给定大小为N的数组a [] 。任务是找到数组b []的元素的最小可能和，以使a [i] * b [i] = a [j] * b [j]对于所有1≤i
例子：
Input: a[] = {2, 3, 4}
Output: 13
b[] = {6, 4, 3}
Input: a[] = {5, 5, 5}
Output: 3
b = {1, 1, 1}
方法：假定满足给定条件的B i被确定。然后令K = A 1 * B 1 ，则对于所有j> 1 ，约束K = A 1 * B 1 = A j * B j成立。因此， K是A 1 ，…，A N的公倍数。
相反，令lcm为A 1 ，…，A N的最小公倍数，令B i = lcm / A i，则这样的B满足条件。
因此，期望的答案是∑lcm / A i 。但是， lcm可以是一个非常大的数字，因此无法直接计算。现在，让我们考虑进行计算，以因式分解的形式持有lcm 。令p i为质数，并假设因式分解由X = ∏p i g i ， Y = ∏p i f i ( g i ，f i均可为0 )给出。然后X和Y的最小公倍数由∏p i max(g i ，f i )给出。通过使用它，可以以因式分解的形式获得A 1 ，…，A N的最小公倍数。因此，可以在总共O(N * sqrt(A))的时间内解决此问题，其中A = max(A i ) 。同样，通过使用适当的预先计算来加速素数分解，也可以在总计O(A + N * logA)的时间内获得答案。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define mod (int)(1e9 + 7)
#define N 1000005
// To store least prime factors
// of all the numbers
int lpf[N];
// Function to find the least prime
// factor of all the numbers
void least_prime_factor()
{
for (int i = 1; i < N; i++)
lpf[i] = i;
for (int i = 2; i < N; i++)
if (lpf[i] == i)
for (int j = i * 2; j < N; j += i)
if (lpf[j] == j)
lpf[j] = i;
}
// Function to return the ((a^m1) % mod)
int power(int a, int m1)
{
if (m1 == 0)
return 1;
else if (m1 == 1)
return a;
else if (m1 == 2)
return (1LL * a * a) % mod;
else if (m1 & 1)
return (1LL * a * power(power(a, m1 / 2), 2)) % mod;
else
return power(power(a, m1 / 2), 2) % mod;
}
// Function to return the sum of
// elements of array B
long long sum_of_elements(int a[], int n)
{
// Find the prime factors of
// all the numbers
least_prime_factor();
// To store each prime count in lcm
map prime_factor;
for (int i = 0; i < n; i++) {
// Current number
int temp = a[i];
// Map to store the prime count
// of a single number
map single_number;
// Basic way to calculate all prime factors
while (temp > 1) {
int x = lpf[temp];
single_number[x]++;
temp /= x;
}
// If it is the first number in the array
if (i == 0)
prime_factor = single_number;
// Take the maximum count of
// prime in a number
else {
for (auto x : single_number)
prime_factor[x.first] = max(x.second,
prime_factor[x.first]);
}
}
long long ans = 0, lcm = 1;
// Calculate lcm of given array
for (auto x : prime_factor)
lcm = (lcm * power(x.first, x.second)) % mod;
// Calculate sum of elements of array B
for (int i = 0; i < n; i++)
ans = (ans + (lcm * power(a[i],
mod - 2)) % mod) % mod;
return ans;
}
// Driver code
int main()
{
int a[] = { 2, 3, 4 };
int n = sizeof(a) / sizeof(int);
cout << sum_of_elements(a, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static int mod = 1000000007;
static int N = 1000005;
// To store least prime factors
// of all the numbers
static int lpf[] = new int[N];
// Function to find the least prime
// factor of all the numbers
static void least_prime_factor()
{
for (int i = 1; i < N; i++)
lpf[i] = i;
for (int i = 2; i < N; i++)
if (lpf[i] == i)
for (int j = i * 2; j < N; j += i)
if (lpf[j] == j)
lpf[j] = i;
}
// Function to return the ((a^m1) % mod)
static long power(long a, long m1)
{
if (m1 == 0)
return 1;
else if (m1 == 1)
return a;
else if (m1 == 2)
return (1l * a * a) % mod;
else if ((m1 & 1) != 0)
return (1l * a * power(power(a, m1 / 2), 2)) % mod;
else
return power(power(a, m1 / 2), 2) % mod;
}
// Function to return the sum of
// elements of array B
static long sum_of_elements(long a[], int n)
{
// Find the prime factors of
// all the numbers
least_prime_factor();
// To store each prime count in lcm
HashMap prime_factor
= new HashMap<>();
for (int i = 0; i < n; i++)
{
// Current number
long temp = a[i];
// Map to store the prime count
// of a single number
HashMap single_number
= new HashMap<>();
// Basic way to calculate all prime factors
while (temp > 1)
{
long x = lpf[(int)temp];
single_number.put(x,(single_number.get(x) ==
null ? 1:single_number.get(x) + 1));
temp /= x;
}
// If it is the first number in the array
if (i == 0)
prime_factor = single_number;
// Take the maximum count of
// prime in a number
else {
for (Map.Entry x : single_number.entrySet() )
prime_factor.put(x.getKey(), Math.max(x.getValue(),
(prime_factor.get(x.getKey()) ==
null ? 0:prime_factor.get(x.getKey()))));
}
}
long ans = 0, lcm = 1;
