write the program to display all numbers between m and n input the keyboard check and display the nos that are prefect square are said t
Answers
Answer:
C program to display all numbers between m and n input from the keyboard (where m0, n>0), check and print the numbers that are a perfect square
Explanation:
1= 1^2; 4= 2^2; 9= 3^2; 16= 4^2; 25= 5^21=12;4=22;9=32;16=42;25=52
Hence, the perfect squares are numbers : 1, 4, 9, 16, 25, etc.
C Program To Find Perfect Squares Between Two Numbers
#include<stdio.h>
#include<math.h>
int main(){
int i,j,m,n;
float ps;
printf("enter the first number\n");
scanf("%d",&m);
printf("enter the last number\n");
scanf("%d",&n);
printf("perfect squares between %d and %d:",m,n);
for(j=m+1;j<n;j++){
ps=sqrt(j);
i=(int)ps;
if(i==ps){
printf("%d\n",j);
}
}
return 0;
}
Output
enter the first number
5
enter the last number
100
perfect squares between 5 and 100: 9
16
25
36
49
64
81
Question:-
Write a program to display all numbers between m and n taking input from keyboard. Check and display the numbers that are perfect squares.
Program:-
import java.util.*;
class Program
{
static boolean isPerfectSquare(int n)
{
double s=Math.sqrt(n);
return (s==(int)s);
}
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
System.out.print("Enter the starting range: ");
int m=sc.nextInt();
System.out.print("Enter the ending range: ");
int n=sc.nextInt();
// if m>n then we will swap them.
if(m>n)
{
int t=m;
m=n;
n=t;
}
for(int i=m;i<=n;i++)
{
if(isPerfectSquare(i))
System.out.println(i+" is a Perfect Square.");
}
}
}