给定数字N。求第一个N阶乘积的乘积为1000000007。
Constraints: 1 ≤ N ≤ 1e6
例子：
Input : 3
Output : 12
Explanation: 1! * 2! * 3! = 12 mod (1e9 + 7) = 12
Input : 5
Output : 34560
先决条件：模乘法
方法：解决此问题的基本思想是仅考虑乘以如此大的数字(即阶乘)时的溢出问题。因此，需要通过递归乘法来解决它，以克服溢出的困难。此外，我们必须在迭代计算阶乘和模数乘积的每一步都取模。
facti = facti-1 * i
where facti is the factorial of ith number
prodi = prodi-1 * facti
where prodi is the product of first i factorials
为了求模下两个大数的乘积，我们使用与模下乘幂相同的方法。在乘法函数，我们用+代替*。
下面是上述方法的实现。
C++
// CPP Program to find the
// product of first N factorials
#include
using namespace std;
// To compute (a * b) % MOD
long long int mulmod(long long int a, long long int b,
long long int mod)
{
long long int res = 0; // Initialize result
a = a % mod;
while (b > 0) {
// If b is odd, add 'a' to result
if (b % 2 == 1)
res = (res + a) % mod;
// Multiply 'a' with 2
a = (a * 2) % mod;
// Divide b by 2
b /= 2;
}
// Return result
return res % mod;
}
// This function computes factorials and
// product by using above function i.e.
// modular multiplication
long long int findProduct(long long int N)
{
// Initialize product and fact with 1
long long int product = 1, fact = 1;
long long int MOD = 1e9 + 7;
for (int i = 1; i <= N; i++) {
// ith factorial
fact = mulmod(fact, i, MOD);
// product of first i factorials
product = mulmod(product, fact, MOD);
// If at any iteration, product becomes
// divisible by MOD, simply return 0;
if (product == 0)
return 0;
}
return product;
}
// Driver Code to Test above functions
int main()
{
long long int N = 3;
cout << findProduct(N) << endl;
N = 5;
cout << findProduct(N) << endl;
return 0;
} Java
// Java Program to find the
// product of first N factorials
class GFG{
// To compute (a * b) % MOD
static double mulmod(long a, long b,
long mod)
{
long res = 0; // Initialize result
a = a % mod;
while (b > 0) {
// If b is odd, add 'a' to result
if (b % 2 == 1)
res = (res + a) % mod;
// Multiply 'a' with 2
a = (a * 2) % mod;
// Divide b by 2
b /= 2;
}
// Return result
return res % mod;
}
// This function computes factorials and
// product by using above function i.e.
// modular multiplication
static long findProduct(long N)
{
// Initialize product and fact with 1
long product = 1, fact = 1;
long MOD = (long)(1e9 + 7);
for (int i = 1; i <= N; i++) {
// ith factorial
fact = (long)mulmod(fact, i, MOD);
// product of first i factorials
product = (long)mulmod(product, fact, MOD);
// If at any iteration, product becomes
// divisible by MOD, simply return 0;
if (product == 0)
return 0;
}
return product;
}
// Driver Code to Test above functions
public static void main(String[] args)
{
long N = 3;
System.out.println(findProduct(N));
N = 5;
System.out.println(findProduct(N));
}
}
// this Code is contributed by mitsPython3
# Python Program to find the
# product of first N factorials
# To compute (a * b) % MOD
def mulmod(a, b, mod):
res = 0 # Initialize result
a = a % mod
while (b > 0):
# If b is odd, add 'a' to result
if (b % 2 == 1):
res = (res + a) % mod
# Multiply 'a' with 2
a = (a * 2) % mod
# Divide b by 2
b //= 2
# Return result
return res % mod
# This function computes factorials and
# product by using above function i.e.
# modular multiplication
def findProduct(N):
# Initialize product and fact with 1
product = 1; fact = 1
MOD = 1e9 + 7
for i in range(1, N+1):
# ith factorial
fact = mulmod(fact, i, MOD)
# product of first i factorials
product = mulmod(product, fact, MOD)
# If at any iteration, product becomes
# divisible by MOD, simply return 0
if not product:
return 0
return int(product)
# Driver Code to Test above functions
N = 3
print(findProduct(N))
N = 5
print(findProduct(N))
# This code is contributed by Ansu KumariC#
// C# Program to find the
// product of first N factorials
using System;
public class GFG{
// To compute (a * b) % MOD
static double mulmod(long a, long b,
long mod)
{
long res = 0; // Initialize result
a = a % mod;
while (b > 0) {
// If b is odd, add 'a' to result
if (b % 2 == 1)
res = (res + a) % mod;
// Multiply 'a' with 2
a = (a * 2) % mod;
// Divide b by 2
b /= 2;
}
// Return result
return res % mod;
}
// This function computes factorials and
// product by using above function i.e.
// modular multiplication
static long findProduct(long N)
{
// Initialize product and fact with 1
long product = 1, fact = 1;
long MOD = (long)(1e9 + 7);
for (int i = 1; i <= N; i++) {
// ith factorial
fact = (long)mulmod(fact, i, MOD);
// product of first i factorials
product = (long)mulmod(product, fact, MOD);
// If at any iteration, product becomes
// divisible by MOD, simply return 0;
if (product == 0)
return 0;
}
return product;
}
// Driver Code to Test above functions
static public void Main (){
long N = 3;
Console.WriteLine(findProduct(N));
N = 5;
Console.WriteLine(findProduct(N));
}
}
//This Code is contributed by ajit.PHP
0)
{
// If b is odd, add 'a' to result
if ($b % 2 == 1)
$res = ($res + $a) % $mod;
// Multiply 'a' with 2
$a = ($a * 2) % $mod;
// Divide b by 2
$b /= 2;
}
// Return result
return $res % $mod;
}
// This function computes factorials and
// product by using above function i.e.
// modular multiplication
function findProduct($N)
{
// Initialize product and fact with 1
$product = 1;
$fact = 1;
$MOD = 1000000000;
for ($i = 1; $i <= $N; $i++)
{
// ith factorial
$fact = mulmod($fact, $i, $MOD);
// product of first i factorials
$product = mulmod($product, $fact, $MOD);
// If at any iteration, product becomes
// divisible by MOD, simply return 0;
if ($product == 0)
return 0;
}
return $product;
}
// Driver Code
$N = 3;
echo findProduct($N),"\n";
$N = 5;
echo findProduct($N),"\n";
// This code is contributed by ajit
?>输出：
12
34560
时间复杂度： O(N * logN)，其中O(log N)是模数乘法的时间复杂度。