PROBLEMS SUBMIT CODE MY SUBMISSIONS STATUS STANDINGS CUSTOM INVOCATIONX. Two intervalstime limit per test: 1 second0memory limit per test: 256 megabytesinput: standard inputoutput: standard outputGiven the boundaries of 2 intervals. Print the boundaries of their intersection.Note: Boundaries mean the two ends of an interval which are the starting numberand the ending number.InputOnly one line contains two intervals [l1, r1], [l2, r2] where(1 < l1,12, r1,r2 < 10º), (1 < r1,l2 < r2).It's guaranteed that l1 < ri and l2 < r2.OutputIf there is an intersection between these 2 intervals print its boundaries,otherwise print -1.ExamplesinputСоpy1 15 5 27outputСоpy5 15inputCopy25 6 12outputСоpy-1NoteFirst Example :L115Intersect from 5 to 15Second Example :62R1There are No intersectionsCodeforces (c) Copyright 2010-2021 Mike MirzayanovThe only proarammina contests Web 2 0 platform.II>

As language was not mentioned, so solving it in java. import java.util.*;public class Solution{…

X. Two intervals time limit per test:1 seconde memory limit per test: 256 megabytes input: standard input output: standard output Given the boundaries of 2 intervals. Print the boundaries of their intersection. Note: Boundaries mean the two ends of an interval which are the starting number and the ending number. Input Only one line contains two intervals [l1,r1], [12, r2] where (1< l1,l2,r1,r2 < 10°), (11 < r1,l2 < r2). It's guaranteed that l1

X. Two intervals time limit per test:1 seconde memory limit per test: 256 megabytes input: standard input output: standard output Given the boundaries of 2 intervals. Print the boundaries of their intersection. Note: Boundaries mean the two ends of an interval which are the starting number and the ending number. Input Only one line contains two intervals [l1,r1], [12, r2] where (1< l1,l2,r1,r2 < 10°), (11 < r1,l2 < r2). It's guaranteed that l1

