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FinnZ [79.3K]
3 years ago
5

Program to calculate series 10+9+8+...+n" in java with output

Computers and Technology
1 answer:
avanturin [10]3 years ago
5 0

Answer:

import java.util.Scanner;

class Main {

 public static int calcSeries(int n) {

   int sum = 0;

   for(int i=10; i>=n; i--) {

     sum += i;

   }

   return sum;

 }

 public static void main(String[] args) {

   Scanner reader = new Scanner(System.in);

   int n = 0;

   do {

     System.out.print("Enter n: ");

     n = reader.nextInt();

     if (n >= 10) {

       System.out.println("Please enter a value lower than 10.");

     }

   } while (n >= 10);

   reader.close();

   System.out.printf("sum: %d\n", calcSeries(n));

 }

}

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____ is a method of querying and reporting that takes data from standard relational databases, calculates and summarizes the dat
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We will use linear interpolation in a C program to estimate the population of NJ between the years of the census, which takes pl
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Answer:

See explaination for program code

Explanation:

program code below:

#include<stdio.h>

int main()

{

//file pointer to read from the file

FILE *fptr;

//array to store the years

int years[50];

//array to store the population

float population[50];

int n = 0, j;

//variables for the linear interpolation formula

float x0,y0,x1,y1,xp,yp;

//opening the file

fptr = fopen("njpopulation.dat", "r");

if (fptr != NULL)

{

//reading data for the file

while(fscanf(fptr, "%d %f", &years[n], &population[n]) == 2)

n++;

//prompting the user

int year = -1;

printf("Enter the years for which the population is to be estimated. Enter 0 to stop.\n\n");

while(year != 0)

{

printf("Enter the year (1790 - 2010) : ");

scanf("%d", &year);

if (year >= 1790 && year <= 2010)

{

//calculating the estimation

xp = year;

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

{

if (year == years[j])

{

//if the year is already in the table no nedd for calculation

yp = population[j];

}

else if (j != n - 1)

{

//finding the surrounding census years

if (year > years[j] && year < years[j + 1])

{

//point one

x0 = years[j];

y0 = population[j];

//point two

x1 = years[j + 1];

y1 = population[j + 1];

//interpolation formula

yp = y0 + ((y1-y0)/(x1-x0)) * (xp - x0);

}

}

}

printf("\nEstimated population for year %d : %.2f\n\n", year, yp);

}

else if (year != 0)

{

printf("\nInvalid chioce!!\n\n");

}

}

printf("\nAborting!!");

}

else

{

printf("File cannot be opened!!");

}

close(fptr);

return 0;

}

OUTPUT :

Enter the years for which the population is to be estimated. Enter 0 to stop.

Enter the year (1790 - 2010) : 1790

Estimated population for year 1790 : 184139.00

Enter the year (1790 - 2010) : 1855

Estimated population for year 1855 : 580795.00

Enter the year (1790 - 2010) : 2010

Estimated population for year 2010 : 8791894.00

Enter the year (1790 - 2010) : 4545

Invalid chioce!!

Enter the year (1790 - 2010) : 1992

Estimated population for year 1992 : 7867020.50

Enter the year (1790 - 2010) : 0

Aborting!!

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