原文:
给定一个数 n，求 n 的所有因子的乘积，由于乘积可以很大，所以求它的模 10^9 7。 例:
input : 12
output : 1728
1 * 2 * 3 * 4 * 6 * 12 = 1728
input : 18
output : 5832
1 * 2 * 3 * 6 * 9 * 18 = 5832
方法 1(天真的方法): 我们可以对 i 运行从 1 到 n 的循环，如果 n 可被 i 整除，则乘以数字。该pg电子试玩链接的解决方案的时间复杂度为 0(n)。 但是这种方法对于大的 n 值是不够的。 方法 2(更好的方法): 更好的方法是运行 i 从 1 到 sqrt(n)的循环。如果数能被 i 整除，用 i 和 n/i 相乘， 这个解的时间复杂度是 o(sqrt(n))。
c
// c program to calculate product
// of factors of number
#include
#define m 1000000007
using namespace std;
// function to product the factors
long long multiplyfactors(int n)
{
long long prod = 1;
for (int i = 1; i * i <= n; i )
{
if (n % i == 0)
{
// if factors are equal,
// multiply only once
if (n / i == i)
prod = (prod * i) % m;
// otherwise multiply both
else {
prod = (prod * i) % m;
prod = (prod * n / i) % m;
}
}
}
return prod;
}
// driver code
int main()
{
int n = 12;
cout << multiplyfactors(n) << endl;
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to calculate product
// of factors of number
class gfg
{
public static final long m = 1000000007;
// function to product the factors
static long multiplyfactors(int n)
{
long prod = 1;
for (int i = 1; i * i <= n; i )
{
if (n % i == 0)
{
// if factors are equal,
// multiply only once
if (n / i == i)
prod = (prod * i) % m;
// otherwise multiply both
else {
prod = (prod * i) % m;
prod = (prod * n / i) % m;
}
}
}
return prod;
}
// driver code
public static void main(string[] args)
{
int n = 12;
system.out.println(multiplyfactors(n));
}
}
计算机编程语言
# python program to calculate product
# of factors of number
m = 1000000007
# function to product the factors
def multiplyfactors(n) :
prod = 1
i = 1
while i * i <= n :
if (n % i == 0) :
# if factors are equal,
# multiply only once
if (n / i == i) :
prod = (prod * i) % m
#otherwise multiply both
else :
prod = (prod * i) % m
prod = (prod * n / i) % m
i = i 1
return prod
# driver code
n = 12
print (multiplyfactors(n))
# this code is contributed by nikita tiwari.
c
//c# program to calculate product
// of factors of number
using system;
class gfg
{
public static long m = 1000000007;
// function to product the factors
static long multiplyfactors(int n)
{
long prod = 1;
for (int i = 1; i * i <= n; i )
{
if (n % i == 0)
{
// if factors are equal,
// multiply only once
if (n / i == i)
prod = (prod * i) % m;
// otherwise multiply both
else {
prod = (prod * i) % m;
prod = (prod * n / i) % m;
}
}
}
return prod;
}
// driver code
public static void main()
{
int n = 12;
console.write(multiplyfactors(n));
}
}
// this code is contributed by nitin mittal.
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
输出:
1728
方法 3(另一种方法): 让我们观察一件事:
all factors of 12 are: 1, 2, 3, 4, 6, 12
1 * 2 * 3 * (2*2) * (2*3) * (2*2*3) = 2^6 * 3^3 = 12^3
and number of factors are 6
所以我们可以观察到因子的乘积将是因子/2 的 n^(number)。但是当因子数是奇数时(这意味着该数是完美平方)，在这种情况下，乘积将是因子/2 的 n^(number)* sqrt(n)。我们可以计算类似上述方法的因素数量。我们可以使用 高效地计算功率
c
// c program to calculate product
// of factors of number
#include
#define m 1000000007
using namespace std;
// iterative function to calculate
// (x^y) in o(log y)
long long power(long long x, long long y)
{
long long res = 1;
while (y > 0)
{
if (y & 1)
res = (res * x) % m;
y = (y >> 1) % m;
x = (x * x) % m;
}
return res;
}
// function to count the factors
int countfactors(int n)
{
int count = 0;
for (int i = 1; i * i <= n; i )
{
if (n % i == 0)
{
// if factors are equal,
// count only once
if (n / i == i)
count ;
// otherwise count both
else
count = 2;
}
}
return count;
}
long long multiplyfactors(int n)
{
int numfactor = countfactors(n);
// calculate product of factors
long long product = power(n, numfactor / 2);
// if numfactor is odd return
// power(n, numfactor/2) * sqrt(n)
if (numfactor & 1)
product = (product * (int)sqrt(n)) % m;
return product;
}
// driver code
int main()
{
int n = 12;
