给定一个由n个整数组成的数组。找到要分配给每个数组元素的最小值x，以使此新数组的所有元素的乘积严格大于初始数组的所有元素的乘积。
例子：
Input: 4 2 1 10 6
Output: 4
Explanation: Product of elements of initial
array 4*2*1*10*6 = 480. If x = 4 then 4*4*
4*4*4 = 480, if x = 3 then 3*3*3*3*3=243.
So minimal element = 4
Input: 3 2 1 4
Output: 3
Explanation: Product of elements of initial
array 3*2*1*4 = 24. If x = 3 then 3*3*3*3
= 81, if x = 2 then 2*2*2*2 = 243. So minimal
element = 3. 简单方法：一个简单的方法是从1开始运行一个循环，直到发现乘积大于初始数组乘积为止。
时间复杂度： O(x ^ n) ，如果使用了pow函数，则O(x * log n)
数学方法：
Let, x^n = a1 * a2 * a3 * a4 *....* an
we have been given n and value of a1, a2, a3, ..., an.
Now take log on both sides with base e
n*logex > loge(a1) + loge(a2) +......+ loge(an)
Lets sum = loge(a1) + loge(a2) + ...... + loge(an)
n*loge x > sum
loge x > sum/n
Then take antilog on both side
x > e^(sum/n) 下面是上述方法的实现。
C++
// CPP program to find minimum element whose n-th
// power is greater than product of an array of
// size n
#include
using namespace std;
// function to find the minimum element
int findMin(int a[], int n)
{
// loop to traverse and store the sum of log
double sum = 0;
for (int i = 0; i < n; i++)
sum += log(a[i]); // computes sum
// calculates the elements according to formula.
int x = exp(sum / n);
// returns the minimal element
return x + 1;
}
// Driver program to test above function
int main()
{
// initialised array
int a[] = { 3, 2, 1, 4 };
// computes the size of array
int n = sizeof(a) / sizeof(a[0]);
// prints out the minimal element
cout << findMin(a, n);
} Java
// JAVA Code to find Minimum element whose
// n-th power is greater than product of
// an array of size n
import java.util.*;
class GFG {
// function to find the minimum element
static int findMin(int a[], int n)
{
// loop to traverse and store the
// sum of log
double sum = 0;
for (int i = 0; i < n; i++)
// computes sum
sum += Math.log(a[i]);
// calculates the elements
// according to formula.
int x = (int)Math.exp(sum / n);
// returns the minimal element
return x + 1;
}
/* Driver program to test above function */
public static void main(String[] args)
{
// initialised array
int a[] = { 3, 2, 1, 4 };
// computes the size of array
int n = a.length;
// prints out the minimal element
System.out.println(findMin(a, n));
}
}
// This code is contributed by Arnav Kr. Mandal.Python3
# Python3 program to find minimum element
# whose n-th power is greater than product
# of an array of size n
import math as m
# function to find the minimum element
def findMin( a, n):
# loop to traverse and store the
# sum of log
_sum = 0
for i in range(n):
_sum += m.log(a[i]) # computes sum
# calculates the elements
# according to formula.
x = m.exp(_sum / n)
# returns the minimal element
return int(x + 1)
# Driver program to test above function
# initialised array
a = [ 3, 2, 1, 4 ]
# computes the size of array
n = len(a)
# prints out the minimal element
print(findMin(a, n))
# This code is contributed by "Abhishek Sharma 44"C#
// C# Code to find Minimum element whose
// n-th power is greater than product of
// an array of size n
using System;
class GFG {
// function to find the minimum element
static int findMin(int []a, int n)
{
// loop to traverse and store the
// sum of log
double sum = 0;
for (int i = 0; i < n; i++)
// computes sum
sum += Math.Log(a[i]);
// calculates the elements
// according to formula.
int x = (int)Math.Exp(sum / n);
// returns the minimal element
return x + 1;
}
/* Driver program to test above function */
public static void Main()
{
// initialised array
int []a = { 3, 2, 1, 4 };
// computes the size of array
int n = a.Length;
// prints out the minimal element
Console.WriteLine(findMin(a, n));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出：
3