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sp2606 [1]
3 years ago
13

A python script to print fibonacci series first 20 elements

Computers and Technology
1 answer:
madreJ [45]3 years ago
5 0

Answer:

   nterms =20

   # first two terms

   n1, n2 = 0, 1

   count = 0

   # check if the number of terms is valid

   if nterms <= 0:

      print("Please enter a positive integer")

   elif nterms == 1:

      print("Fibonacci sequence upto",nterms,":")

      print(n1)

   else:

      print("Fibonacci sequence:")

      while count < nterms:

          print(n1)

          nth = n1 + n2

          # update values

          n1 = n2

          n2 = nth

          count += 1

Explanation:

Here, we store the number of terms in nterms. We initialize the first term to 0 and the second term to 1.

If the number of terms is more than 2, we use a while loop to find the next term in the sequence by adding the preceding two terms. We then interchange the variables (update it) and continue on with the process.

<em>output:</em>

Fibonacci sequence:

0

1

1

2

3

5

8

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import java.util.*;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

class GFG

{

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      for (int i = 0; i < n; i++)

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  public static double findMode(double a[], int n)

{

// The output array b[] will

// have sorted array

//int []b = new int[n];

 

// variable to store max of

// input array which will

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double max = Arrays.stream(a).max().getAsDouble();

 

// auxiliary(count) array to

// store count. Initialize

// count array as 0. Size

// of count array will be

// equal to (max + 1).

double t = max + 1;

double[] count = new double[(int)t];

for (int i = 0; i < t; i++)

{

count[i] = 0;

}

 

// Store count of each element

// of input array

for (int i = 0; i < n; i++)

{

count[(int)(10*a[i])]++;

}

 

// mode is the index with maximum count

double mode = 0;

double k = count[0];

for (int i = 1; i < t; i++)

{

if (count[i] > k)

{

k = count[i];

mode = i;

}

}

return mode;

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public static double findSmallest(double [] A, int total){

Arrays.sort(A);

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}

 

public static void printAboveAvg(double arr[], int n)

{

 

// Find average

double avg = 0;

for (int i = 0; i < n; i++)

avg += arr[i];

avg = avg / n;

 

// Print elements greater than average

for (int i = 0; i < n; i++)

if (arr[i] > avg)

System.out.print(arr[i] + " ");

System.out.println();

}

 

public static void printrand(double [] A, int n){

Arrays.sort(A);

for(int i=0;i<n;i++){

System.out.print(A[0]+"/t");

}

System.out.println();

}

 

public static void printHist(double [] arr, int n) {

 

for (double i = 1.0; i >= 0; i-=0.1) {

System.out.print(i+" | ");

for (int j = 0; j < n; j++) {

 

// if array of element is greater

// then array it print x

if (arr[j] >= i)

System.out.print("x");

 

// else print blank spaces

else

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System.out.println();

}

// print last line denoted by ----

for(int l = 0; l < (n + 3); l++){    

System.out.print("---");

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System.out.println();

System.out.print(" ");

 

for (int k = 0; k < n; k++) {

System.out.print(arr[k]+" ");

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}

  // Driver program

  public static void main(String args[]) throws IOException

  {

      //Enter data using BufferReader

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

double [] A = new double[100];

int i=0;

System.out.println("Enter the numbers(0.0-1.0) /n Enter 9 if u have entered the numbers. /n");

do

{

A[i++]=Double.parseDouble(br.readLine());

}while(A[i-1]==9);

      i--;

      System.out.println("Average = " + findMean(A,i) );

      System.out.println("Median = " + findMedian(A,i));

      System.out.println("Element that occured most frequently = " + findMode(A,i));

      System.out.println("number closest to 0.0 =" + findSmallest(A,i));

      System.out.println("Numbers that are greater than the average are follows:");

      printAboveAvg(A,i);

      System.out.println("Numbers in random order are as follows:");

      printrand(A,i);

      System.out.println("Histogram is bellow:");

      printHist(A,i);

  }

}

Explanation:

3 0
3 years ago
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