Given a range [L, R], and a prime number P. We are required to find the sum of the highest power of P in all numbers from L to R.
Examples:
Input : L = 1, R = 10, P = 2 Output : 8 There are 10 integers in the range, and: In 1 the highest power of 2 is 0 In 2 the highest power of 2 is 1 In 3 the highest power of 2 is 0 Similarly the highest powers of 2 in 4, 5, 6, 7, 8, 9, 10 are 2, 0, 1, 0, 3, 0, 1 respectively. Input : L = 10, R = 20, P = 7 Output : 1 There are 11 integers in the range, and: In 10 the highest power of 7 is 0 In 11 the highest power of 7 is 0 In 12 the highest power of 7 is 0 Similarly the highest power of 7 in 13, 14, 15, 16, 17, 18, 19, 20 and 10 is 0, 1, 0, 0, 0, 0, 0 and 0 respectively.
Native Approach : Run a loop from L to R. For every number, check if it is divisible by P. If yes, then find the largest power of P that divides current number. If yes, then add this value to result.
C++
// Simple CPP Program to find sum of largest // divisible powers of P in [L, R] #include <bits/stdc++.h> using namespace std;
int sumOfDivisblePowers(int L, int R, int P) {
// Traverse through all numbers from L to R
int res = 0;
for (int i = L; i <= R; i++)
{
// If P divides current number x, add
// largest power of P that divides x.
int x = i;
while (x % P == 0)
{
res++;
x /= P;
}
}
return res;
} // Driver code int main() {
int L = 1, R = 10, P = 2;
cout << sumOfDivisblePowers(L, R, P);
return 0;
} |
Java
// Simple Java Program to find sum of largest // divisible powers of P in [L, R] class GFG
{ // Utility Function
static int sumOfDivisblePowers(int L, int R, int P)
{
// Traverse through all numbers from L to R
int res = 0;
for (int i = L; i <= R; i++)
{
// If P divides current number x, add
// largest power of P that divides x.
int x = i;
while (x % P == 0)
{
res++;
x /= P;
}
}
return res;
}
// Driver code
public static void main (String[] args)
{
int L = 1, R = 10, P = 2;
System.out.println(sumOfDivisblePowers(L, R, P));
}
} // This code is contributed by Anant Agarwal. |
Python3
# Simple Python 3 Program to find sum of largest # divisible powers of P in [L, R] def sumOfDivisblePowers(L, R, P):
# Traverse through all numbers from L to R
res = 0
for i in range(L, R + 1):
# If P divides current number x, add
# largest power of P that divides x.
x = i
while (x % P == 0):
res += 1
x /= P
return res
# Driver code L = 1
R = 10
P = 2
print(sumOfDivisblePowers(L, R, P))
# This code is contributed by Smitha Dinesh Semwal |
C#
// Simple C# Program to find sum of // largest divisible powers of P in // [L, R] using System;
class GFG
{ // Utility Function
static int sumOfDivisblePowers(int L, int R, int P)
{
// Traverse through all numbers
// from L to R
int res = 0;
for (int i = L; i <= R; i++)
{
// If P divides current number,
// x, add largest power of P
// that divides x.
int x = i;
while (x % P == 0)
{
res++;
x /= P;
}
}
return res;
}
// Driver code
public static void Main ()
{
int L = 1, R = 10, P = 2;
Console.WriteLine(sumOfDivisblePowers(L, R, P));
}
} // This code is contributed by vt_m. |
PHP
<?php // Simple PHP Program to // find sum of largest // divisible powers of // P in [L, R] // function return result function sumOfDivisblePowers($L, $R, $P)
{ // Traverse through all
// numbers from L to R
$res = 0;
for ($i = $L; $i <= $R; $i++)
{
// If P divides current
// number x, add
// largest power of
// P that divides x.
$x = $i;
while ($x % $P == 0)
{
$res++;
$x /= $P;
}
} return $res;
} // Driver Code
$L = 1;
$R = 10;
$P = 2;
echo sumOfDivisblePowers($L, $R, $P);
// This code is contributed by ajit ?> |
Javascript
<script> // Simple Javascript Program to find sum of
// largest divisible powers of P in
// [L, R]
// Utility Function
function sumOfDivisblePowers(L, R, P)
{
// Traverse through all numbers
// from L to R
let res = 0;
for (let i = L; i <= R; i++)
{
// If P divides current number,
// x, add largest power of P
// that divides x.
let x = i;
while (x % P == 0)
{
res++;
x = parseInt(x / P, 10);
}
}
return res;
}
let L = 1, R = 10, P = 2;
document.write(sumOfDivisblePowers(L, R, P));
// This code is contributed by divyeshrabadiya07.
</script> |
Output:
8
Efficient approach: The idea is to use to Legendre’s formula.