cout << multiplyfactors(n) << endl;
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to calculate product
// of factors of number
class gfg
{
public static final long m = 1000000007;
// iterative function to calculate
// (x^y) in o(log y)
static long power(long x, long y)
{
long res = 1;
while (y > 0)
{
if (y % 2 == 1)
res = (res * x) % m;
y = (y >> 1) % m;
x = (x * x) % m;
}
return res;
}
// function to count the factors
static int countfactors(int n)
{
int count = 0;
for (int i = 1; i * i <= n; i )
{
if (n % i == 0)
{
// if factors are equal,
// count only once
if (n / i == i)
count ;
// otherwise count both
else
count = 2;
}
}
return count;
}
static long multiplyfactors(int n)
{
int numfactor = countfactors(n);
// calculate product of factors
long product = power(n, numfactor / 2);
// if numfactor is odd return
// power(n, numfactor/2) * sqrt(n)
if (numfactor % 2 == 1)
product = (product * (int)math.sqrt(n)) % m;
return product;
}
// driver code
public static void main(string[] args)
{
int n = 12;
system.out.println(multiplyfactors(n));
}
}
计算机编程语言
# python program to calculate product
# of factors of number
m = 1000000007
# iterative function to calculate
# (x^y) in o(log y)
def power(x, y) :
res = 1
while (y > 0) :
if (y % 2 == 1) :
res = (res * x) % m
y = (y >> 1) % m
x = (x * x) % m
return res
# function to count the factors
def countfactors(n) :
count = 0
i = 1
while i * i <= n :
if (n % i == 0) :
# if factors are equal,
# count only once
if (n / i == i) :
count = count 1
# otherwise count both
else :
count = count 2
i = i 1
return count
def multiplyfactors(n) :
numfactor = countfactors(n)
# calculate product of factors
product = power(n, numfactor / 2)
# if numfactor is odd return
# power(n, numfactor/2) * sqrt(n)
if (numfactor % 2 == 1) :
product = (product *
(int)(math.sqrt(n))) % m
return product
# driver code
n = 12
print multiplyfactors(n)
# this code is contributed by nikita tiwari.
c
// c# program to calculate product
// of factors of number
using system;
class gfg {
public static long m = 1000000007;
// iterative function to calculate
// (x^y) in o(log y)
static long power(long x, long y)
{
long res = 1;
while (y > 0)
{
if (y % 2 == 1)
res = (res * x) % m;
y = (y >> 1) % m;
x = (x * x) % m;
}
return res;
}
// function to count the factors
static int countfactors(int n)
{
int count = 0;
for (int i = 1; i * i <= n; i )
{
if (n % i == 0)
{
// if factors are equal,
// count only once
if (n / i == i)
count ;
// otherwise count both
else
count = 2;
}
}
return count;
}
static long multiplyfactors(int n)
{
int numfactor = countfactors(n);
// calculate product of factors
long product = power(n, numfactor / 2);
// if numfactor is odd return
// power(n, numfactor/2) * sqrt(n)
if (numfactor % 2 == 1)
product = (product *
(int)math.sqrt(n)) % m;
return product;
}
// driver code
public static void main()
{
int n = 12;
console.write(multiplyfactors(n));
}
}
// this code is contributed by nitin mittal
服务器端编程语言（professional hypertext preprocessor 的缩写）
0)
{
if ($y & 1)
$res = ($res * $x) % $m;
$y = ($y >> 1) % $m;
$x = ($x *$x) % $m;
}
return $res;
}
// function to count the factors
function countfactors( $n)
{
$count = 0;
for ($i = 1; $i * $i <= $n; $i )
{
if ($n % $i == 0)
{
// if factors are equal,
// count only once
if ($n / $i == $i)
$count ;
// otherwise count both
else
$count = 2;
}
}
return $count;
}
function multiplyfactors( $n)
{
$numfactor = countfactors($n);
// calculate product of factors
$product = power($n, $numfactor / 2);
// if numfactor is odd return
// power(n, numfactor/2) * sqrt(n)
if ($numfactor & 1)
$product = ($product * sqrt($n)) % $m;
return $product;
}
// driver code
$n = 12;
echo multiplyfactors($n);
// this code is contributed by anuj_67.
?>
java 描述语言
输出:
1728
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