Given N, and ff(N) = f(1) + f(2) + …… + f(N), where f(k) is the sum of all subsets of a set formed by first k natural numbers. The task is to find ff(N) modulo 1000000007.
Examples:
Input: 2
Output: 7
f(1) + f(2)
f(1) = 1 = 1
f(2) = 1 + 2 + {1 + 2} = 6Input: 3
Output: 31
f(1) + f(2) + f(3)
f(1) = 1 = 1
f(2) = 1 + 2 + {1 + 2} = 6
f(3) = 1 + 2 + 3 + {1 + 2} + {2 + 3} + {1 + 3} + {1 + 2 + 3} = 24
Approach: Find a pattern of the sequence that will form. The values of f(1), f(2), f(3) are 1, 6 and 31 respectively. Let’s find f(4).
f(4) = 1 + 2 + 3 + 4 + {1 + 2} + {1 + 3} + {1 + 4} + {2 + 3} + {2 + 4} + {3 + 4} + {1 + 2 + 3} + {1 + 2 + 4} + {1 + 3 + 4} + {2 + 3 + 4} + {1 + 2 + 3 + 4} = 80.
Hence ff(N) will be
ff(1) = f(1) = 1 ff(2) = f(1) + f(2) = 7 ff(3) = f(1) + f(2) + f(3) = 31 ff(4) = f(1) + f(2) + f(3) + f(4) = 111 . . .
The series formed is 1, 7, 31, 111… There exists a formula for it which is 2^n*(n^2 + n + 2) – 1. where, N is starting from zero.
Below is the implementation of the above approach.
// C++ program to find Sum of all // subsets of a set formed by // first N natural numbers | Set-2 #include <bits/stdc++.h> using namespace std;
// modulo value #define mod (int)(1e9 + 7) // Iterative Function to calculate (x^y)%p in O(log y) int power( int x, int y, int p)
{ int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with the result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
} // function to find ff(n) int check( int n)
{ // In formula n is starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod * ans % mod) % mod;
// adding modulo while subtraction is very necessary
// otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
} // Driver code int main()
{ int n = 4;
// function call
cout << check(n) << endl;
return 0;
} |
// C program to find Sum of all // subsets of a set formed by // first N natural numbers | Set-2 #include <stdio.h> // modulo value #define mod (int)(1e9 + 7) // Iterative Function to calculate (x^y)%p in O(log y) int power( int x, int y, int p)
{ int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with the result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
} // function to find ff(n) int check( int n)
{ // In formula n is starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod * ans % mod) % mod;
// adding modulo while subtraction is very necessary
// otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
} // Driver code int main()
{ int n = 4;
// function call
printf ( "%d\n" ,check(n));
return 0;
} // This code is contributed by kothavvsaakash. |
// Java program to find Sum of all // subsets of a set formed by // first N natural numbers | Set-2 class Geeks {
// Iterative Function to calculate // (x^y)%p in O(log y) static int power( int x, int y, int p)
{ // Initialize result
int res = 1 ;
// Update x if it is more
// than or equal to p
x = x % p;
while (y > 0 ) {
// If y is odd, multiply x
// with the result
if (y != 0 )
res = (res * x) % p;
// y must be even now
// y = y / 2
y = y >> 1 ;
x = (x * x) % p;
}
return res;
} // function to find ff(n) static int check( int n)
{ // modulo value
int mod = ( int )(1e9 + 7 );
// In formula n is
// starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2 ;
if (ans >= mod)
ans %= mod;
ans = (power( 2 , n, mod) % mod *
ans % mod) % mod;
// adding modulo while subtraction
// is very necessary otherwise it
// will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
} // Driver Code public static void main(String args[])
{ int n = 4 ;
// function call
System.out.println(check(n));
} } // This code is contributed by ankita_saini |
#Python3 program to find Sum of all # subsets of a set formed by # first N natural numbers | Set-2 # modulo value mod = ( int )( 1e9 + 7 )
# Iterative Function to calculate (x^y)%p in O(log y) def power(x,y,p):
res = 1 # Initialize result
x = x % p # Update x if it is more than or
# equal to p
while (y > 0 ):
