给定两个给定的数字a和b,其中1 <= a <= b,找到a和b之间的完美平方数(包括a和b)。
例子
Input : a = 3, b = 8
Output : 1
The only perfect in given range is 4.
Input : a = 9, b = 25
Output : 3
The three squares in given range are 9,
16 and 25
方法1 :一种幼稚的方法是检查a和b之间的所有数字(包括a和b),并在遇到完美正方形时将计数加一。
下面是上述想法的实现:
C++
// A Simple Method to count squares between a and b
#include
using namespace std;
int countSquares(int a, int b)
{
int cnt = 0; // Initialize result
// Traverse through all numbers
for (int i = a; i <= b; i++)
// Check if current number 'i' is perfect
// square
for (int j = 1; j * j <= i; j++)
if (j * j == i)
cnt++;
return cnt;
}
// Driver code
int main()
{
int a = 9, b = 25;
cout << "Count of squares is "
<< countSquares(a, b);
return 0;
}
Java
// Java program to count squares between a and b
class CountSquares {
static int countSquares(int a, int b)
{
int cnt = 0; // Initialize result
// Traverse through all numbers
for (int i = a; i <= b; i++)
// Check if current number 'i' is perfect
// square
for (int j = 1; j * j <= i; j++)
if (j * j == i)
cnt++;
return cnt;
}
}
// Driver Code
public class PerfectSquares {
public static void main(String[] args)
{
int a = 9, b = 25;
CountSquares obj = new CountSquares();
System.out.print("Count of squares is " + obj.countSquares(a, b));
}
}
Python
# Python program to count squares between a and b
def CountSquares(a, b):
cnt = 0 # initialize result
# Traverse through all numbers
for i in range (a, b + 1):
j = 1;
while j * j <= i:
if j * j == i:
cnt = cnt + 1
j = j + 1
i = i + 1
return cnt
# Driver Code
a = 9
b = 25
print "Count of squares is:", CountSquares(a, b)
C#
// C# program to count squares
// between a and b
using System;
class GFG {
// Function to count squares
static int countSquares(int a, int b)
{
// Initialize result
int cnt = 0;
// Traverse through all numbers
for (int i = a; i <= b; i++)
// Check if current number
// 'i' is perfect square
for (int j = 1; j * j <= i; j++)
if (j * j == i)
cnt++;
return cnt;
}
// Driver Code
public static void Main()
{
int a = 9, b = 25;
Console.Write("Count of squares is " + countSquares(a, b));
}
}
// This code is contributed by Sam007
PHP
Javascript
// A Simple Method to count squares
//between a and b
function countSquares(a, b)
{
let cnt = 0;
// Traverse through all numbers
for (let i = a; i <= b; i++)
// Check if current number
// 'i' is perfect square
for (let j = 1; j * j <= i;j++)
if (j * j == i)
cnt++;
return cnt;
}
// Driver code
let a = 9;
let b = 25;
document.write( "Count of squares is ",
countSquares(a, b));
// This code is contributed by sravan.
Count of squares is 3
C++
// An Efficient Method to count squares between a and b
#include
using namespace std;
// An efficient solution to count square between a
// and b
int countSquares(int a, int b)
{
return (floor(sqrt(b)) - ceil(sqrt(a)) + 1);
}
// Driver code
int main()
{
int a = 9, b = 25;
cout << "Count of squares is "
<< countSquares(a, b);
return 0;
}
Java
// An Efficient method to count squares between
// a and b
class CountSquares {
double countSquares(int a, int b)
{
return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1);
}
}
// Driver Code
public class PerfectSquares {
public static void main(String[] args)
{
int a = 9, b = 25;
CountSquares obj = new CountSquares();
System.out.print("Count of squares is " + (int)obj.countSquares(a, b));
}
}
Python
# An Efficient Method to count squares between a
# and b
import math
def CountSquares(a, b):
return (math.floor(math.sqrt(b)) - math.ceil(math.sqrt(a)) + 1)
# Driver Code
a = 9
b = 25
print "Count of squares is:", int(CountSquares(a, b))
C#
// C# program for efficient method
// to count squares between a & b
using System;
class GFG {
// Function to count squares
static double countSquares(int a, int b)
{
return (Math.Floor(Math.Sqrt(b)) - Math.Ceiling(Math.Sqrt(a)) + 1);
}
// Driver Code
public static void Main()
{
int a = 9, b = 25;
Console.Write("Count of squares is " + (int)countSquares(a, b));
}
}
// This code is contributed by Sam007.
PHP
Javascript
// A Simple Method to count squares
//between a and b
function countSquares(a, b)
{
return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1);
}
// Driver code
let a = 9;
let b = 25;
document.write( "Count of squares is ",
countSquares(a, b));
// This code is contributed by sravan.
输出 :
Count of squares is 3
该解决方案的时间复杂度上限为O((ba)* sqrt(b))。
方法2(有效)我们可以简单地取’a’的平方根和’b’的平方根,然后使用来计算它们之间的理想平方
floor(sqrt(b)) - ceil(sqrt(a)) + 1
We take floor of sqrt(b) because we need to consider
numbers before b.
We take ceil of sqrt(a) because we need to consider
numbers after a.
For example, let b = 24, a = 8. floor(sqrt(b)) = 4,
ceil(sqrt(a)) = 3. And number of squares is 4 - 3 + 1
= 2. The two numbers are 9 and 16.
下面是上述想法的实现:
C++
// An Efficient Method to count squares between a and b
#include
using namespace std;
// An efficient solution to count square between a
// and b
int countSquares(int a, int b)
{
return (floor(sqrt(b)) - ceil(sqrt(a)) + 1);
}
// Driver code
int main()
{
int a = 9, b = 25;
cout << "Count of squares is "
<< countSquares(a, b);
return 0;
}
Java
// An Efficient method to count squares between
// a and b
class CountSquares {
double countSquares(int a, int b)
{
return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1);
}
}
// Driver Code
public class PerfectSquares {
public static void main(String[] args)
{
int a = 9, b = 25;
CountSquares obj = new CountSquares();
System.out.print("Count of squares is " + (int)obj.countSquares(a, b));
}
}
Python
# An Efficient Method to count squares between a
# and b
import math
def CountSquares(a, b):
return (math.floor(math.sqrt(b)) - math.ceil(math.sqrt(a)) + 1)
# Driver Code
a = 9
b = 25
print "Count of squares is:", int(CountSquares(a, b))
C#
// C# program for efficient method
// to count squares between a & b
using System;
class GFG {
// Function to count squares
static double countSquares(int a, int b)
{
return (Math.Floor(Math.Sqrt(b)) - Math.Ceiling(Math.Sqrt(a)) + 1);
}
// Driver Code
public static void Main()
{
int a = 9, b = 25;
Console.Write("Count of squares is " + (int)countSquares(a, b));
}
}
// This code is contributed by Sam007.
的PHP
Java脚本
// A Simple Method to count squares
//between a and b
function countSquares(a, b)
{
return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1);
}
// Driver code
let a = 9;
let b = 25;
document.write( "Count of squares is ",
countSquares(a, b));
// This code is contributed by sravan.
输出 :
Count of squares is 3
该解决方案的时间复杂度为O(Log b)。对于n的平方根的典型实现需要花费等于O(Log n)的时间。