done! this will be seen next week

Hello everyone! I have attached my application for VP Curricula, make sure to vote! Thank you for watching and I hope you have a great day!

hi guys

import java.util.Scanner; import java.io.*; public class Main { public static void main(String[] args) throws IOException { Scanner in = new Scanner(new FileReader("input.txt")); in.useDelimiter(", |\\s|,"); int cards, reference, best_value; String suits, best_suit; int[] card_numbers = new int[6]; String[] card_suits = new String[6]; for (int repeating_times = 0; repeating_times < 5; repeating_times++){ reference = 0; best_suit = "NONE"; best_value = 10; cards = in.nextInt(); suits = in.next(); card_numbers[reference] = cards; card_suits[reference] = suits; reference++; cards = in.nextInt(); suits = in.next(); card_numbers[reference] = cards; card_suits[reference] = suits; reference++; cards = in.nextInt(); suits = in.next(); card_numbers[reference] = cards; card_suits[reference] = suits; reference++; cards = in.nextInt(); suits = in.next(); card_numbers[reference] = cards; card_suits[reference] = suits; reference++; cards = in.nextInt(); suits = in.next(); card_numbers[reference] = cards; card_suits[reference] = suits; reference++; cards = in.nextInt(); suits = in.next(); card_numbers[reference] = cards; card_suits[reference] = suits; // This is the first case, does nothing to best_value if it does not exist. for (int i = 1; i < 6; i++){ if (card_suits[0].equals(card_suits[i])){ best_suit = card_suits[0]; if (card_numbers[i] > card_numbers[0] && best_value > card_numbers[i]){ best_value = card_numbers[i]; } } } //If the first case is not satisfied, then this calculates the smallest of the same suit. if (best_value == 10) { for (int i = 1; i < 6; i++){ if (card_suits[0].equals(card_suits[i])){ if (card_numbers[i] < card_numbers[0] && card_numbers[i] < best_value){ best_value = card_numbers[i]; } } } } //If no cases are satisfied, then this does either NONE or the best choice of card. if (best_suit.equals("NONE")){ System.out.println("NONE"); } else { System.out.println(best_value + ", " + best_suit); } } } } 2. 22

1. 101000011 2. 11101 3. 1000000100 4. 10110 5. 9

1. Decimal: 319 2. Binary: 100111111 3. Octal: 477 4. Hexadecimal: 13F 5. Conversion 6. 280 7. 544 8. 2805 9. Bonus: 1188 -hastin ツ

f(18, 6) = f(17, 8) - 10 = f(16, 10) - 20 = f(15, 12) - 30 = f(14, 14) - 40 = f(13, 16) - 50 = f(12, 18) - 60 = f(11, 20) - 70 = f(10, 22) - 80 = f(9, 24) - 90 = f(6, 31) - 81 = f(3, 38) - 72 = f(0, 45) - 63 = 2(0) + 45 - 63 = -18 NOTE: I wasn't included in the call yesterday, I might be wrong... And can you please include me in the class emails? I didn't get the zoom links, and I tried all the previous links, and none of them worked. I would really appreciate that :) thanks!

1. 5 * 8 / 2 ^ 2 + 4 * 6 2. PREFIX NOTATION: + / * 5 8 ^ 2 2 * 4 6 3. POSTFIX NOTATION: 5 8 * 2 2 ^ / 4 6 * + 4. 0 5. 60 BONUS: X = 9

1. 10101 2. 11000 EDIT: (this bottom portion is only for X AND 011010 = 001000) BONUS: (assuming 01000 is 001000 since its not 6 digits like 011010...) 1. 001000 2. 001001 3. 001100 4. 001101 5. 101000 6. 101001 7. 101100 8. 101101 EDIT: this is the correct bonus: BONUS: 1. 01000 2. 01001 3. 01100 4. 01101

MY DECIMAL NUMBER: 420 BINARY: 111011001 420 to binary is: 110100100 1 1 01 001 00 + 1 1 01 01 1 1 1 0 1 1 001 OCTAL: 445 420 to octal is: 644 644 - 1 77 445 DECIMAL: 2274 420 * 21 = 8820 8820 to hexadecimal is: 2274 HEXADECIMAL: A R 0 420 to hexadecimal is: 1A4 _______ → A R 0 2 A ) 1 A 4

My Number: 420 BINARY: 110100100 420 ÷ 2 = 210 R 0 210 ÷ 2 = 105 R 0 105 ÷ 2 = 52 R 1 52 ÷ 2 = 26 R 0 26 ÷ 2 = 13 R 0 13 ÷ 2 = 6 R 1 6 ÷ 2 = 3 R 0 3 ÷ 2 = 1 R 1 1 ÷ 2 = 0 R 1 OCTAL: 644 420 ÷ 8 = 52 R 4 52 ÷ 8 = 6 R 4 6 ÷ 8 = 0 R 6 HEXADECIMAL: 1A4 420 ÷ 16 = 26 R 4 26 ÷ 16 = 1 R 10 (A) 1 ÷ 16 = 0 R 1

Set int r = number of rows / Set int c = number of columns/ For the the first row, print c blocks. Then "\n" or skip a line. Determine x, y, m, and n./ If c is even, then int m = 2/ If c is odd, then int m = 3/ If r is even, then int n = 2/ If r is odd, then int n = 1/ If c is even, then x = c/2 - 2/ If c is odd, then x = c/2 - 1.5/ If r is even, then y = r/2 -2./ If r is odd, then y = r/2-1.5/ Print one block, (c-2) blank spaces, another block, and skip a line. Repeat y times. Then, print one block, x spaces, m blocks, x spaces, another block, and skip a line. Repeat n times. Subsequently, print one block, (c-2) blank spaces, another block, skip a line, and repeat that y times. Finally, print c blocks. Then System.out.println("THIS IS BETTER THAN WHAT HENRY DID!")

Make robot face south. Move forward until impossible and sense the material. Then robot turn right and detect window or wall. If robot reaches a corner. turn right and continue detecting. Set detecting = turn left and determine the material. When material is wall, turn back right and move forward by one unit. When the material is a window, turn back right and move forward by 2 units. If the robot detects a window, print (i) If another window is detected, print (i + 1). Int i = 1, i = i + 1 Detect the room exactly 35 times. The last number the robot prints is the number of windows in the room.