# If y is odd, multiply x with the result
if (y & 1 ):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
# function to find ff(n) def check(n):
# In formula n is starting from zero
n = n - 1
# calculate answer using
# formula 2^n*(n^2 + n + 2) - 1
ans = n * n
# whenever answer is greater than
# or equals to mod then modulo it.
if (ans > = mod):
ans % = mod
ans + = n + 2
if (ans > = mod):
ans % = mod
ans = ( pow ( 2 , n, mod) % mod * ans % mod) % mod
# adding modulo while subtraction is very necessary
# otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod
return ans
#Driver code if __name__ = = '__main__' :
n = 4
# function call print (check(n))
# This code is contributed by ash264 |
// C# program to find Sum // of all subsets of a set // formed by first N natural // numbers | Set-2 using System;
class GFG
{ // Iterative Function // to calculate (x^y)%p // in O(log y) static int power( int x, int y,
int p)
{ // Initialize result
int res = 1;
// Update x if it is more
// than or equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply
// x with the result
if (y != 0)
res = (res * x) % p;
// y must be even
// now y = y / 2
y = y >> 1;
x = (x * x) % p;
}
return res;
} // function to find ff(n) static int check( int n)
{ // modulo value
int mod = ( int )(1e9 + 7);
// In formula n is
// starting from zero
n--;
// calculate answer
// using formula
// 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is
// greater than or equals
// to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod *
ans % mod) % mod;
// adding modulo while
// subtraction is very
// necessary otherwise it
// will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
} // Driver Code public static void Main(String []args)
{ int n = 4;
// function call
Console.WriteLine(check(n));
} } // This code is contributed // by ankita_saini |
<?php // PHP program to find Sum // of all subsets of a set // formed by first N natural // numbers | Set-2 // modulo value // Iterative Function to // calculate (x^y)%p in O(log y) function power( $x , $y , $p )
{ $res = 1; // Initialize result
$x = $x % $p ; // Update x if it
// is more than or
// equal to p
while ( $y > 0)
{
// If y is odd, multiply
// x with the result
if ( $y & 1)
$res = ( $res * $x ) % $p ;
// y must be even now
$y = $y >> 1; // y = y/2
$x = ( $x * $x ) % $p ;
}
return $res ;
} // function to find ff(n) function check( $n )
{ $mod = 1e9+7;
// In formula n is
// starting from zero
$n --;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
$ans = $n * $n ;
// whenever answer is greater
// than or equals to mod then
// modulo it.
if ( $ans >= $mod )
$ans %= $mod ;
$ans += $n + 2;
if ( $ans >= $mod )
$ans %= $mod ;
$ans = (power(2, $n , $mod ) %
$mod * $ans %
$mod ) % $mod ;
// adding modulo while subtraction
// is very necessary otherwise it
// will cause wrong answer
$ans = ( $ans - 1 + $mod ) % $mod ;
return $ans ;
} // Driver code $n = 4;
// function call echo check( $n ) . "\n" ;
// This code is contributed // by Akanksha Rai(Abby_akku) ?> |
<script> // javascript program to find Sum of all // subsets of a set formed by // first N natural numbers | Set-2 // modulo value const mod = (1e9 + 7); // Iterative Function to calculate (x^y)%p in O(log y) function power( x, y, p)
{ let res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with the result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
} // function to find ff(n) function check( n)
{ // In formula n is starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
let ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod * ans % mod) % mod;
// adding modulo while subtraction is very necessary
// otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
} // Driver code let n = 4;
// function call
document.write(check(n));
// This code is contributed by aashish1995 </script> |
111
Time complexity: O(log n).
Auxiliary Space: O(1)