The largest possible power of P that divides factorial of a number x is ?x/p? + ?x/(p2)? + ?x/(p3)? + ……
Using above, we can find the largest power of P that divides R! (or 1 * 2 * 3…. R).
Sum of largest powers from L to R = Largest Power of P that divides R! (or 1 * 2 … L-1 * L * … R) – Largest Power of P that divides (L-1)! (or 1 * 2 … L-1)
C++
// Efficient CPP Program to find sum of // largest divisible powers of P in [L, R] #include <bits/stdc++.h> using namespace std;
// Implements Lagrange's theorem (Finds largest // power of P that divides x! int largestPower(int x, int P)
{ // Calculate x/p + x/(p^2) + x/(p^3) + ....
int res = 0;
while (x)
{
x /= P;
res += x;
}
return res;
} // Returns sum of largest powers of P that divide // numbers in range from L to R. int sumOfDivisblePowers(int L, int R, int P) {
return largestPower(R, P) - largestPower(L-1, P);
} // Driver code int main() {
int L = 1, R = 10, P = 2;
cout << sumOfDivisblePowers(L, R, P);
return 0;
} |
Java
// Efficient CPP Program to find sum of // largest divisible powers of P in [L, R] class GFG
{ // Implements Lagrange's theorem
// (Finds largest power of P that
// divides x!
static int largestPower(int x, int P)
{
// Calculate x/p + x/(p^2) + x/(p^3) + ....
int res = 0;
while(x != 0)
{
x /= P;
res += x;
}
return res;
}
// Returns sum of largest powers
// of P that divide numbers in
// range from L to R.
static int sumOfDivisblePowers(int L, int R, int P)
{
return largestPower(R, P) - largestPower(L-1, P);
}
// Driver code
public static void main (String[] args)
{
int L = 1, R = 10, P = 2;
System.out.println(sumOfDivisblePowers(L, R, P));
}
} // This code is contributed by Anant Agarwal. |
Python3
# Efficient Python 3 Program to find sum of # largest divisible powers of P in [L, R] # Implements Lagrange's theorem (Finds largest # power of P that divides x! def largestPower(x, P):
# Calculate x/p + x/(p^2) + x/(p^3) + ....
res = 0
while (x):
x = int(x / P)
res += x
return res
# Returns sum of largest powers of P that divide # numbers in range from L to R. def sumOfDivisblePowers( L, R, P) :
return largestPower(R, P) - largestPower(L-1, P)
# Driver code L = 1
R = 10
P = 2
print(sumOfDivisblePowers(L, R, P))
# This code is contributed by Smitha Dinesh Semwal |
C#
// Efficient C# Program to find sum of // largest divisible powers of P in [L, R] using System;
class GFG {
// Implements Lagrange's theorem (Finds
// largest power of P that divides x!
static int largestPower(int x, int P)
{
// Calculate x/p + x/(p^2) +
// x/(p^3) + ....
int res = 0;
while (x>0)
{
x /= P;
res += x;
}
return res;
}
// Returns sum of largest powers of
// P that divide numbers in range
// from L to R.
static int sumOfDivisblePowers(int L,
int R, int P)
{
return largestPower(R, P)
- largestPower(L-1, P);
}
// Driver Program to test above
// function
static void Main()
{
int L = 1, R = 10, P = 2;
Console.Write(
sumOfDivisblePowers(L, R, P));
}
} // This code is contributed by Anuj_67 |
PHP
<?php // Efficient php Program to find sum of // largest divisible powers of P in [L, R] // Implements Lagrange's theorem (Finds largest // power of P that divides x! function largestPower($x, $P)
{ // Calculate x/p + x/(p^2) + x/(p^3) + ....
$res = 0;
while ($x)
{
$x = floor($x/$P);
$res += $x;
}
return $res;
} // Returns sum of largest powers of P that divide // numbers in range from L to R. function sumOfDivisblePowers($L, $R, $P) {
return largestPower($R, $P) - largestPower($L-1, $P);
} // Driver code $L = 1;
$R = 10;
$P = 2;
echo sumOfDivisblePowers($L, $R, $P);
//This code is contributed by mits. ?> |
Javascript
<script> // Efficient Javascript Program to find sum of
// largest divisible powers of P in [L, R]
// Implements Lagrange's theorem (Finds
// largest power of P that divides x!
function largestPower(x, P)
{
// Calculate x/p + x/(p^2) +
// x/(p^3) + ....
let res = 0;
while (x>0)
{
x = parseInt(x / P, 10);
res += x;
}
return res;
}
// Returns sum of largest powers of
// P that divide numbers in range
// from L to R.
function sumOfDivisblePowers(L, R, P)
{
return largestPower(R, P) - largestPower(L-1, P);
}
let L = 1, R = 10, P = 2;
document.write(sumOfDivisblePowers(L, R, P));
// This code is contributed by mukesh07.
</script> |
Output:
8