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GenaCL600 [577]
3 years ago
9

. Consider the problem of finding the largest element in a list of n elements. What will be the basic operation of an algorithm

to solve this problem?
assignment

addition

looping

comparesn
Computers and Technology
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

Comparison.

Explanation:

When we have to find a largest in a list of n elements.First we have to iterate over the list so we can access all the elements of the list in one go.Then to find the largest element in the list we have to initialize a variable outside the loop with the minimum value possible and in the loop compare each element with this value,if the element is greater than the variable assign the element to the variable.Then the loop will find the largest element and it will be the variable.

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Polygon transform (25 points). Write a library of static methods that performs various geometric transforms on polygons. Mathema
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Answer:

import java.lang.*;               //for using Math.cos() and Math.sin() and Math.toRadians() funcitons in rotate method

public class PolygonTransform

{

  public static double[] copy(double[] array)

  {

      double[] newArray = new double[array.length];

      for(int i=0;i<array.length;i++)

      {

          newArray[i] = array[i];

          return newArray;

      }

  }

  public static void scale(double[]x,double[]y,double alpha)

  {

      for(int i=0;i<x.length;i++)

      {

          x[i] *= alpha;

          y[i] *= alpha;

      }

  }

  public static void translate(double[]x,double[]y,double dx,double dy)

  {

      for(int i=0;i<x.length;i++)

      {

          x[i] += dx;

          y[i] += dy;

      }

  }

  public static void rotate(double[] x, double[]y,double theta)

  {

      double rad = Math.toRadians(theta);

      double temp;

      for(int i=0;i<x.length;i++)

      {

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  public static void main(String args[])

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      //This is just implementing you testcase discripted.

      double[]x = new double[4];

      double[]y = new double[4];

      StdDraw.setScale(-5.0, +5.0);

      x = { 0, 1, 1, 0 };

      y = { 0, 0, 2, 1 };

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      StdDraw.polygon(x, y);

      scale(x, y, alpha);

      StdDraw.setPenColor(StdDraw.BLUE);

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      StdDraw.setScale(-5.0, +5.0);

      x = { 0, 1, 1, 0 };

      y = { 0, 0, 2, 1 };

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      StdDraw.setPenColor(StdDraw.BLUE);

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      StdDraw.setScale(-5.0, +5.0);

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      y = { 0, 0, 2, 1 };

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      rotate(x, y, theta);

      StdDraw.setPenColor(StdDraw.BLUE);

      StdDraw.polygon(x, y);

  }

}

Explanation:

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Explanation:

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Amplifier:

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Answer:

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The dis-advantages of exposure value bracketing are:

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kolezko [41]
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Answer:

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Explanation:

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