code to create the product for each row, which we call a “score”. If a cell is not filled in, we assume that its value is 1, so that multiplying it has a neutral effect (any number multiplied by 1 is the number itself). Or, as in the code below, just do not multiply the value of that cell at all.
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code to create the product for each row, which we call a “score”. If a
cell is not filled in, we assume that its value is 1, so that multiplying it has a neutral
effect (any number multiplied by 1 is the number itself). Or, as in the code below, just
do not multiply the value of that cell at all.
Step by step
Solved in 2 steps
- 3. Card Flipper: You walk into a room, and see a row of n cards. Each one has a number x; written on it, where i ranges from 1 to n. However, initially all the cards are face down. Your goal is to find a local minimum: that is, a card i whose number is less than or equal to those of its neighbors, xj-1 = X; <= Xj+1. The first and last cards can also be local minima, and they only have one neighbor to compare to. There can be many local minima, but you are only responsible for finding one of them. Obviously you can solve this problem by turning over all n cards, and scanning through them. However, show that you can find such a minimum by turning over only O(log n) cards.Write a section of code that creates the pairings for two sets of numbers a = {0 – 5} and b = {9 - 15}. Each number must be paired with all numbers from the other set. This is a cross product, denoted as a x b. That is, for every number in a, it should be listed with every number in b, such as (0, 9), … (0, 15), and so on. Make sure there are no spaces between numbers and the (, ), or comma.Assignment 5B: Keeping Score. Now that we know about arrays, we can keep track of when events occur during our program. To prove this, let's create a game of Rock-Paper-Scissors. For those not familiar with this game, the basic premise is that each player says "Rock-Paper- Scissors!" and then makes their hand into the shape of one of the three objects. Winners are determined based on the following rules: Rock beats Scissors Scissors beats Paper Paper beats Rock At the start of the program, it will ask the player how many rounds of Rock-Paper- Scissors they want to play. After this, the game will loop for that many number of times. Each loop, it will ask the player what item they want to use – Rock, Paper, or Scissors. The computer will randomly generate its own item, and a winner will be determined. The game will then save the result as an element of an array, and the next round will begin. Once all the rounds have been played, the program will say "Game Over" and display a list of who…
- One way to put a list of people into teams is by counting off. For example, if the list of people is: ['paul', 'francois', 'andrew', 'sue', 'steve', 'arnold', 'tom', 'danny', 'nick', 'anna', 'dan', 'diane', 'michelle', 'jeremy', 'karen'] and the number of teams desired is four, then you assign the people to teams 0, 1, 2, 3, 0, 1, 2, 3, etc. like this: [['paul', 'steve', 'nick', 'michelle'], ['francois', 'arnold', 'anna', 'jeremy'], ['andrew', 'tom', 'dan', 'karen'], ['sue', 'danny', 'diane']] Write the code for the following function. Do not assume that the list of people will divide evenly into the desired number of teams. If the list of people does not divide evenly into the desired number of teams, the teams will simply have unequal sizes. Do not assume that the number of people is at least as large as the number of teams. If there aren't enough people to cover the teams, some (or even all) teams will simply be empty. Hint: The range function will be very helpful. def form_teams…You have N boxes in Samosas, where each box is a cube. To pack the box, you need to use a rubber band (pseudo-circular, elastic band) by placing it next to the box (4 faces of a cube). The rubber band A (R1, R2) has an initial radius of R1 and can be expanded in size up to R2 without breaking. You can pack a cubic box of side length L using a 4* L round rubber band (see Notes for clarity). Given the M-band bands and their first radius and the main radius, we need to match (provide) some rubber bands and boxes. The box needs at least one rubber band to pack and of course, each rubber band can be used to pack at least one box. Find the maximum number of boxes we can pack. Answer this in python programming language. Input Output 2 1 4 10 20 34 55 4 714 7 21 14 21 7:35You have N boxes in Samosas, where each box is a cube. To pack the box, you need to use a rubber band (pseudo-circular, elastic band) by placing it next to the box (4 faces of a cube). The rubber band A (R1, R2) has an initial radius of R1 and can be expanded in size up to R2 without breaking. You can pack a cubic box of side length L using a 4 * L round rubber band (see Notes for clarity). Given the M-band bands and their first radius and the main radius, we need to match (provide) some rubber bands and boxes. The box needs at least one rubber band to pack and of course, each rubber band can be used to pack at least one box. Find the maximum number of boxes we can pack. Answer this in java programming language. Input Output 2 1 4 10 20 34 55 4 714 7 21 14 21 7 35