EBK JAVA PROGRAMMING

9th Edition

ISBN:9781337671385

Author:FARRELL

Publisher:FARRELL

Chapter9: Advanced Array Concepts

Section: Chapter Questions

Problem 2GZ

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text file 80 1 2 3 100 100 100 1001 0 2 100 3 4 100 1002 2 0 4 4 100 5 1003 100 4 0 100 100 4 100100 3 4 100 0 3 3 3100 4 100 100 3 0 100 1100 100 5 4 3 100 0 2100 100 100 100 3 1 2 0 My code below. I am getting an error when trying to create my adjacency matrix. i dont know what i am doing wrong def readMatrix(inputfilename): ''' Returns a two-dimentional array created from the data in the given file. Pre: 'inputfilename' is the name of a text file whose first row contains the number of vertices in a graph and whose subsequent rows contain the rows of the adjacency matrix of the graph. ''' # Open the file f = open(inputfilename, 'r') # Read the number of vertices from the first line of the file n = int(f.readline().strip()) # Read the rest of the file stripping off the newline characters and splitting it into # a list of intger values rest = f.read().strip().split() # Create the adjacency matrix adjMat = []…
given ASCIIOFSET .FILL x0030 NegASCIIOFSET .FILL xFFD0 ASCIINewline .FILL x000d ; Newline ascii code ASCIISpace .FILL x0020 ; Space ascii code ASCIIComma .FILL x002C ; Comma ascii code Use LC-3 I/O device to print number, commas, and spaces in printCommaSpace subroutine, also check code to see if it follows instructions and modify accordingly
If the user chooses to generate a Pentagon o This is a combination of a triangular shape and a square shape -- a symmetric triangle sits on top of a rectangle. o The triangular shape must be symmetric about the vertical axis. There must be exactly one character in the first line, and the number of characters increases by 2 in every successive lineo The rectangle shape must have LENGTH lines of height and all lines are 13 characters wide. o Make sure to add delay to view a progressively growing pentagon, otherwise you will just see the final shape. Code for triangle: for (int i = 1; i <= length; i++) { sleep (1); for (int j = length - i; j > 0; j--) cout << " "; for (int b = 1; b <= i; b++) cout << char(rdm) << " "; cout << endl; } Code for square: for (int row = 1; row <= length; row++)…
1. Create a vector of time ranging from 0 to 41 seconds with an increment of 0.01 seconds between points. Suppress output to the command window. 2. Evaluate the function ℎ(??) at the points in the time vector using element-by-element math. Suppress output to the command window. 3. Create a new figure window and plot the height versus time of the rocket. Be sure to include appropriate labels and units for the x and y axes. 4. Since we are concerned with approximating the derivative of the height at the critical time of ???? = 7.5 seconds, define this time value as a scalar variable and evaluate the height at this point. DO NOT suppress output to the command window. 5. In order to approximate the derivative at the critical time, we should move forward in time some increment ∆?? away from the critical point as previously demonstrated in Figure 1. We would like to explore the effect of moving forward by different amounts. Create a vector that contains increments corresponding to 20, 15,…
Q1. Srikanth is a school student. He studies in class 8th. One day his Math’s teacher came to the class and told them that he is going to start a new chapter “Measurement”. In order to give them a demo of that chapter, he told the students to find the volume of the class room using Volume( ). Input Constraint 15<=Length<=25 10<=Breadth<=20 20<=Height<=30 Note:- It needs to be solved in Dev C++ using #include<stdio.h>
link to MINST code: https://www.dropbox.com/s/i21pedejv0sknnq/MINST.pdf?dl=0 1) Add a confusion matrix to the MNIST code example from the link provided. Set the epochs parameter to 5. Explain which entry of the matrix yields the number of ones (the number 1) that are falsely recognized as eights (the number 8). (If you use indices in your explanation, use the Python index notation, i.e., start at zero.) Run your code a few times. Is the confusion matrix always the same? Briefly explain.
In Dist.txt it will add vaccine in storage like how many vaccines are needed and from where is coming and Add search function in which from user it will check how many vaccines are available in storage this will be done in Vaccine.txt somewhat like in image given #include <stdio.h>#include <stdlib.h>#include <string.h> struct vacc{ char vaccName[15]; char vaccCode[2]; char country[15]; int qty; float population; }v[10]; // Function Declarations void create_inventory(); void update_vacc_qty(); int search_vaccine(); void display_vaccine(); // Main Function starts here int main() { //create_inventory(); //display_vaccine(); //search_vaccine(); update_vacc_qty(); return 0; } //Function to Create Vaccine.txt as per the given table void create_inventory() { int option = 1; // variables to collect data as per table given char vaccName[15]; char vaccCode[2]; char country[15]; int qty; float populaion; //File definition FILE *infile; infile = fopen("Vaccine.txt","w");…
Modify LiveExample 12.12 EvaluateExpression.cpp to add operators ^ for exponent and % for modulus. For example, 3 ^ 2 is 9 and 3 % 2 is 1. The ^ operator has the highest precedence and the % operator has the same precedence as the * and / operators. For simplicity, assume the expression has at most one ^ operator. Sample Run: Enter an expression: (5 * 2 ^ 3 + 2 * 3 % 2) * 4 (5 * 2 ^ 3 + 2 * 3 % 2) * 4 = 160 Code posted below, c++ please! #include <vector>#include <iostream>#include <cmath>using namespace std; #ifndef STACK_H#define STACK_H template<typename T>class Stack{public: Stack(); bool empty() const; T peek() const; void push(T value); T pop(); int getSize() const; private: vector<T> elements;}; template<typename T>Stack<T>::Stack(){} template<typename T>bool Stack<T>::empty() const{ return elements.size() == 0;} template<typename T>T Stack<T>::peek() const{ return…
Instrument FrequencyCounter to use Stopwatch and StdDraw to make a plot where the x-axis is the number of calls on get() or put() and the y-axis is the total running time, with a point plotted of the cumulative time after each call. Run your program for Tale of Two Cities using SequentialSearchST and again using BinarySearchST and discuss the results. Note : Sharp jumps in the curve may be explained by caching, which is beyond the scope of this question.
Write an application (including algorithm) that uses the predefined functions,“exp” (exponential), “log10” (common logarithm), “fmax” (maximum value), “fmin”(minimum value), and “div” (integer division). The declaration of thesefunctions are contained in either “cmath” or “cstdlib” header files. The application run in a loop until user makes a selection, using a switch statement to determine the choice.
1. You are given N number of intervals, where each interval contains two integers denoting the start time and the end time for the interval. The task is to merge all the overlapping intervals and return the list of merged intervals sorted by increasing order of their start time. Two intervals [A,B] and [C,D] are said to be overlapping with each other if there is at least one integer that is covered by both of them. Input Format: The first line of input contains an integer N, the number of intervals. The second line of input contains N integers, i.e. all the start times of the N intervals. The third line of input contains N integers, i.e. all the end times of the N intervals. Output Format: Print the final list of merged intervals Sample Input 1: 5 1 3 6 8 10 4 5 8 9 12 Sample Output 1: 1 5 6 9 10 12 Explanation: For the given 5 intervals - [1, 4], [3, 5], [6, 8], [8, 9], [10, 12] Since intervals [1, 4] and [3, 5] overlap with each other, we will merge them into a…
Teacher ask different version of this code please provide that '''Problem A DNA: CG PairsPurpose:This function will take as an argument a string containing a dna strand Parameter(s):dnaReturn Value: fraction of dinucleotides ''' def cg_pairs(dna): dna_simple = 'AGCTTTCATTCTGAC' dna= dna.lower() for i in range (0,len(dna)): if(dna[i]=='a'or dna[i]=='c'or dna[i]=='g' or dna[i]=='t' ): continue else: print("Error in DNA strand ") return 0.0 count = 0 for i in range(0,len(dna)-1): var = dna[i:i + 2] if(var=="cg"): count+=1 else: continue return (float(count/(len(dna)-1)