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Blababa [14]
3 years ago
8

You are given an integer N, followed by N lines of input (1 <= N <= 100). Each line of input contains one or several words

separated with single spaces. Each word is a sequence of letters of English alphabet containing between 1 and 10 characters, inclusive. The total number of words in the input is between 1 and 100, inclusive
Your task is to reverse the order of words in each line of input, while preserving the words themselves. The lines of your output should not have any trailing or leading spaces.
Computers and Technology
1 answer:
Fittoniya [83]3 years ago
3 0

Answer:

Python 3 code:

n = int(input())

rev_str = [ ]

for i in range(n):

   s = str(input())

   s.split()

   words = s.split(' ')

   string = [ ]

     

   for word in words:

       string.insert(0, word)

 

   rev_str.append(" ".join(string))

     

   #print(" ".join(string))

for i in range(len(rev_str)):

   print(rev_str[i])

Explanation:

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What country is associated with the tor relay using the ip address 104. 244. 74. 97?.
Lelu [443]

Answer:

104.28.63.45 is an IP address located in Santos, Sao Paulo.

Explanation:

3 0
2 years ago
Exercise1 : Defining Circle type and Creating a Driver Circle class private members double radius double xPos double yPos public
asambeis [7]

Answer:

Aee explaination

Explanation:

1) class_circle.h

#include<iostream>

#include<cmath>

#include<iomanip>

#define PI 3.14159

using namespace std;

class circle

{

private:

double radius,xPos,yPos;

public:

double diameter();

double area();

double circumference();

double getRadius();

double getX();

double getY();

void setX(double x);

void setY(double y);

double distanceToOrigin();

bool intersect(const circle& otherCircle);

bool setRadius(double r);

};

class CircleDriver:public circle

{

private:

circle circ1;

circle circ2;

void obtainCircles();

void printCircleInfo();

public:

void run();

};

2) class_circle.cpp

#include "class_circle.h"

double circle::diameter()

{

return (radius * 2);

}

double circle::area()

{

return PI * radius * radius;

}

double circle::circumference()

{

return 2 * PI * radius;

}

double circle::distanceToOrigin()

{

return sqrt((xPos * xPos) + (yPos * yPos));

}

double circle::getRadius()

{

return radius;

}

double circle::getX()

{

return xPos;

}

double circle::getY()

{

return yPos;

}

void circle::setX(double x)

{

xPos = x;

}

void circle::setY(double y)

{

yPos = y;

}

bool circle::setRadius(double r)

{

if(r > 0)

{

radius = r;

return true;

}

else

{

radius = 0;

return false;

}

}

bool circle::intersect(const circle& otherCircle)

{

double dist1 = (radius - otherCircle.radius) * (radius - otherCircle.radius);

double dist2 = ((xPos - otherCircle.xPos) * (xPos - otherCircle.xPos)) + ((yPos - otherCircle.yPos)*(yPos - otherCircle.yPos));

double dist3 = (radius + otherCircle.radius) * (radius + otherCircle.radius);

if((dist1 <= dist2 ) && (dist2 <= dist3) && (dist1 < dist3))

return true;

else

return false;

}

void CircleDriver::obtainCircles()

{

double x,y,rad;

cout<<"Enter xPos yPos circle 1 :";

cin>>x>>y;

do

{

cout<<"\n\nEnter radius for circle 1 :";

cin>>rad;

}while(rad <= 0);

circ1.setX(x);

circ1.setY(y);

circ1.setRadius(rad);

cout<<"\n\nEnter xPos yPos circle 2 :";

cin>>x>>y;

do

{

cout<<"\n\nEnter radius for circle 2 :";

cin>>rad;

}while(rad <= 0);

circ2.setX(x);

circ2.setY(y);

circ2.setRadius(rad);

}

void CircleDriver::printCircleInfo()

{

cout<<setprecision(6)<<endl;

cout<<"\n\nInformation for Circle 1:"<<endl;

cout<<"location:( "<<circ1.getX()<<" ,"<<circ1.getY()<<" )"<<endl;

cout<<"diameter: "<<circ1.diameter()<<endl;

cout<<"area: "<<circ1.area()<<endl;

cout<<"circumference: "<<circ1.circumference()<<endl;

cout<<"distance from the origin: "<<circ1.distanceToOrigin()<<endl;

cout<<"\n\nInformation for Circle 2:"<<endl;

cout<<"location:( "<<circ2.getX()<<" ,"<<circ2.getY()<<" )"<<endl;

cout<<"diameter: "<<circ2.diameter()<<endl;

cout<<"area: "<<circ2.area()<<endl;

cout<<"circumference: "<<circ2.circumference()<<endl;

cout<<"distance from the origin: "<<circ2.distanceToOrigin()<<endl<<endl;

if(circ1.intersect(circ2))

cout<<"The circles intersect."<<endl;

else

cout<<"The circles does not intersect."<<endl;

}

void CircleDriver::run()

{

obtainCircles();

printCircleInfo();

}

3) main.cpp

#include "class_circle.h"

int main()

{

CircleDriver myDriver;

myDriver.run();

return(0);

}

Please refer below output for reference

Enter xPos yPos circle 1 :0.0 50.0

Enter radius for circle 1 :100.0

Enter xPos yPos circle 2 :50.0 0.0

Enter radius for circle 2 :100.0

Information for Circle 1:

location:( 0 ,50 )

diameter: 200

area: 31415.9

circumference: 628.318

distance from the origin: 50

Information for Circle 2:

location:( 50 ,0 )

diameter: 200

area: 31415.9

circumference: 628.318

distance from the origin: 50

The circles intersect.

Process returned 0 (0x0) execution time : 18.245 s

Press any key to continue.

8 0
4 years ago
A B-tree can be ____ in three ways: inorder, preorder, and postorder.
Harman [31]

Answer:

traversed

Explanation:

B-tree is a self balancing tree and it is a data structure which allows insertion, deletion operations.

inorder, preorder and post order, they all are traversing algorithms in the tree data structure.

inorder: first print left, then root and then right.

preorder: first print root, then left and then right.

postorder: first print left, then right and then root.

Therefore, the answer is traversed.

4 0
3 years ago
Sales Prediction A company has determined that its annual profit is typically 23 percent of total sales. Design a program that a
Brut [27]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

 This question is about calculating the profit from the annual projected sales of a company.      

You can calculate the profit by using the following formula:

profit = annual projected sales x annual profit.

So, the code (solution) is given below, however, it is noted that this program is written in c++ language.

#include <iostream>

using namespace std;

int main()

{

   long double annualProfitPercentage=0.23;// annual profit 23%.

   long double projectedSales,profit;/* declare variable to store the annual projected sales and profit */

   cout<<"*********Sales Prediction*********\n"; //output heading

   cout<<"\nEnter the project amount of total sales: ";// take input from user

   cin>>projectedSales;// store input into variable projectedSales

   

   profit = projectedSales *  annualProfitPercentage;//calculate profit

   cout<<"Profit is: "<<profit;//print result

   

   

   

   return 0;

}

3 0
3 years ago
If AX contains hex information 2111 and BX contains hex information 4333, what is the content in BX in hex format after performi
il63 [147K]

Answer:

BX will hold 6444 in hex format.

Explanation:

From the given information:

We need to understand that AX is 64-bit hex number.

When performing ASM instruction (Assembly language instruction),

The addition of 2111 + 4333 = 6444 will be in BX.

This is because ADD, BX, AX will add bx and ax; and the final outcome will be stored in BX.

∴

BX will hold 6444 in hex format.

7 0
3 years ago
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