- CCC '13 J1 - Next in line Canadian Computing Competition: 2013 Stage 1, Junior #1 You know a family with three children. Their ages form an arithmetic sequence: the difference in ages between the middle child and youngest child is the same as the difference in ages between the oldest child and the middle child. For example, their ages could be 5, 10 and 15, since both adjacent pairs have a difference of 5 years. Given the ages of the youngest and middle children, what is the age of the oldest child? Input Specification The input consists of two integers, each on a separate line. The first line is the age Y of the youngest child (0Idiot’s Delight is a fairly simple game of solitaire, yet it is difficult to win. The goal is to draw all of the cards from the deck, and end up with no cards left in your hand. You will run through the deck of cards one time. Start by dealing 4 cards to your hand. You will always look at the last 4 cards in your hand. If the ranks of the “outer” pair (1st and 4th) are the same, discard all four cards. Otherwise, if the suits of the “inner” pair (2nd and 3rd) are the same, discard those 2 cards only. If you have less than 4 cards, draw enough to have 4 cards in your hand. If the deck is empty, the game is over. Your score will be the number of cards that remain in your hand. Like in golf, the lower the score the better. Create a new Python module in a file named “idiots_delight.py”. Add a function called deal_hand that creates a standard deck of cards, deals out a single hand of 4 cards and returns both the hand and the deck. Remember that for the last assignment, you created several…Nationality. Prove the function works by testing it 2. BLACKJACK: Sample two cards from a "deck of cards" (ace, 2-10, jack, queen, king). Find the total of the two cards (ace counts as 11 and facecards count as 10). If the total is 21 print "BLACKJACK!" otherwise print "Try again". Keep drawing pairs of two cards until you get blackjack. Set your seed to 30 at the start of this problem ? Write 2 that ohe to frame for NAe If NA is found it should replace NA with theWrite a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10), print: 7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = NUM_VALS - 1. Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element array (int courseGrades[4]). Also note: If the submitted code tries to access an invalid array element, such as courseGrades[9] for a 4-element array, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. 1 #include HNm in 10 m DO G 2 3 int main(void) { 4 5 6 const int NUM_VALS = 4; int courseGrades [NUM_VALS]; int i; 7 8 for (i = 0; i < NUM_VALS; ++i) { scanf("%d", &(courseGrades[i])); 9 } 10 11 12 13 14 15 } /*Your solution…Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = NUM_VALS - 1. Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element array (int courseGrades[4]). Also note: If the submitted code tries to access an invalid array element, such as courseGrades[9] for a 4-element array, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. #include <iostream>using namespace std; int main() { const int NUM_VALS = 4; int courseGrades[NUM_VALS]; int i; for (i = 0; i < NUM_VALS; ++i) { cin >> courseGrades[i]; } /* Your solution goes…Create the following variables • a = 2.3; • b = -87.3; • A = [1,2; 4,5]; Create a matrix 2 × 2 B using the rand() function. Create two random complex numbers, call them z1 and z, using the randi() function. Both the real and complex components should be random integers in the interval [-5,5]. Make sure to include the semicolons. Otherwise, the numbers are printed to the screen. For intensive applications, constantly printing to the screen slows the software down. Add the following lines of code. You should notice the workspace fill up with variables after running your file. I want you to print to the screen for this part. Normally, you wouldn’t unless you were looking for mistakes. • a + b a – b a * A – 2 * b * B • 3 * 21 – 4 * Z2SEE MORE QUESTIONS