在给定的序列中找到数字乘积的总和

给定两个数字N和T ， $1\leq N\leq 10000000000$和$1\leq T \leq 1000$ 。任务是找到…的价值$sum = \sum_{i=1}^{i=N}\prod_{j=1}^{j=T} (i+j)$ 。
由于总和可能很大，请以10 ⁹ +7为模输出。

例子：

Input : 3 2
Output : 38
2*3 + 3*4 + 4*5 = 38

Input : 4 2
Output : 68

In the Given Sample Case n = 3 and t = 2.
sum = 2*3+3*4+4*5.
Notice that:
$1*2 = \frac{2!}{0!}$
$2*3 = \frac{3!}{1!}$
$3*4 = \frac{4!}{2!}$
$4*5 = \frac{5!}{3!}$

所以每个术语的形式$\frac{x!}{(x-t)!}$

如果我们乘以除以t！它成为了$t!*\frac{x!}{(x-t)!*t!}$

那不过是什么$t!*\;_{t}^{x}\textrm{C}$

所以， $sum = t!\;*\;\sum_{x=t+1}^{n+t}\; _{t}^{x}\textrm{C}$

但是我们知道$\sum_{x=t}^{N}\;_{t}^{x}\textrm{C}\;=\;_{t+1}^{N+1}\textrm{C}$

所以$\sum_{x=t+1}^{n+t} _{k}^{x}\textrm{C}\; =\; _{t+1}^{n+t+1}\textrm{C}-1$

所以最终表达出来是$t!*\;_{t+1}^{n+t+1}\textrm{C}-t!$

但是由于n太大，我们不能直接计算它，因此我们必须简化上面的表达式。

在简化我们得到$\frac{\prod_{i=1}^{t+1}*(n+i)}{t+1} - t!$ 。

下面是上述方法的实现

C++

// C++ program to find sum of product 
// of number in given series
#include 
using namespace std;
  
typedef long long ll;
const long long MOD = 1000000007;
  
// function to calculate (a^b)%p
ll power(ll x, unsigned long long y, ll p)
{
    ll res = 1; // Initialize result
  
    // Update x if it is more than or equal to p
    x = x % p;
  
    while (y > 0) {
  
        // If y is odd, multiply x with 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 return required answer
ll sumProd(ll n, ll t)
{
    // modulo inverse of denominator
    ll dino = power(t + 1, MOD - 2, MOD);
  
    // calculating commentator part
    unsigned long long ans = 1;
    for (ll i = n + t + 1; i > n; --i)
        ans = (ans % MOD * i % MOD) % MOD;
  
    // calculating t!
    ll tfact = 1;
    for (int i = 1; i <= t; ++i)
        tfact = (tfact * i) % MOD;
  
    // accumulating the final answer
    ans = ans * dino - tfact + MOD;
  
    return ans % MOD;
}
int main()
{
    ll n = 3, t = 2;
  
    // function call to print required sum
    cout << sumProd(n, t);
  
    return 0;
}

Java

// Java program to find sum of product 
// of number in given series
  
public class GFG {
  
     static long MOD = 1000000007;
       
    //function to calculate (a^b)%p
     static long power(long x, long y, long p)
     {
      long res = 1; // Initialize result
  
      // Update x if it is more than or equal to p
      x = x % p;
  
      while (y > 0) {
  
          // If y is odd, multiply x with result
          if ((y & 1)!= 0)
              res = (res * x) % p;
  
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
      }
  
      return res;
     }
  
     //function to return required answer
     static long sumProd(long n, long t)
     {
      // modulo inverse of denominator
      long dino = power(t + 1, MOD - 2, MOD);
  
      // calculating commentator part
      long ans = 1;
      for (long i = n + t + 1; i > n; --i)
          ans = (ans % MOD * i % MOD) % MOD;
  
      // calculating t!
      long tfact = 1;
      for (int i = 1; i <= t; ++i)
          tfact = (tfact * i) % MOD;
  
      // accumulating the final answer
      ans = ans * dino - tfact + MOD;
  
      return ans % MOD;
     }
  
     // Driver program
    public static void main(String[] args) {
          
        long n = 3, t = 2;
  
         // function call to print required sum
         System.out.println(sumProd(n, t));
    }
}

Python3

# Python 3 program to find sum of product 
# of number in given series 
  
MOD = 1000000007
  
# function to calculate (a^b)%p
def power(x, y, p) :
  
    # Initialize result
    res = 1
  
    # Update x if it is more than or equal to p
    x = x % p
  
    # If y is odd, multiply x with result 
    while y > 0 :
  
        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 return required answer
def sumProd(n, t) :
  
    # modulo inverse of denominator 
    dino = power(t + 1, MOD - 2, MOD)
  
    ans = 1
  
    # calculating commentator part 
    for i in range(n + t + 1 , n, -1) :
        ans = (ans % MOD * i % MOD) % MOD
  
    # calculating t! 
    tfact = 1
    for i in range(1, t+1) :
        tfact = (tfact * i) % MOD
  
    # accumulating the final answer 
    ans = ans * dino - tfact + MOD
  
    return ans % MOD
              
      
# Driver Code
if __name__ == "__main__" :
  
    n, t = 3, 2
  
    # function call to print required sum 
    print(sumProd(n, t))
  
# This code is contributed by ANKITRAI1

C#

// C# program to find sum of product 
// of number in given series
using System;
class GFG 
{
static long MOD = 1000000007;
  
// function to calculate (a^b)%p
static long power(long x, long y, 
                  long p)
{
    long res = 1; // Initialize result
      
    // Update x if it is more
    // than or equal to p
    x = x % p;
      
    while (y > 0) 
    {
      
        // If y is odd, multiply x 
        // with result
        if ((y & 1) != 0)
            res = (res * x) % p;
      
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
  
    return res;
}
  
// function to return required answer
static long sumProd(long n, long t)
{
      
// modulo inverse of denominator
long dino = power(t + 1, MOD - 2, MOD);
  
// calculating commentator part
long ans = 1;
for (long i = n + t + 1; i > n; --i)
    ans = (ans % MOD * i % MOD) % MOD;
  
// calculating t!
long tfact = 1;
for (int i = 1; i <= t; ++i)
    tfact = (tfact * i) % MOD;
  
// accumulating the final answer
ans = ans * dino - tfact + MOD;
  
return ans % MOD;
}
  
// Driver Code
public static void Main() 
{
    long n = 3, t = 2;
  
    // function call to print required sum
    Console.WriteLine(sumProd(n, t));
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

PHP

 0) 
    {
  
        // If y is odd, multiply 
        // x with 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 return required answer
function sumProd($n, $t)
{
    $MOD = 1000000007;
      
    // modulo inverse of denominator
    $dino = power($t + 1, $MOD - 2, $MOD);
  
    // calculating commentator part
    $ans = 1;
    for ($i = $n + $t + 1; $i > $n; --$i)
        $ans = ($ans % $MOD * $i % 
                       $MOD) % $MOD;
  
    // calculating t!
    $tfact = 1;
    for ($i = 1; $i <= $t; ++$i)
        $tfact = ($tfact * $i) % $MOD;
  
    // accumulating the final answer
    $ans = $ans * $dino - $tfact + $MOD;
  
    return $ans % $MOD;
}
  
// Driver code
$n = 3;
$t = 2;
  
// function call to print
// required sum
echo sumProd($n, $t);
  
// This code is contributed
// by Shivi_Aggarwal
?>

输出：

38

时间复杂度： O(T)