You have a five-quart jug, a three-quart jug, and an unlimited supply of water (but no measuring cups). How would you come up with exactly four quarts of water? Note that the jugs are oddly shaped, such that filling up exactly "half" of the jug would be impossible.
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You have a five-quart jug, a three-quart jug, and an unlimited supply of water (but
no measuring cups). How would you come up with exactly four quarts of water? Note that the jugs are oddly shaped, such that filling up exactly "half" of the jug would be impossible.
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- Let l be a line in the x-yplane. If l is a vertical line, its equation is x = a for some real number a. Suppose l is not a vertical line and its slope is m. Then the equation of l is y = mx + b, where b is the y-intercept. If l passes through the point (x₀, y₀), the equation of l can be written as y - y₀ = m(x - x₀). If (x₁, y₁) and (x₂, y₂) are two points in the x-y plane and x₁ ≠ x₂, the slope of line passing through these points is m = (y₂ - y₁)/(x₂ - x₁). Instructions Write a program that prompts the user for two points in the x-y plane. Input should be entered in the following order: Input x₁ Input y₁ Input x₂You and your friends decided to hold a “Secret Santa” gift exchange, where each person buys a gift for someone else. To see how this whole thing works, let’s consider the following example. Suppose there are 7 people A, B, C, D, E, F, and G. We denote x → y to mean “x gives a gift to y.” If the gift exchange starts with person A, then they give a gift to E. Then E gives a gift to B. And it is entirely possible that B gives a gift to A; in such a case we have completed a “cycle.” In case a cycle occurs, the gift exchange resumes with another person that hasn’t given their gift yet. If the gift exchange resumes with person D, then they give a gift to G. Then G gives a gift to F. Then F gives a gift to C. Then finally C gives a gift to D, which completes another cycle. Since all of the people have given their gifts, the giftexchange is done, otherwise the gift exchange resumes again with another person. All in all, there are two cycles that occurred during the gift exchange: A → E → B → A…C++ : solve the following question in the picture.
- To cut an 'n' centimeter-long gold bar into 2 pieces costs $n. When a gold bar is cut into many pieces, the order in which the cuts occur can affect the total amount of costs. For example, to cut a 20 centimeter gold bar at length marks 2, 8, and 10 (numbering the length marks in ascending order from the left-hand end, starting from 1). If the cuts to occur in left-to-right order, then the first cut costs $20, the second cut costs $18 (cutting the remaining 18 centimeter bar at originally length mark 8), and the third cut costs $12, totaling $50. If the cuts to occur in right-to-left order, however, then the first cut costs $20 time, the second cut costs $10, and the third cut costs $8, totaling $38. In yet another order, the first cut is at 8 (costing $20), then the 2nd cut is at 2 (costing $8), and finally the third cut is at 10 (costing $12), for a total cost of $40. Given an 'n' centimeter-long gold bar G and an array C[1..m] containing the cutting points in ascending order): a.…Suppose, you have been given a non-negative integer which is the height of a ‘house of cards’. To build such a house you at-least require 8 cards. To increase the level (or height) of that house, you would require four sides and a base for each level. Therefore, for the top level, you would require 8 cards and for each of the rest of the levels below you would require 5 extra cards. If you were asked to build level one only, you would require just 8 cards. Of course, the input can be zero; in that case, you do not build a house at all. Complete the recursive method below to calculate the number of cards required to build a ‘house of cards’ of specific height given by the parameter. there is a picture given in that section. public int hocBuilder (int height){ // TO DO } OR def hocBuilder(height): #TO DOSuppose, you have been given a non-negative integer which is the height of a ‘house of cards’. To build such a house you at-least require 8 cards. To increase the level (or height) of that house, you would require four sides and a base for each level. Therefore, for the top level, you would require 8 cards and for each of the rest of the levels below you would require 5 extra cards. If you were asked to build level one only, you would require just 8 cards. Of course, the input can be zero; in that case, you do not build a house at all. Complete the recursive method below to calculate the number of cards required to build a ‘house of cards’ of specific height given by the parameter. public int hocBuilder (int height){ // TO DO } OR def hocBuilder(height):
