as a floating-pointvalue.Your program must do the following:1. Prompt the user to input a positive integer value.2. Validate the input (i.e., check that it is nonnegative). Notify the user and terminate the program if aninvalid value is entered.3. For each iteration of the algorithm loop, output the value of the guess. (This is a useful debuggingtechnique and will help you understand what your code is doing!)4. Output the square root of the value as determined by the Babylonian algorithm.5. Output the square root of the value as determined by the cmath function sqrt( ).Demonstrate your program to a TA using the following input (at a minimum):225423Workout1) Crime Scene InvestigationNewtons Law o代写CSci 1113留学生作业、C/C++编程作业帮做、C/C++语言调试、C/C++程序/课程设计代做 代做R语言编f Heating and Cooling states that the rate of change of the temperature function T(t) withrespect to time t of an object is proportional to the difference between the object’s temperature and itssurrounding’s temperature, Ts. Here assume Ts is a constant. We can write this as ?T/?t = k (Ts – T ) where ?Tis the change in temperature over time step ?t and k is the growth rate (heating) or decay rate (cooling) perunit time. We can now simulate the heating or cooling of an object usingT(t) = T(t - ?t) + k [Ts - T(t - ?t)] (?t), ornew_temperature = old_temperature + change_in_temperature,where T(t) is the new temperature at time t and T(t-?t) is the temperature at time t - ?t .For example, suppose cold water at 6 degrees Celsius is placed in a room that has temperature 25 degreesCelsius. If the time step ?t is 0.1 hour and k is 1.335, then the temperature after the first time step is T(0.1) = 6 + 1.335 (25 – 6)(0.1) = 8.5365 degrees Celsius.Write a program to simulate the heating or cooling of an object over a time period (i.e., the simulation length).Include the following:· Prompt the user, and have him or her input values of the initial temperature, the temperature of thesurroundings, the growth (decay) rate, the length of the time step, and the simulation length.· Using a while loop do the following:· In each iteration of the loop, calculate the time t and the new temperature T(t).· In each iteration, display t and T(t).· Continue the loop as long as time t is less than the simulation length.Example:Enter the initial temperature: 6Enter the temperature of the surroundings: 25Enter the growth (decay) rate: 1.335Enter the time step in hours: .1Enter the simulation length in hours: 20.100000 8.5365000.200000 10.7343770.300000 12.6388380.400000 14.2890530.500000 15.7189640.600000 16.9579830.700000 18.0315920.800000 18.9618740.900000 19.7679641.000000 20.4664411.100000 21.0716711.200000 21.5961031.300000 22.05052341.400000 22.4442781.500000 22.7854671.600000 23.0811071.700000 23.3372801.800000 23.5592531.900000 23.7515922.000000 23.918255Suppose someone whose body temperature was originally 37O C is murdered in a room that has a constanttemperature of 25O C. The body was discovered at 10:00pm with a temperature of 28O C. Use your programto determine at what time the murder was committed. Assume k = 0.4055 and use a time step of 0.1 hours.2) Decimal to Roman Numeral ConversionThe ancient Roman numbering system employed a sequence of letters to represent numeric values. Valueswere obtained by summing individual letter values in a left-to-right fashion. The value of individual Romannumerals is provided in the following list:I = 1V = 5X = 10L = 50C = 100D = 500M = 1000In Roman numbers, larger numeral values generally appear before smaller values, and no single numeral maybe repeated more than 3 times in a row. There are a few numbers that cannot be represented with theserestrictions, so an additional rule was provided to deal with these cases: if a higher value numeral is precededby a single smaller numeral, the smaller value is subtracted from the larger one. This rule is only applied to asmall number of cases: IV (4), IX (9), XL (40), XC (90), CD (400) and CM (900). Using these simple rules,we can construct a method to convert an integer (decimal) value to a Roman number as follows: Step 1:If the value of the number is in [900,999] output “CM”If the value is in [500,899] output “D”If the value is in [400,499] output “CD”If the value is in [100,399] output “C”If the value is in [90,99] output “XC”If the value is in [50,89] output “L”If the value is in [40,49] output “XL”If the value is in [10,39] output “X”If the value is 9, output “IX”If the value is in [5,8] output “V”If the value is 4, output “IV”If the value is in [1,3], output “I”Step 2:Subtract the value of the 1 or 2 letter sequence that was just output from the integer number value,producing a remainder. For example, if the original number was 93, the program would have printed“XC” in Step 1, and then, in this step, subtracted 90 to get a remainder of 3, and then continued with3 as the number.5Step 3:Repeat the process until the remainder value is zero.Using this method, write a program that does the following:· Reads in an integer value from 1 to 999, inclusive.· Checks to make sure the input value is > 0 and program without doing anything.· Converts the input value to a Roman number and output it as a string of consecutive letters withoutany spaces (ie., XCIII)Example:Enter an integer value from 1 to 999: 0Invalid input. Program terminated.Enter an integer value from 1 to 999: 42Roman numeral equivalent: XLIIEnter an integer value from 1 to 999: 3Roman numeral equivalent: IIIEnter an integer value from 1 to 999: 9Roman numeral equivalent: IXEnter an integer value from 1 to 999: 999Roman numeral equivalent: CMXCIXCheckYou and your partner should individually list (i) one important thing you learned from today’s lab, and (ii) onequestion you still have. When you have done this, share with your partner what you have written.Have a TA check your work before going on to the challenge questions.ChallengeHere are two extra challenge problems. You should be able to complete the warm-up, stretch, and workoutproblems in the lab. Try these problems if you have extra time or would like additional practice outside thelab. Pay particularly attention not only to solving these problems, but also to solving them efficiently, i.e.,with as few calculations as possible.1) Reverse the DigitsWrite a C++ program that will read in any positive integer value (not string) and print the digits out in reverseorder.6Examples:Input a positive integer: 1234Reversed: 4321Input a positive integer: 1Reversed: 1Input a positive integer: 624578Reversed: 8754262) Factors of an IntegerWrite a C++ program that will read in any positive integer value and display it as a product of its smallestinteger factors in increasing order.Examples:Input a positive integer: 8Factors: 2*2*2Input a positive integer: 25Factors: 5*5Input a positive integer: 80Factors: 2*2*2*2*5Input a positive integer: 17Factors: 17转自：http://ass.3daixie.com/2018092722237702.html
