由前N个自然数组成的集合的所有子集的总和

给定N，并且ff(N)= f(1)+ f(2)+……+ f(N)，其中f(k)是由前k个自然数组成的集合的所有子集的总和。任务是找到ff(N)模1000000007。
例子：

Input: 2 
Output: 7 
f(1) + f(2) 
f(1) = 1 = 1 
f(2) = 1 + 2 + {1 + 2} = 6
Input: 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 
 

方法：找到将要形成的序列模式。 f(1)，f(2)，f(3)的值分别为1、6和31。让我们找到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.

因此ff(N)将是

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
.
.
.

形成的级数为1、7、31、111…。有一个公式为2 ^ n *(n ^ 2 + n + 2)– 1 。其中，N从零开始。
下面是上述方法的实现。

C++

// C++ program to find Sum of all
// subsets of a set formed by
// first N natural numbers | Set-2
#include 
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 substraction 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;
}

Java

// 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 substraction
    // 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

#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 substraction 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#

// 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
    // substraction 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

 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 substraction
    // 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)
?>

Javascript

// 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 substraction 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

输出：

111

时间复杂度： O(log n)。