- Consider the problem of making change for n cents using the fewest number of coins. Assume that we live in a country where coins come in k dierent denominations c1, c2, . . . , ck, such that the coin values are positive integers, k ≥ 1, and c1 = 1, i.e., there are pennies, so there is a solution for every value of n. For example, in case of the US coins, k = 4, c1 = 1, c2 = 5, c3 = 10, c4 = 25, i.e., there are pennies, nickels, dimes, and quarters. To give optimal change in the US for n cents, it is sufficient to pick as many quarters as possible, then as many dimes as possible, then as many nickels as possible, and nally give the rest in pennies. Design a bottom-up (non-recursive) O(nk)-time algorithm that makes change for any set of k different coin denominations. Write down the pseudocode and analyze its running time. Argue why your choice of the array and the order in which you fill in the values is the correct one. Notice how it is a lot easier to analyze the running time of…Consider the problem of making change for n cents using the fewest number of coins. Assume that we live in a country where coins come in k dierent denominations c1, c2, . . . , ck, such that the coin values are positive integers, k ≥ 1, and c1 = 1, i.e., there are pennies, so there is a solution for every value of n. For example, in case of the US coins, k = 4, c1 = 1, c2 = 5, c3 = 10, c4 = 25, i.e., there are pennies, nickels, dimes, and quarters. To give optimal change in the US for n cents, it is sufficient to pick as many quarters as possible, then as many dimes as possible, then as many nickels as possible, and nally give the rest in pennies. Design a bottom-up (non-recursive) O(nk)-time algorithm that makes change for any set of k different coin denominations. Write down the pseudocode and analyze its running time. Argue why your choice of the array and the order in which you ll in the values is the correct one.Consider the problem of making change for n cents using the fewest number of coins. Assume that we live in a country where coins come in k dierent denominations c1, c2, . . . , ck, such that the coin values are positive integers, k ≥ 1, and c1 = 1, i.e., there are pennies, so there is a solution for every value of n. For example, in case of the US coins, k = 4, c1 = 1, c2 = 5, c3 = 10, c4 = 25, i.e., there are pennies, nickels, dimes, and quarters. To give optimal change in the US for n cents, it is sufficient to pick as many quarters as possible, then as many dimes as possible, then as many nickels as possible, and nally give the rest in pennies. Prove that the coin changing problem exhibits optimal substructure. Design a recursive backtracking (brute-force) algorithm that returns the minimum number of coins needed to make change for n cents for any set of k different coin denominations. Write down the pseudocode and prove that your algorithm is correct.
- [Medium] Suppose, you have been given a non-negative integer which is the height of a ‘house of cards’. To build such a house you at-least require 8 cards. To increase the level (or height) of that house, you would require four sides and a base for each level. Therefore, for the top level, you would require 8 cards and for each of the rest of the levels below you would require 5 extra cards. If you were asked to build level one only, you would require just 8 cards. Of course, the input can be zero; in that case, you do not build a house at all. Complete the recursive method below to calculate the number of cards required to build a ‘house of cards’ of specific height given by the parameter. public int hocBuilder (int height){ // TO DO } OR def hocBuilder(height):May you please solve quick within 20-40 minutes ? There are two types of personnel working in a government office, workers and civil servants. The salary of the worker personnel is calculated by multiplying the number of days worked and the daily wage, and the salary of the civil servant personnel is calculated by multiplying the title score (0-1000), the score he will get according to the language level (100 if A is entered, 80 if B is entered, 60 if C is entered, the rest is 0) by 30. Write the C program, in which the personnel type can be entered as upper or lower case letters as W(Worker), O(Officer) and which calculates the wage to be paid to the person according to the values that are the basis for the calculation, and displays it on the screen. Also it is required that, to check the employee type you need define checkEmployee function. To calculate the title score you need to define checkTitleScore function. And finally to get the language level you need to define…We examine a problem in which we are handed a collection of coins and are tasked with forming a sum of money n out of the coins. The currency numbers are coins = c1, c2,..., ck, and each coin can be used as many times as we want. What is the bare amount of money required?If the coins are the euro coins (in euros) 1,2,5,10,20,50,100,200 and n = 520, we need at least four coins. The best option is to choose coins with sums of 200+200+100+20.