// Calculate lcm of given array
for (Map.Entry x : prime_factor.entrySet())
lcm = (lcm * power(x.getKey(), x.getValue())) % mod;
// Calculate sum of elements of array B
for (int i = 0; i < n; i++)
ans = (ans + (lcm * power(a[i],
mod - 2)) % mod) % mod;
return ans;
}
// Driver code
public static void main(String args[])
{
long a[] = { 2, 3, 4 };
int n = a.length;
System.out.println(sum_of_elements(a, n));
}
}
// This code is contributed by Arnab Kundu Python3
# Python3 implementation of the approach
mod = 10 ** 9 + 7
N = 1000005
# To store least prime factors
# of all the numbers
lpf = [0 for i in range(N)]
# Function to find the least prime
# factor of all the numbers
def least_prime_factor():
for i in range(1, N):
lpf[i] = i
for i in range(2,N):
if (lpf[i] == i):
for j in range(i * 2, N, i):
if (lpf[j] == j):
lpf[j] = i
# Function to return the sum of
# elements of array B
def sum_of_elements(a, n):
# Find the prime factors of
# all the numbers
least_prime_factor()
# To store each prime count in lcm
prime_factor=dict()
for i in range(n):
# Current number
temp = a[i]
# Map to store the prime count
# of a single number
single_number = dict()
# Basic way to calculate all prime factors
while (temp > 1):
x = lpf[temp]
single_number[x] = single_number.get(x, 0) + 1
temp //= x
# If it is the first number in the array
if (i == 0):
prime_factor = single_number
# Take the maximum count of
# prime in a number
else:
for x in single_number:
if x in prime_factor:
prime_factor[x] = max(prime_factor[x],
single_number[x])
else:
prime_factor[x] = single_number[x]
ans, lcm = 0, 1
# Calculate lcm of given array
for x in prime_factor:
lcm = (lcm * pow(x, prime_factor[x],mod)) % mod
# Calculate sum of elements of array B
for i in range(n):
ans = (ans + (lcm * pow(a[i],
mod - 2,mod)) % mod) % mod
return ans
# Driver code
if __name__ == '__main__':
a = [2, 3, 4]
n = len(a)
print(sum_of_elements(a, n))
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
static int mod = 1000000007;
static int N = 1000005;
// To store least prime factors
// of all the numbers
static int []lpf = new int[N];
// Function to find the least prime
// factor of all the numbers
static void least_prime_factor()
{
for(int i = 1; i < N; i++)
lpf[i] = i;
for(int i = 2; i < N; i++)
if (lpf[i] == i)
for(int j = i * 2;
j < N; j += i)
if (lpf[j] == j)
lpf[j] = i;
}
// Function to return the ((a^m1) % mod)
static long power(long a, long m1)
{
if (m1 == 0)
return 1;
else if (m1 == 1)
return a;
else if (m1 == 2)
return (a * a) % mod;
else if ((m1 & 1) != 0)
return (a * power(power(a, m1 / 2),
2)) % mod;
else
return power(power(a, m1 / 2), 2) % mod;
}
// Function to return the sum of
// elements of array B
static long sum_of_elements(long []a, int n)
{
// Find the prime factors of
// all the numbers
least_prime_factor();
// To store each prime count in lcm
Dictionary prime_factor = new Dictionary();
for(int i = 0; i < n; i++)
{
// Current number
long temp = a[i];
// Map to store the prime count
// of a single number
Dictionary single_number = new Dictionary();
// Basic way to calculate all prime factors
while (temp > 1)
{
long x = lpf[(int)temp];
if (single_number.ContainsKey(x))
{
single_number[x]++;
}
else
{
single_number[x] = 1;
}
temp /= x;
}
// If it is the first number in the array
if (i == 0)
prime_factor = single_number;
// Take the maximum count of
// prime in a number
else
{
foreach(KeyValuePair ele in single_number)
{
if (prime_factor.ContainsKey(ele.Key))
{
prime_factor[ele.Key] = Math.Max(
ele.Value, prime_factor[ele.Key]);
}
else
{
prime_factor[ele.Key] = Math.Max(
ele.Value, 0);
}
}
}
}
long ans = 0, lcm = 1;
// Calculate lcm of given array
foreach(KeyValuePair x in prime_factor)
{
lcm = (lcm * power(x.Key, x.Value)) % mod;
}
// Calculate sum of elements of array B
for(int i = 0; i < n; i++)
ans = (ans + (lcm * power(a[i],
mod - 2)) % mod) % mod;
return ans;
}
// Driver Code
public static void Main (string[] args)
{
long []a = { 2, 3, 4 };
int n = a.Length;
Console.Write(sum_of_elements(a, n));
}
}
// This code is contributed by rutvik_56 输出：
13